Multivariable Calculus Review Problems With Solutions

MULTI-VARIABLE CALCULUS REVIEW PROBLEMS
NOTES WRITTEN BY SILVIUS KLEIN
1. The geometry of the space with three dimensions
Problem 1.1. Find the volume of the parallelepiped with adjacent edges PQ, PR and PS, where
P(2, 0, −1), Q(4, 1, 0), R(3, −1, 1), S(2, −2, 2)
Problem 1.2. Verify that the vectors
a = 2i + 3j +k
b = i −j
c = 7i + 3j + 2k
are coplanar. Explain your answer.
Hint: Use the scalar triple product.
Problem 1.3. Find the area of the triangle with vertices
A(1, 2, 3), B(2, 1, 5), C(3, 10, 6)
Show that ABC is a right triangle, with right angle ∠A.
Problem 1.4. Show that the curve with parametric equations
x = t cos t, y = t sin t, z = t
lies on the cone z
2
= x
2
+y
2
, and use this fact to help sketch the curve.
Find parametric equations for the tangent line to this curve at the point (0,
π
2
,
π
2
).
Problem 1.5. Consider the curve C defined by the vector function
r(t) =< t sin πt, t cos πt, t
2
>
where t ≥ 0.
a) Show that the curve C lies on a paraboloid and use this fact to help sketch the curve.
b) Find symmetric equations for the tangent line to the curve C at the point (0, −1, 1).
Problem 1.6. Find an equation of the plane that passes through the points
P(0, 1, 1), Q(1, 0, 1), R(1, 1, 0)
Find parametric equations for the line of intersection of this plane with the coordinate plane z = 0.
1
2. Functions of two variables: domain, graph, level curves
Problem 2.1. Sketch a contour map of the function
f(x, y) =

4x
2
+y
2
Then sketch the graph of this function. How are they related to each other?
Problem 2.2. Consider the function f(x, y) =

x
2
+ 2y
2
+ 1.
a) Find the domain and the range of f(x, y).
b) Sketch a contour map of f(x, y). Explain its relationship with the graph of f(x, y).
Problem 2.3. Consider the function f(x, y) =

x
2
+y
2
−1.
a) Find the domain and the range of f(x, y).
b) Sketch a contour map of f(x, y). Explain its relationship with the graph of f(x, y).
3. Limits of functions of two variables
Find the limit, if it exists (and prove that it does) or show that the limit does not exist:
Problem 3.1.
lim
(x,y)→(1,0)
y
2
ln x −3
sin πx +e
y
Problem 3.2.
lim
(x,y)→(0,0)
x · y
x
3
+y
2
Problem 3.3.
lim
(x,y)→(0,0)
x
2
ln(1 +y)
x
2
+ 3y
2
Problem 3.4.
lim
(x,y)→(0,0)
x
2
sin(x −y)
x
3
+y
2
Problem 3.5.
lim
(x,y)→(0,0)
x
4
sin(y)
2x
4
+y
2
Problem 3.6.
lim
(x,y)→(0,0)
e
1
x
2
· y
2
2
4. Partial and directional derivatives, gradient vector, linear approximation
Problem 4.1. Let f(x, y) = ln(x −3y).
a) Compute f
x
(7, 2), f
y
(7, 2) and explain their geometric interpretation.
b) Find the tangent plane to the graph of f(x, y) at the point (7, 2, 0).
c) Find the linearization of f(x, y) at (7, 2) and use it to approximate f(6.9, 2.06). Explain why
you can use its linearization to approximate f.
Problem 4.2. Consider the function f(x, y) = sin(πx
2
+πy).
a) What are the domain and the range of f(x, y) ? Why ?
b) Find the tangent plane to the graph of f(x, y) at the point (1, 0, 0).
c) Find the linear approximation of f(x, y) at (1, 0) and use it to approximate f(1.1, −0.1).
Problem 4.3. Find the maximum rate of change of the function
f(x, y, z) = ln(xy
2
) +z
at (1, 1, 0).
Find the direction in which this maximum rate of change occurs.
Problem 4.4. The temperature (in degrees Fahrenheit) at a point (x, y) on a hot plate is given
by the function
T(x, y) = ln(9 +x
2
y)
An ant crawls on the plate so that its position after t seconds is
x = 10 −t, y = t
2
What is the rate of change of the temperature on the ant’s path after 9 seconds?
Hint: You may want to interpret the rate of change as a derivative and use the chain rule.
5. Local and absolute minima and maxima
Problem 5.1. Find the absolute maximum and the absolute minimum values of the function
f(x, y) = 2x + 6y −xy −y
2
on the set: D = {(x, y) : 0 ≤ x ≤ 4, 0 ≤ y ≤ 3}.
Problem 5.2. Find the absolute maximum and the absolute minimum values of
f(x, y) = x
2
+y
2
+xy −x −y
on the region enclosed by the triangle with vertices (0, 0), (1, 0), (0, 2).
Indicate where these extreme values are attained.
Explain why these extreme values have to exist.
Problem 5.3. Find the local minimum and the local maximum points (and values) and the saddle
points (if any) of the function
f(x, y) = 3x
3
+ 6yx −y
2
Sketch the graph of the function around the points found above.
Note: You do not have to sketch the graph of the whole function. Just show how the graph of the
function looks like around the local minimum, the local maximum or the saddle points determined.
3
6. Constrained optimization: the method of Lagrange multipliers
Problem 6.1. Find the distance from the origin O to the line L with equation 3x −2y = 13
Problem 6.2. The plane x +y + 2z = 2 intersects the paraboloid z = x
2
+y
2
in an ellipse.
Find the highest and the lowest points on this ellipse.
Hint: The highest (or the lowest) point on the ellipse is the point with the largest (or the lowest)
z coordinate. Use Lagrange multipliers with two constraints.
Problem 6.3. Find the absolute maximum and the absolute minimum values of the function
f(x, y) = 2x
3
+y
4
on the domain: 3x
2
+ 2y
2
≤ 1.
Indicate where these extreme values are attained.
Explain why these extreme values have to exist.
Problem 6.4. Find the absolute minimum and the absolute maximum values of the function
f(x, y) = x
4
+y
4
on the closed unit disk x
2
+y
2
≤ 1.
Indicate where these extreme values are attained.
Explain why these extreme values have to exist.
7. Double and triple integrals
Problem 7.1. Compute

D
xe
y
dA
where D is the region that lies between the curves y = x and y = x
2
.
Problem 7.2. Consider a lamina that occupies the part of the disk x
2
+ y
2
≤ 4 that lies in the
first quadrant.
Find the center of mass assuming that the density at any point on the lamina is given by the
function ρ(x, y) = x ·

x
2
+y
2
.
Problem 7.3. Compute the following double integral by making an appropriate change of variables:

R
xy dA
where R is the square with vertices (0, 0), (1, 1), (2, 0), (1, −1).
Problem 7.4. Find the volume and the area of the sphere of radius 1.
4
8. Line integrals of functions and of vector fields, Green’s theorem
Problem 8.1. Let f(x, y) = x
3
and let C be the curve that consists of the arc C
1
of the parabola
y = x
2
from (0, 0) to (1, 1), followed by the straight line C
2
from (1, 1) to (4, 1).
Sketch the curve C. Then find the area of the surface between the curve C and the graph of f(x, y)
along C. That is, find the area of the ”fence” whose base is C and whose height above (x, y) is
f(x, y).
Problem 8.2. Let f(x, y) = ln(4x
2
+y
2
+ 1).
a) Find the gradient vector field ∇f.
b) Sketch the gradient vector field ∇f together with a contour map of f. Explain how they are
related to each other.
c) Compute

C
∇f · dr
where C is a piecewise - smooth curve from (0, 0) to (1, 2).
Problem 8.3. Find the work done by the force field
F(x, y, z) = zi +xj +yk
in moving a particle from the point (3, 0, 0) to the point (0,
π
2
, 3) along:
a) A straight line.
b) The helix x = 3 cos t, y = t, z = 3 sin t. It would be nice if you also drew the helix.
Problem 8.4. Let F(x, y) =< 3x
2
+ ln y,
x
y
+e
7y
> be a vector field.
a) Show that F is a conservative vector field. Indicate the domain of F.
b) Find a function f(x, y) such that F = ∇f.
c) Compute

C
F · dr
where C is a curve from (0, 1) to (1, e).
Problem 8.5. Compute

C
x
2
y dx −xy
2
dy
where C is the circle x
2
+y
2
= 4 with counterclockwise orientation.
Problem 8.6. Compute

C
e
x
2
dx + (x
3
+ sin y) dy
where C is the boundary of the region between the circles x
2
+y
2
= 1 and x
2
+y
2
= 4 in the first
quadrant.
5
SOLUTIONS TO THE MULTI-VARIABLE CALCULUS REVIEW PROBLEMS
WRITTEN BY SILVIUS KLEIN
1. The geometry of the space with three dimensions
Problem 1.1. Find the volume of the parallelepiped with adjacent edges PQ, PR and PS, where
P(2, 0, −1), Q(4, 1, 0), R(3, −1, 1), S(2, −2, 2)
Solution: The volume of the parallelepiped with adjacent edges PQ, PR, PS is the absolute
value of the scalar triple product of the vectors
−−→
PQ,
−→
PR,
−→
PS.
That is,
V = |
−−→
PQ· (
−→
PR ×
−→
PS)|
Since P(2, 0, −1), Q(4, 1, 0), R(3, −1, 1), S(2, −2, 2), we have:
−−→
PQ =< 4 −2, 1 −0, 0 −(−1) >=< 2, 1, 1 >
−→
PR =< 3 −2, −1 −0, 1 −(−1) >=< 1, −1, 2 >
−→
PS =< 2 −2, −2 −0, 2 −(−1) >=< 0, −2, 3 >
Then
−→
PR ×
−→
PS =
¸
¸
¸
¸
¸
¸
i j k
1 −1 2
0 −2 3
¸
¸
¸
¸
¸
¸
=
=
¸
¸
¸
¸
−1 2
−2 3
¸
¸
¸
¸
· i −
¸
¸
¸
¸
1 2
0 3
¸
¸
¸
¸
· j +
¸
¸
¸
¸
1 −1
0 −2
¸
¸
¸
¸
· k =
(−3 −(−4))i −(3 −0)j + (−2 −0)k = i −3j −2k =< 1, −3, −2 >
Hence
−→
PR ×
−→
PS =< 1, −3, −2 >
Then
V = |
−−→
PQ· (
−→
PR ×
−→
PS)| = | < 2, 1, 1 > · < 1, −3, −2 > | =
= |2 · 1 + 1 · (−3) + 1 · (−2)| = |2 −3 −2| = | −3| = 3,
so
V = 3
Problem 1.2. Verify that the vectors
a = 2i + 3j +k
b = i −j
c = 7i + 3j + 2k
are coplanar. Explain your answer.
Hint: Use the scalar triple product.
Date: March 2012.
1
Solution: We compute the scalar triple product of these three vectors: (a ×b) · c.
We know that this is the volume of the parallelepiped with edges a, b, c.
a ×b =
¸
¸
¸
¸
¸
¸
i j k
2 3 1
1 −1 0
¸
¸
¸
¸
¸
¸
=
=
¸
¸
¸
¸
3 1
−1 0
¸
¸
¸
¸
· i −
¸
¸
¸
¸
2 1
1 0
¸
¸
¸
¸
· j +
¸
¸
¸
¸
2 3
1 −1
¸
¸
¸
¸
· k =
= (0 −(−1))i −(0 −1)j + (−2 −3)k = i +j −5k =< 1, 1, −5 >
Hence
a ×b =< 1, 1, −5 >
Then
V = (a ×b) · c =< 1, 1, −5 > · < 7, 3, 2 >= 7 + 3 −10 = 0,
so V = 0.
Since the volume V of the parallelepiped with edges a, b, c is 0, the parallelepiped must be flat,
therefore its edges a, b, c should all be in the same plane.
Problem 1.3. Find the area of the triangle with vertices
A(1, 2, 3), B(2, 1, 5), C(3, 10, 6)
Show that ABC is a right triangle, with right angle ∠A.
Solution: The area of the triangle with vertices A(1, 2, 3), B(2, 1, 5), C(3, 10, 6), is one-half the
area of the parallelogram with edges
−−→
AB and
−→
AC, which can be computed as the magnitude of the
cross product of these two vectors.
We have:
−−→
AB =< 2 −1, 1 −2, 5 −3 >=< 1, −1, 2 >
−→
AC =< 3 −1, 10 −2, 6 −3 >=< 2, 8, 3 >
Then
−−→
AB ×
−→
AC =
¸
¸
¸
¸
¸
¸
i j k
1 −1 2
2 8 3
¸
¸
¸
¸
¸
¸
=
=
¸
¸
¸
¸
−1 2
8 3
¸
¸
¸
¸
· i −
¸
¸
¸
¸
1 2
2 3
¸
¸
¸
¸
· j +
¸
¸
¸
¸
1 −1
2 8
¸
¸
¸
¸
· k =
(−3 −16)i −(3 −4)j + (8 −(−2))k = −19i +j + 10k =< −19, 1, 10 >
| < −19, 1, 10 > | =
_
(−19)
2
+ 1
2
+ 10
2
=
√
462
Therefore, the area of the triangle with vertices A(1, 2, 3), B(2, 1, 5), C(3, 10, 6) is
1
2
·
√
462.
To show that ∠A is a right angle, we need to show that
−−→
AB ⊥
−→
AC
To do that, we compute the dot product of these two vectors:
−−→
AB ·
−→
AC =< 1, −1, 2 > · < 2, 8, 3 >= 2 −8 + 6 = 0
This shows that the two vectors are perpendicular, so ∠A is a right angle.
2
Problem 1.4. Show that the curve with parametric equations
x = t cos t, y = t sin t, z = t
lies on the cone z
2
= x
2
+y
2
, and use this fact to help sketch the curve.
Find parametric equations for the tangent line to this curve at the point (0,
π
2
,
π
2
).
Solution: Any point on this curve C has coordinates (x, y, z) such that
x = t cos t, y = t sin t, z = t for some parameter t
If (x, y, z) is such a point on the curve C, then
x
2
+y
2
= t
2
cos
2
t +t
2
sin
2
t = t
2
(cos
2
t + sin
2
t) = t
2
· 1 = t
2
and
z
2
= t
2
so
x
2
+y
2
= z
2
Hence the point (x, y, z) is on the cone x
2
+y
2
= z
2
.
Therefore, the whole curve lies on this cone.
When t = 0 we have x = 0, y = 0, z = 0, so the curve C passes through the origin.
When t > 0, the curve intersects a circle of radius t in the plane z = t, so the curve spirals upward
around the cone as t increases.
When t < 0, the curve intersects a circle of radius |t| in the plane z = t, so the curve spirals
downward around the cone as t decreases.
Here is the sketch of the curve C:
To find the equation of the tangent line to the curve C at the point (0,
π
2
,
π
2
), first notice that this
point corresponds to the value t =
π
2
of the parameter t.
Then the direction v of the tangent line is the derivative at the point t =
π
2
of the vector function
< x(t), y(t), z(t) >.
We have x

(t) = cos t −t sin t, y

(t) = sin t +t cos t, z

(t) = 1.
Put t =
π
2
, so the vector v = < x

(
π
2
), y

(
π
2
), z

(
π
2
) >=< −
π
2
, 1, 1 >.
The parametric equation of the line passing through the point (0,
π
2
,
π
2
) and parallel to the vector
< −
π
2
, 1, 1 > is
x = 0 +s · (−
π
2
) = −
π
2
s
y =
π
2
+s · 1 =
π
2
+s
z =
π
2
+s · 1 =
π
2
+s
Therefore, the equation of the tangent line to the curve C at the point (0,
π
2
,
π
2
) is
x = −
π
2
s, y =
π
2
+s, z =
π
2
+s
3
Problem 1.5. Consider the curve C defined by the vector function
r(t) =< t sin πt, t cos πt, t
2
>
where t ≥ 0.
a) Show that the curve C lies on a paraboloid and use this fact to help sketch the curve.
b) Find symmetric equations for the tangent line to the curve C at the point (0, −1, 1).
Solution: a) A point on this curve C has coordinates (x, y, z) such that
x = t sin πt, y = t cos πt, z = t
2
for some parameter t.
If (x, y, z) is any such a point on the curve C, then
x
2
+y
2
= t
2
sin
2
πt +t
2
cos
2
πt = t
2
(sin
2
πt + cos
2
πt) = t
2
· 1 = t
2
and
z = t
2
so
x
2
+y
2
= z
Hence the point (x, y, z) is on the quadric surface x
2
+y
2
= z.
This quadric surface is an upward elliptic (circular) paraboloid along the z axis. That is because
the x and y traces are parabolas and the z-traces are circles, while z ≥ 0.
Therefore, the whole curve lies on this paraboloid.
When t = 0 we have x = 0, y = 0, z = 0, so the curve C passes through the origin.
When t ≥ 0, the curve intersects a circle of radius t in the plane z = t
2
, so the curve spirals upward
around the paraboloid as t increases to ∞.
Below is the sketch of the curve C.
b) To find the equation of the tangent line to the curve C at the point (0, −1, 1), first notice that
this point corresponds to the value t = 1 of the parameter t.
Then the direction v of the tangent line is the derivative at the point t = 1 of the vector function
r(t).
We have r

(t) =< sin πt +πt cos πt, cos πt −πt sin πt, 2t >.
Put t = 1, so the vector
v = r

(1) =< 0 +π(−1), −1 −0, 2 >=< −π, −1, 2 > .
The symmetric equations of the line passing through the point (0, −1, 1) and parallel to the vector
< −π, −1, 2 > are
x −0
−π
=
y + 1
−1
=
z −1
2
Therefore, the symmetric equations of the tangent line to the curve C at the point (0, −1, 1) are
−
x
π
= −(y + 1) =
z −1
2
4
Problem 1.6. Find an equation of the plane that passes through the points
P(0, 1, 1), Q(1, 0, 1), R(1, 1, 0)
Find parametric equations for the line of intersection of this plane with the coordinate plane z = 0.
Solution: We have:
−−→
PQ =< 1 −0, 0 −1, 1 −1 >=< 1, −1, 0 >
−→
PR =< 1 −0, 1 −1, 0 −1 >=< 1, 0, −1 >
Then a normal vector to the plane that contains the vectors
−−→
PQ,
−→
PR (and therefore the points P,
Q, R) is
n =
−−→
PQ×
−→
PR =
¸
¸
¸
¸
¸
¸
i j k
1 −1 0
1 0 −1
¸
¸
¸
¸
¸
¸
=
=
¸
¸
¸
¸
−1 0
0 −1
¸
¸
¸
¸
· i −
¸
¸
¸
¸
1 0
1 −1
¸
¸
¸
¸
· j +
¸
¸
¸
¸
1 −1
1 0
¸
¸
¸
¸
· k =
(1 −0)i −(−1 −0)j + (0 −(−1))k = i +j +k =< 1, 1, 1 >
The equation of the plane passing through the point P(0, 1, 1) and with normal vector n =< 1, 1, 1 >
is
1 · (x −0) + 1 · (y −1) + 1 · (z −1) = 0
which becomes
x +y +z = 2
To find parametric equations of the line of intersection between the plane x + y + z = 2 and the
pane z = 0, note that if we put z = 0 in the equation of the first plane, we get x + y = 2, or
y = 2 −x.
Let x = t. Then y = 2 −t.
A parametric equation for the line of intersection between the two planes is then:
x = t, y = 2 −t, z = 0
5
2. Functions of two variables: domain, graph, level curves
Problem 2.1. Sketch a contour map of the function
f(x, y) =
_
4x
2
+y
2
Then sketch the graph of this function. How are they related to each other?
Solution. The contour map consists of level curves f(x, y) = c for c ≥ 0 (since c should belong to
the range of f(x, y)).
The equation for each level curve at level c is then:
_
4x
2
+y
2
= c, c ≥ 0
4x
2
+y
2
= c
2
, c ≥ 0
If c = 0, the corresponding level ’curve’ is just the point (0, 0).
If c > 0, the level curve is an ellipse with major axis y.
The picture below is the contour map of f(x, y).
1
2
3
4
4
5
5
6
6
6
3 2 1 0 1 2 3
3
2
1
0
1
2
3
The graph of f(x, y) is the surface of equation z = f(x, y), or
z =
_
4x
2
+y
2
Putting the level curves on top of each other (according to their levels), we will get an idea of how
the graph looks like. However, it will not be clear whether we get a cone or a surface with a rounder
tip (such as a paraboloid or a hyperboloid).
To decide what kind of surface this is, one would have to consider all its cross-sections. The
cross-sections will show that the graph is in fact (the upper part of) a cone, pictured below.
2
0
2
x
2
0
2
y
0
1
2
3
z
Note that the level curves (at every level c), correspond to horizontal cross - sections of the graph
(at height c).
6
Problem 2.2. Consider the function f(x, y) =
_
x
2
+ 2y
2
+ 1.
a) Find the domain and the range of f(x, y).
b) Sketch a contour map of f(x, y). Explain its relationship with the graph of f(x, y).
Solution. a) The domain of f(x, y) is the set of all pairs of real numbers (x, y) for which f(x, y)
is well defined.
We know that the square root is defined for all nonnegative numbers.
But x
2
+ 2y
2
+ 1 ≥ 0 + 0 + 1 = 1, so x
2
+ 2y
2
+ 1 ≥ 1, which shows that the number under the
square root in the formula for f is always nonnegative.
The domain of f is then the set of all pairs of real numbers, that is, R
2
.
The range of f(x, y) is the set of all values of f(x, y), that is, the set of all numbers z, such that
z = f(x, y) =
_
x
2
+ 2y
2
+ 1 for some (x, y).
Hence z =
_
x
2
+ 2y
2
+ 1 ≥
√
1 = 1, for any pair (x, y), and the range of f(x, y) is then the interval
[1, ∞) of all real numbers greater then or equal to 1.
b) The level curves of f(x, y) are the curves defined by an equation f(x, y) = c for some c ≥ 1 (c
has to be in the range of f). We have:
f(x, y) = c, c ≥ 1
_
x
2
+ 2y
2
+ 1 = c
x
2
+ 2y
2
+ 1 = c
2
x
2
+ 2y
2
= c
2
−1
When c = 1, this is the point (0, 0).
When c > 1, this is the equation of an ellipse with major axis x.
Below is the sketch of the the contour map.
1.5 2
2.5
3
3.5
3.5
4
4
4.5
4.5
4.5
4.5
5
5 5
5
3 2 1 0 1 2 3
3
2
1
0
1
2
3
The graph of f(x, y) is the surface of equation z = f(x, y), that is:
z =
_
4x
2
+y
2
+ 1
z
2
= 4x
2
+y
2
+ 1
The graph is a (the upper part) of a hyperboloid (with two sheets).
2
0
2 x
2
0
2
y
1.0
1.5
2.0
2.5
3.0
z
2
0
2
x
2
0
2
y
1.0
1.5
2.0
2.5
3.0
z
The level curves at level c correspond to horizontal cross - sections of the graph at height c.
7
Problem 2.3. Consider the function f(x, y) =
_
x
2
+y
2
−1.
a) Find the domain and the range of f(x, y).
b) Sketch a contour map of f(x, y). Explain its relationship with the graph of f(x, y).
Solution. a) The domain of f(x, y) is the set of all pairs of real numbers (x, y) for which f(x, y)
is well defined. We know that the square root is defined only for nonnegative numbers.
We should then require that x
2
+y
2
−1 ≥ 0, or x
2
+y
2
≥ 1. This is the outside of the unit circle,
including the unit circle. The domain of f is then the set of all pairs of real numbers that lie on,
or outside of the circle of radius 1.
The range of f(x, y) is the set of all possible values of f(x, y), that is, the set of all numbers z,
such that z = f(x, y) =
_
x
2
+y
2
−1 for some (x, y). Because a square root is always nonnegative,
z ≥ 0, so the range of f(x, y) is the interval [0, ∞) of all real numbers greater then or equal to 0.
b) The level curves of f(x, y) are the curves defined by an equation f(x, y) = c for some c ≥ 0 (c
has to be in the range of f). We have:
f(x, y) = c, c ≥ 0
_
x
2
+y
2
−1 = c
x
2
+y
2
−1 = c
2
x
2
+y
2
= c
2
+ 1
For any number c (greater that or equal to 0), the level curve at level c is then a circle of radius
√
c
2
+ 1.
Note that the radius will be at least 1, (corresponding to the level c = 0), so the contour plot will
consist of concentric circles or larger and larger radii, starting with a circle of radius 1 - that is why
there is a whole in the contour plot below:
0.5
1
1.5
2
2.5
3
3
3
3
3.5 3.5
3.5
3.5
4 4
4
4
3 2 1 0 1 2 3
3
2
1
0
1
2
3
The graph of f(x, y) is the surface of equation z = f(x, y), that is:
z =
_
x
2
+y
2
−1
z
2
= x
2
+y
2
−1
This is (the upper part of) a hyperboloid (with one sheet).
2
0
2
x 2
0
2
y
0.0
0.5
1.0
1.5
2.0
z
The level curves at level c correspond to horizontal cross - sections of the graph at height c.
8
3. Limits of functions of two variables
Find the limit, if it exists (and prove that it does) or show that the limit does not exist:
Problem 3.1.
lim
(x,y)→(1,0)
y
2
ln x −3
sin πx +e
y
Solution:
lim
(x,y)→(1,0)
y
2
ln x −3
sin πx +e
y
=
0 · ln 1 −3
sin π +e
0
=
−3
1
= −3
Therefore, the limit exists and it equals −3.
Problem 3.2.
lim
(x,y)→(0,0)
x · y
x
3
+y
2
Solution: Let us take the path C
1
: x = 0, y →0, y = 0. Then
x · y
x
3
+y
2
=
0 · y
0 +y
2
=
0
y
2
= 0 →0 as (x, y) →(0, 0),
so the limit on this path is L
1
= 0.
Let us take the path C
2
: x = y →0. Then
x · y
x
3
+y
2
=
x
2
x
3
+x
2
=
1
x + 1
→
1
0 + 1
= 1, as (x, y) →(0, 0),
so the limit on this path is L
2
= 1.
L
1
= L
2
, so the limit does not exist.
Problem 3.3.
lim
(x,y)→(0,0)
x
2
ln(1 +y)
x
2
+ 3y
2
Solution: We have:
x
2
ln(1 +y)
x
2
+ 3y
2
= ln(1 +y) ·
x
2
x
2
+ 3y
2
But x
2
≤ x
2
+ 3y
2
, since 3y
2
≥ 0, therefore
x
2
x
2
+3y
2
≤ 1.
Then
0 ≤ | ln(1 +y) ·
x
2
x
2
+ 3y
2
| = | ln(1 +y)| · |
x
2
x
2
+ 3y
2
| ≤ | ln(1 +y)| · 1 →ln(1 + 0) = ln 1 = 0
as (x, y) →(0, 0).
Therefore, by the ’Squeeze Theorem’, the limit exists and equals 0.
9
Problem 3.4.
lim
(x,y)→(0,0)
x
2
sin(x −y)
x
3
+y
2
Solution: Let’s take the path C
1
: x = y →0, y = 0. Then
x
2
sin(x −y)
x
3
+y
2
=
x
2
· sin 0
x
3
+x
2
=
0
x
3
+x
2
= 0 →0
as (x, y) →(0, 0).
Then the limit on this path is L
1
= 0.
Let’s take the path C
2
: y = 0, x →0, x = 0. Then
x
2
sin(x −y)
x
3
+y
2
=
x
2
· sin x
x
3
=
sin x
x
→1
as (x, y) →(0, 0).
Then the limit on this path is L
2
= 1.
L
1
= L
2
, so the limit does not exist.
Problem 3.5.
lim
(x,y)→(0,0)
x
4
sin(y)
2x
4
+y
2
Solution: We can write
x
4
sin y
2x
4
+y
2
=
x
4
2x
4
+y
2
· sin y
But
x
4
2x
4
+y
2
≤ 1 since x
4
≤ 2x
4
≤ 2x
4
+y
2
.
Also sin y →0 as (x, y) →(0, 0).
Therefore, by the ’Squeeze Theorem’, the limit exists and equals 0.
Problem 3.6.
lim
(x,y)→(0,0)
e
1
x
2
· y
2
Solution: On the path x = y →0, we have:
e
1
x
2
· y
2
= e
1
x
2
· x
2
=
e
1
x
2
1
x
2
=
e
t
t
,
where t =
1
x
2
, so t →∞ as x →0.
But
e
t
t
→∞ as t →∞ (by L’Hˆopital’s rule, for instance).
Therefore, on this path the limit is L
1
= ∞.
Now let’s take the path y = e
−
1
x
2
, x →0.
Note that as x →0, we have y = e
−
1
x
2
→0.
On this path, we have:
e
1
x
2
· y
2
= e
1
x
2
· e
−2
1
x
2
= e
−
1
x
2
→0
as x →0.
Therefore, on this path the limit is L
2
= 0.
Since L
1
= L
2
, the limit does not exist.
10
4. Partial and directional derivatives, gradient vector, linear approximation
Problem 4.1. Let f(x, y) = ln(x −3y).
a) Compute f
x
(7, 2), f
y
(7, 2) and explain their geometric interpretation.
b) Find the tangent plane to the graph of f(x, y) at the point (7, 2, 0).
c) Find the linearization of f(x, y) at (7, 2) and use it to approximate f(6.9, 2.06). Explain why
you can use its linearization to approximate f.
Solution: a) We have:
f
x
=
1
x −3y
and f
y
= −
3
x −3y
Hence
f
x
(7, 2) =
1
7 −6
= 1 and f
y
(7, 2) = −
3
7 −6
= −3
Therefore, the slope of the tangent to the curve of intersection between the plane y = 2 and the
surface z = ln(x −3y) is 1.
Similarly, the slope of the tangent to the curve of intersection between the plane x = 7 and the
surface z = ln(x −3y) is −3.
b) The equation of the tangent plane to a surface z = f(x, y) at the point (x
0
, y
0
, z
0
) is
z −z
0
= f
x
(x
0
, y
0
) · (x −x
0
) +f
y
(x
0
, y
0
) · (y −y
0
)
The equation of the tangent plane to the graph of our function f(x, y) at (7, 2, 0) is then
z −0 = 1 · (x −7) −3 · (y −2)
which becomes: z = x −7 −3y + 6, or
x - 3 y - z = 1
b) The linearization of f(x, y) at (7, 2) is the linear function
L(x, y) = f(7, 2) +f
x
(7, 2) · (x −7) +f
y
(7, 2) · (y −2) =
= 0 + (x −7) −3(y −2) = x −7 −3y + 6 = x −3y −1
so
L(x, y) = x −3y −1
But f(x, y) is differentiable at (7, 2), since f
x
, f
y
are continuous at this point. Therefore
f(x, y) ≈ L(x, y)
when (x, y) is near (7, 2).
In particular, f(6.9, 2.06) ≈ L(6.9, 2.06) = 6.9 −3 · 2.06 −1 = −0.28, so
f(6.9, 2.06) ≈ −0.28
Problem 4.2. Consider the function f(x, y) = sin(πx
2
+πy).
a) What are the domain and the range of f(x, y) ? Why ?
b) Find the tangent plane to the graph of f(x, y) at the point (1, 0, 0).
c) Find the linear approximation of f(x, y) at (1, 0) and use it to approximate f(1.1, −0.1).
Solution: a) The domain of f(x, y) is the set of all pairs of real numbers (x, y) for which
f(x, y) is well defined. Since the sin function is defined for all real numbers, it follows that the
domain of f(x, y) is the set R
2
of all pairs of real numbers.
The range of f(x, y) is the set of all values of f(x, y). Since the range of the sin function is the
interval [−1, 1], the range of f(x, y) is [−1, 1] as well.
11
b) The equation of the tangent plane to the surface z = f(x, y) at the point (x
0
, y
0
, z
0
) is
z −z
0
= f
x
(x
0
, y
0
) · (x −x
0
) +f
y
(x
0
, y
0
) · (y −y
0
)
We have: f
x
= cos(πx
2
+πy) · 2πx, f
y
= cos(πx
2
+πy) · π
Then f
x
(1, 0) = cos π · 2π = −2π and f
y
(1, 0) = cos π · π = −π
The equation of the tangent plane to the graph of f(x, y) at (1, 0, 0) is then
z −0 = −2π · (x −1) −π · (y −0)
which becomes:
z = −2πx −πy + 2π
c) The linearization of f(x, y) at (1, 0) is the linear function
L(x, y) = z = −2πx −πy + 2π
But f(x, y) is differentiable at (1, 0), since f
x
, f
y
are continuous at this point. Therefore
f(x, y) ≈ L(x, y)
when (x, y) is near (1, 0).
In particular,
f(1.1, −0.1) ≈ L(1.1, −0.1) = −2π · 1.1 +π · 0.1 + 2π =
= (−2.2 + 0.1 + 2) · π = −0.1 · π, so
f(1.1, −0.1) ≈ −0.1 · π
Problem 4.3. Find the maximum rate of change of the function
f(x, y, z) = ln(xy
2
) +z
at (1, 1, 0).
Find the direction in which this maximum rate of change occurs.
Solution: The maximum rate of change of f(x, y, z) at the point (1, 1, 0) is the magnitude of its
gradient vector at that point, namely |∇f(1, 1, 0)|. This maximum rate of change occurs in the
direction of the gradient vector ∇f(1, 1, 0).
We have:
f
x
(x, y, z) =
1
xy
2
· y
2
=
1
x
, so f
x
(1, 1, 0) = 1
f
y
(x, y, z) =
1
xy
2
· x · 2y =
2
y
, so f
x
(1, 1, 0) = 2
f
z
(x, y, z) = 1 so f
z
(1, 1, 0) = 1
Then ∇f(1, 1, 0) =< 1, 2, 1 > and | < 1, 2, 0 > | =
√
1
2
+ 2
2
+ 1
2
=
√
6.
Therefore, the maximum rate of change of f at the point (1, 1, 0) is
√
6
and it occurs in the direction of the vector < 1, 2, 1 >.
12
Problem 4.4. The temperature (in degrees Fahrenheit) at a point (x, y) on a hot plate is given
by the function
T(x, y) = ln(9 +x
2
y)
An ant crawls on the plate so that its position after t seconds is
x = 10 −t, y = t
2
What is the rate of change of the temperature on the ant’s path after 9 seconds?
Hint: You may want to interpret the rate of change as a derivative and use the chain rule.
Solution: The temperature T is a function of two variables x and y (where x and y describe the
position on the plate) The ant crawls on a path so that its position after t seconds is given by
x = 10 − t, y = t
2
. Then the temperature T on the ant’s path is indirectly a function of t. The
rate of change of a quantity at a point is the derivative of that quantity at the given point.
Therefore, the rate of change of the temperature function T on the ant’s path after 9 seconds is
dT
dt
|
t=9
We will compute this derivative using the chain rule (case 1).
dT
dt
=
∂T
∂x
·
dx
dt
+
∂T
∂y
·
dy
dt
We have:
∂T
∂x
=
1
9 +x
2
y
· 2xy
∂T
∂y
=
1
9 +x
2
y
· x
2
dx
dt
= −1,
dy
dt
= 2t
When t = 9, x = 10 −t = 1 and y = t
2
= 81.
Then
∂T
∂x
|
t=9
=
1
9 + 81
· 2 · 81 =
2 · 81
90
=
18
10
= 1.8
∂T
∂y
|
t=9
=
1
90
· 1 =
1
90
Then
dT
dt
|
t=9
= 1.8 · (−1) +
1
90
· 2 · 9 = −1.8 + 0.2 = −1.6
The rate of change of the temperature on the ant’s path after 9 seconds is then
- 1.6
degrees Fahrenheit
seconds
13
5. Local and absolute minima and maxima
Problem 5.1. Find the absolute maximum and the absolute minimum values of the function
f(x, y) = 2x + 6y −xy −y
2
on the set: D = {(x, y) : 0 ≤ x ≤ 4, 0 ≤ y ≤ 3}.
Solution: i.) We first find the critical points of the function f(x, y) inside the domain, and evaluate
f at these critical points. To find the critical points we solve the system
f
x
(x, y) = 0
f
y
(x, y) = 0
We have: f
x
= 2 −y, f
y
= 6 −x −2y, so
2 −y = 0
6 −x −2y = 0
Then y = 2 from the first equation, and if we plug in 2 for y in the second equation, we get x = 2.
So the only critical point is (2, 2), and f (2, 2) = 8 .
ii.) We find the extreme values (absolute minimum and maximum) of f on the boundary of the
domain D. The boundary of D (which is a rectangle) consists of four line segments that we denote
(counterclockwise) by L
1
, L
2
, L
3
, L
4
. We will find the extreme values of f on each of these four
line segments.
→ I) The equation of L
1
is: y = 0 and 0 ≤ x ≤ 4.
Then on L
1
we have: f(x, y) = f(x, 0) = 2x, which is an increasing function on [0, 4] so its extreme
values are attained at the endpoints 0, 4.
The extreme values of f on L
1
are f (0, 0) = 0 and f (4, 0) = 8 .
→ II) The equation of L
2
is: x = 4 and 0 ≤ y ≤ 3.
Then on L
2
we have: f(x, y) = f(4, y) = 8 + 6y −4y −y
2
= 8 + 2y −y
2
.
Hence we have to find the absolute minimum and maximum values of the one - variable function
g(y) = 8 + 2y −y
2
on the interval [0, 3].
We proceed according to the method in one - variable calculus.
The critical points of g(y) are obtained by solving g

(y) = 0.
g

(y) = 2 −2y, then 2 −2y = 0 so the only critical point is y = 1 and g(1) = 9.
Now we evaluate g(y) at the endpoints of the interval [0, 3] and get g(0) = 8 and g(3) = 5.
Then the maximum of g on [0, 3] is max {9, 8, 5} = 9 at y = 1 and the minimum of g on [0, 3] is
min {9, 8, 5} = 5 at y = 3.
Therefore, the extreme values of f(x, y) on L
2
are f (4, 1) = 9 and f (4, 3) = 5 .
→ III) The equation of L
3
is: y = 3 and 0 ≤ x ≤ 4.
Then on L
3
we have: f(x, y) = f(x, 3) = 2x + 18 −3x −9 = 9 −x, which is a decreasing function
on [0, 4] so its extremes values are attained at the endpoints x = 0 and x = 4.
The extreme values of f on L
3
are f (0, 3) = 9 and f (4, 3) = 5 .
→ IV) The equation of L
4
is: x = 0 and 0 ≤ y ≤ 3.
Then on L
2
we have: f(x, y) = f(0, y) = 0 + 6y −0 −y
2
= 6y −y
2
.
We proceed like in II). Let h(y) = 6y −y
2
on [0, 3].
Then h

(y) = 6 −2y so 6 −2y = 0 and get y = 3.
Then h(3) = 9, h(0) = 0 are the maximum and minimum of h on [0, 3].
Therefore, the extreme values of f(x, y) on L
4
are f (0, 3) = 9 and f (0, 0) = 0 .
iii.) Now we take the maximum and minimum of all values obtained in the previous two steps and
conclude that:
The absolute maximum value of f(x, y) on the domain D is 9, and it is attained at (4, 1) and (0, 3).
The absolute minimum value of f on D is 0, attained at (0, 0) .
14
Problem 5.2. Find the absolute maximum and the absolute minimum values of
f(x, y) = x
2
+y
2
+xy −x −y
on the region enclosed by the triangle with vertices (0, 0), (1, 0), (0, 2).
Indicate where these extreme values are attained.
Explain why these extreme values have to exist.
Solution. f(x, y) is a continuous function and its domain is bounded and it contains its boundary
(the sides of the triangle are part of the domain.) Therefore, the extreme (minimum and maximum)
values of f have to exist on this domain.
L
1
L
3
L
2
y = -2 x + 2
y = 0
x = 0
1
x
1
2
y
i.) We first find the critical points of the function f(x, y) inside the domain, and evaluate f at
these critical points. To find the critical points we solve the system
f
x
= 0
f
y
= 0
We have: f
x
= 2x +y −1, f
y
= 2y +x −1, so
2x +y = 1
x + 2y = 1
Solving this linear system of equations, we get x =
1
3
and y =
1
3
.
So the only critical point is (
1
3
,
1
3
), this point is inside the domain (inside the triangle) and
f(
1
3
,
1
3
) = −
1
3
ii.) We find the extreme values (absolute minimum and maximum) of f on the boundary of the
domain.
The boundary of the domain (which is a triangle) consists of three line segments that we denote
by L
1
, L
2
and L
3
.. We will find the critical points of f on each of these three line segments, and
evaluate f at these critical points. Then, we will evaluate f at the three vertices of the triangle.
→ I) The equation of L
1
is: y = 0 and 0 ≤ x ≤ 1.
Then on L
1
we have: f(x, y) = f(x, 0) = x
2
−x.
We find the critical points on this line: (x
2
−x)

= 2x −1 = 0, so x =
1
2
, which is between 0 and 1.
We get the point (
1
2
, 0) and f(
1
2
, 0) = −
1
4
15
→ II) L
2
is the line segment joining the points (1, 0) and (0, 2). The equation of L
2
is then :
y = −2x + 2, for 0 ≤ x ≤ 1.
Then on L
2
we have:
f(x, y) = f(x, −2x + 2) = x
2
+ (−2x + 2)
2
+x · (−2x + 2) −x −(−2x + 2) =
= (after multiplying out and collecting like terms) 3x
2
−5x + 2.
We find the critical points on this line: (3x
2
−5x + 2)

= 6x −5 = 0, so x =
5
6
, which is between 0
and 1.
Then y = −2 ·
5
6
+ 2 =
1
3
and we get the point (
5
6
,
1
3
) for which f(
5
6
,
1
3
) = −
1
12
→ III) The equation of L
3
is: x = 0 and 0 ≤ y ≤ 2.
Then on L
3
we have: f(x, y) = f(0, y) = y
2
−y.
We find the critical points on this line: (y
2
−y)

= 2y −1 = 0, so y =
1
2
, which is between 0 and 2.
We get the point (0,
1
2
) and f(0,
1
2
) = −
1
4
We now evaluate f at the vertices of the triangle:
f (0, 0) = 0 , f (1, 0) = 0 and f (0, 2) = 2 .
iii.) Now we take the maximum and the minimum of all the values obtained in the previous two
steps (they were all put in boxes) and conclude that:
The absolute maximum value of f(x, y) on the given domain is 2,
and it is attained at (0, 2).
The absolute minimum value of f on the given domain is −
1
3
, attained at (
1
3
,
1
3
) .
Maximum
Minimum
1
3
1
2
5
6
1
x
1
3
1
2
2
y
16
Problem 5.3. Find the local minimum and the local maximum points (and values) and the saddle
points (if any) of the function
f(x, y) = 3x
3
+ 6yx −y
2
Sketch the graph of the function around the points found above.
Note: You do not have to sketch the graph of the whole function. Just show how the graph of the
function looks like around the local minimum, the local maximum or the saddle points determined.
Solution: i.) We find the critical points first, by solving the system of equations:
f
x
(x, y) = 0
f
y
(x, y) = 0
We have: f
x
= 9x
2
+ 6y, f
y
= 6x −2y, so
9x
2
+ 6y = 0
6x −2y = 0
From the second equations it follows that y = 3x. Substitute 3x for y in the first equation to get:
9x
2
+ 6 · 3x = 0
9x
2
+ 18x = 0
9x · (x + 2) = 0
Then x = 0 or x = −2.
But y = 3x. Hence y = 0 when x = 0 and y = −6 when x = −2.
The critical points of f(x, y) are then: (0, 0) and (−2, −6).
ii.) We use the second derivative test to classify these critical points.
We have:
f
xx
= 18x, f
yy
= −2, f
xy
= f
yx
= 6
Then
D(x, y) =
¸
¸
¸
¸
f
xx
f
xy
f
yx
f
yy
¸
¸
¸
¸
=
¸
¸
¸
¸
18x 6
6 −2
¸
¸
¸
¸
= 18x · (−2) −6 · 6 = −36x −36
Therefore D(x, y) = −36x −36.
D(0, 0) = −36 < 0 so (0, 0) is a saddle point and the graph around (0, 0) looks like (surprise !) a saddle.
D(−2, −6) = −36 · (−2) −36 = +36 > 0
Moreover, f
xx
(−2, −6) = 18 · (−2) = −36 < 0, so
(−2, −6) is a local maximum point and the graph around (−2, −6) looks like a downward elliptic paraboloid.
17
6. Constrained optimization: the method of Lagrange multipliers
Problem 6.1. Find the distance from the origin O to the line L with equation 3x −2y = 13
Solution: The problem can be rephrased as follows: find the minimum distance from the origin O
to a point P(x, y) that lies on the line L with equation 3x −2y = 13.
The distance from O(0, 0) to the point P(x, y) is d =
_
x
2
+y
2
. Clearly this quantity will attain
its minimum at the same point as the quantity d
2
= x
2
+y
2
attains its minimum.
Consider then the functions f(x, y) = x
2
+y
2
and g(x, y) = 3x −2y.
Our problem then becomes: find the minimum of f(x, y) subject to the constraint g(x, y) = 13.
We will solve it using the method of Lagrange multipliers, which says that at any solution to this
problem, the gradients of f and g have to be collinear.:
∇f = λ · ∇g
so
f
x
= λ · g
x
f
y
= λ · g
y
g(x, y) = 13
This becomes
2x = λ · 3
2y = λ · (−2)
3x −2y = 13
Solving for x and y in the first two equations in terms of λ, and substituting into the third equation
leads to
9
2
λ + 2λ = 13
We solve this equation and get λ = 2. Then we obtain x = 3 and y = −2.
This shows that the closest point on the line L to the origin O is (3, −2).
Its distance to the origin is
_
3
2
+ (−2)
2
=
√
13.
Problem 6.2. The plane x +y + 2z = 2 intersects the paraboloid z = x
2
+y
2
in an ellipse.
Find the highest and the lowest points on this ellipse.
Hint: The highest (or the lowest) point on the ellipse is the point with the largest (or the lowest)
z coordinate. Use Lagrange multipliers with two constraints.
Solution: The problem can be reformulated as follows: find the point (x, y, z) that lies both on
the paraboloid x
2
+ y
2
− z = 0 and on the plane x + y + 2z = 2, for which its z - coordinate is
maximized (or minimized respectively).
Let f(x, y, z) = z, g(x, y, z) = x
2
+y
2
−z and h(x, y, z) = x +y + 2z.
The problem becomes: find the maximum and the minimum values (and the points where they are
attained) of the function f(x, y, z) subject to the following two constraints:
g(x, y, z) = 0, h(x, y, z) = 2
We use the method of Lagrange multipliers with two constraints:
f
x
= λ · g
x
+µ · h
x
f
y
= λ · g
y
+µ · h
y
f
z
= λ · g
z
+µ · h
z
g(x, y, z) = 0
h(x, y, z) = 2
18
We have:
f
x
= 0, f
y
= 0, f
z
= 1
g
x
= 2x, g
y
= 2y, g
z
= −1
h
x
= 1, h
y
= 1, h
z
= 2
Then the system of equations becomes:
0 = λ · 2x +µ · 1
0 = λ · 2y +µ · 1
1 = λ · (−1) +µ · 2
x
2
+y
2
−z = 0
x +y + 2z = 2
From the first two equations we get: λ · 2x = λ · 2y, so 2λ · (x −y) = 0.
Then either λ = 0 or x −y = 0.
If λ = 0, from the 1st (or 2nd) equation we get µ = 0. But from the
3rd equation 2µ = 1 +λ = 1 so µ =
1
2
, which is a contradiction.
Therefore, λ cannot be 0, and then x −y = 0 which implies that x = y.
The 4th and 5th equations become:
2x
2
−z = 0
2x + 2z = 2
Then x +z = 1, or z = 1 −x, so
2x
2
−(1 −x) = 2x
2
+x −1 = 0
Solving this quadratic equation (using, for instance, the quadratic formula), we get two solutions:
x =
1
2
and x = −1.
Since z = 1 − x and x = y, we conclude that the original system has two solutions: (
1
2
,
1
2
,
1
2
) and
(−1, −1, 2).
f(
1
2
,
1
2
,
1
2
) =
1
2
, f(−1, −1, 2) = 2
Therefore, the lowest point on the ellipse is (
1
2
,
1
2
,
1
2
) at height
1
2
, and the highest point the ellipse
is (-1, -1, 2) at height 2 .
Problem 6.3. Find the absolute maximum and the absolute minimum values of the function
f(x, y) = 2x
3
+y
4
on the domain: 3x
2
+ 2y
2
≤ 1.
Indicate where these extreme values are attained.
Explain why these extreme values have to exist.
Solution: f(x, y) is a continuous function and its domain(an ellipse together with its interior) is
a closed and bounded set. Therefore, due to a (known) theorem, the extreme values of f exist on
this domain.
i.) We first find the critical points of the function f(x, y) inside the domain, and evaluate f at
these critical points.
To find the critical points we solve the system
f
x
(x, y) = 0
f
y
(x, y) = 0
19
We have: f
x
= 6x
2
, f
y
= 4y
3
, so
6x
2
= 0
4y
3
= 0
Then x = 0, y = 0 and the only critical point is (0, 0).
The value of f at this critical point is f (0, 0) = 0
ii.) We find the extreme values (absolute minimum and maximum) of f on the boundary of our
domain, which is the ellipse: 3x
2
+ 2y
2
= 1.
Let g(x, y) = 3x
2
+ 2y
2
.
The problem of finding the extreme values of f on the boundary of our domain becomes: find the
extreme values of f(x, y) subject to the constraint g(x, y) = 1. We will use the method of Lagrange
multipliers to solve this problem.
f
x
= λ · g
x
f
y
= λ · g
y
g(x, y) = 1
We have: f
x
= 6x
2
, f
y
= 4y
3
, g
x
= 6x, g
y
= 4y, so:
6x
2
= λ · 6x
4y
3
= λ · 4y
3x
2
+ 2y
2
= 1
Then
6x · (x −λ) = 0
4y · (y
2
−λ) = 0
3x
2
+ 2y
2
= 1
From the 1st equation we get x = 0 or x −λ = 0.
Let us consider these two cases separately:
→ I) If x = 0, then by using the third equation we get 2y
2
= 1 so y = ±
1
√
2
and f(0, ±
1
√
2
) =
1
4
.
→ II) If x −λ = 0 then x = λ.
Let’s look at the second equation now. We get: y = 0 or y
2
−λ = 0
II.1) If y = 0, using the third equation, 3x
2
= 1 so x = ±
1
√
3
so f(
1
√
3
, 0) =
2
√
27
and
f(−
1
√
3
, 0) = −
2
√
27
II.2) If y
2
−λ = 0 then y
2
= λ. We also have x = λ.
Using the third equation: 3x
2
+ 2y
2
= 1 we get 3λ
2
+ 2λ −1 = 0.
Solving this quadratic equation we get λ = −1 and λ =
1
3
.
Since y
2
= λ, λ = −1 does not give any solutions.
For λ =
1
3
we get x =
1
3
and y = ±
1
√
3
Then f(
1
3
, ±
1
√
3
) =
2
27
+
1
9
=
5
27
.
iii.) We conclude, comparing all the values obtained in the previous two steps that:
The absolute maximum value of f(x, y) on 3x
2
+ 2y
2
≤ 1 is
2
√
27
and it is attained at (
1
√
3
, 0).
The absolute minimum value of f(x, y) is −
2
√
27
, attained at (−
1
√
3
, 0).
20
Problem 6.4. Find the absolute minimum and the absolute maximum values of the function
f(x, y) = x
4
+y
4
on the closed unit disk x
2
+y
2
≤ 1.
Indicate where these extreme values are attained.
Explain why these extreme values have to exist.
Solution. f(x, y) is a continuous function and its domain(the closed unit disk) is bounded and
it contains its boundary (the circle is part of the domain.) Therefore, the extreme (minimum and
maximum) values of f have to exist on this domain.
i.) We first find the critical points of the function f(x, y) and evaluate f at these critical points.
To find the critical points we solve the system
f
x
(x, y) = 0
f
y
(x, y) = 0
We have: f
x
= 4x
3
, f
y
= 4y
3
, so 4x
3
= 0 and 4y
3
= 0.
Then x = 0, y = 0 and the only critical point is (0, 0).
The value of f at this critical point is f (0, 0) = 0
ii.) We find the extreme values (absolute minimum and maximum) of f on the boundary of our
domain. The boundary of the closed unit disk is the unit circle: x
2
+y
2
= 1.
Let g(x, y) = x
2
+y
2
.
The problem of finding the extreme values of f on the boundary of the unit disk becomes: find the
extreme values of f(x, y) subject to the constraint g(x, y) = 1. We will use the method of Lagrange
multipliers to solve this problem.
f
x
= λ · g
x
f
y
= λ · g
y
g(x, y) = 1
We have: f
x
= 4x
3
, f
y
= 4y
3
, g
x
= 2x, g
y
= 2y, so:
4x
3
= λ · 2x
4y
3
= λ · 2y
x
2
+y
2
= 1
Then
2x · (2x
2
−λ) = 0
2y · (2y
2
−λ) = 0
x
2
+y
2
= 1
From the first equation we get: x = 0 or 2x
2
−λ = 0
Let us consider these two cases separately:
→ I) If x = 0, using the third equation, y
2
= 1 so y = ±1 and then f(0, ±1) = 1 .
→ II) 2x
2
−λ = 0 then x
2
=
λ
2
.
Let’s look at the second equation now. We get: y = 0 or 2y
2
−λ = 0
II.1) If y = 0, using the third equation, x
2
= 1 so x = ±1 and then f(±1, 0) = 1 .
II. 2) If 2y
2
−λ = 0 then y
2
=
λ
2
We also have x
2
=
λ
2
.
Using the third equation: x
2
+y
2
= 1 we get
λ
2
+
λ
2
= 1 so λ = 1.
But x
2
=
λ
2
=
1
2
and y
2
=
λ
2
=
1
2
, so x = ±
1
√
2
and y = ±
1
√
2
Then f(±
1
√
2
, ±
1
√
2
) =
1
2
.
iii.) We conclude, taking the maximum and the minimum of all the values obtained in the previous
two steps that:
The absolute maximum value of f(x, y) on x
2
+y
2
≤ 1 is 1 and it is attained at (0, ±1) and (±1, 0).
The absolute minimum value of f is 0, attained at (0, 0) .
21
7. Double and triple integrals
Problem 7.1. Compute
__
D
xe
y
dA
where D is the region that lies between the curves y = x and y = x
2
.
Solution:
Note that D is a domain of type I in R
2
(it is also a domain of type II, so the double integral above
can be computed differently). It lies between the curves y = x
2
and y = x (since the straight line
through the points (0, 0) and (1, 1) is y = x) and 0 ≤ x ≤ 1. Hence:
D = {(x, y) : 0 ≤ x ≤ 1, x
2
≤ y ≤ x}
Then
__
D
xe
y
dA =
_
1
0
(
_
x
x
2
xe
y
dy) dx =
=
_
1
0
xe
y
y=x
y=x
2
dx =
_
1
0
xe
x
dx −
_
1
0
xe
x
2
dx
We compute the two single integrals separately.
For the first one we use integration by parts.
Let u = x, v = e
x
, so dv = e
x
dx and du = dx. Then
_
1
0
xe
x
dx = xe
x
1
0
−
_
1
0
e
x
dx = e −e
x
1
0
= e −e + 1 = 1
For the second one we use the substitution u = x
2
so du = 2xdx. Then
_
1
0
xe
x
2
dx =
1
2
_
1
0
e
u
du =
1
2
e
u
1
0
=
1
2
(e −1)
Therefore,
__
D
xe
y
dA = 1 −
1
2
(e −1) =
3
2
−
1
2
e
Problem 7.2. Consider a lamina that occupies the part of the disk x
2
+ y
2
≤ 4 that lies in the
first quadrant.
Find the center of mass assuming that the density at any point on the lamina is given by the
function ρ(x, y) = x ·
_
x
2
+y
2
.
22
Solution: The domain D occupied by the lamina is a polar rectangle (pictured below).
D = {(r, θ) : 0 ≤ r ≤ 2 and 0 ≤ θ ≤
π
2
}
We use polar coordinates: x = r cos θ, y = r sin θ, so
_
x
2
+y
2
=
√
r
2
= r (since r ≥ 0).
The total mass of the lamina is given by:
m =
__
D
ρ(x, y) dA =
__
D
x ·
_
x
2
+y
2
dA =
=
_ π
2
0
_
2
0
r cos θ · r · r dr dθ =
_ π
2
0
_
2
0
r
3
cos θ dr dθ =
=
_
_ π
2
0
cos θ dθ
_
·
__
2
0
r
3
dr
_
=
_
sin θ
π/2
0
_
·
_
r
4
4
2
0
_
= 1 · 4 = 4
Then m = 4 .
The moment M
y
about the y - axis is given by:
M
y
=
__
D
x · ρ(x, y) dA =
__
D
x · x ·
_
x
2
+y
2
dA =
=
_ π
2
0
_
2
0
r
2
cos
2
θ · r · r dr dθ =
_ π
2
0
_
2
0
r
4
cos
2
θ dr dθ =
=
_
_ π
2
0
cos
2
θ dθ
_
·
__
2
0
r
4
dr
_
_
2
0
r
4
dr =
r
5
5
2
0
=
32
5
To compute
_
π
2
0
cos
2
θ dθ we use the trigonometric formula
cos
2
θ =
1 + cos 2θ
2
=
1
2
+
1
2
cos 2θ
Then
_ π
2
0
cos
2
θ dθ =
_ π
2
0
1
2
dθ +
1
2
_ π
2
0
cos 2θ dθ =
=
1
2
·
π
2
+
1
2
·
sin 2θ
2
π/2
0
=
π
4
+ 0 =
π
4
Then M
y
=
32
5
·
π
4
=
8π
5
, so
M
y
=
8π
5
The moment M
x
about the x - axis is given by:
M
x
=
__
D
y · ρ(x, y) dA =
__
D
y · x ·
_
x
2
+y
2
dA =
=
_ π
2
0
_
2
0
r sin θ · r cos θ · r · r dr dθ =
_ π
2
0
_
2
0
r
4
sin θ · cos θ dr dθ =
23
=
_
_ π
2
0
sin θ · cos θ dθ
_
·
__
2
0
r
4
dr
_
_
2
0
r
4
dr =
r
5
5
2
0
=
32
5
To compute
_
π
2
0
sin θ · cosθ dθ we can use either the trigonometric formula
sin 2θ = 2 sin θ cos θ, so
sin θ · cos θ =
1
2
sin 2θ
or the substitution t = sin θ, so dt = cos θ dθ.
If θ = 0 then t = sin 0 = 0 and if θ =
π
2
then t = sin
π
2
= 1.
Then
_ π
2
0
sin θ · cosθ dθ =
_
1
0
t dt =
t
2
2
1
0
=
1
2
Then M
x
=
1
2
·
32
5
=
16
5
, so
M
x
=
16
5
The center of mass (¯ x, ¯ y) is given by:
¯ x =
M
y
m
=
2π
5
¯ y =
M
x
m
=
4
5
The center of mass of the lamina is then (
2π
5
,
4
5
) .
Problem 7.3. Compute the following double integral by making an appropriate change of variables:
__
R
xy dA
where R is the square with vertices (0, 0), (1, 1), (2, 0), (1, −1).
Solution:
Let
u = x +y
v = x −y
Then the images of the points (0, 0), (1, 1), (2, 0), (1, −1) in the xy - plane are the points (0, 0),
(2, 0), (2, 2), (0, 2) in the uv - plane. Therefore, the image of the given square R in the xy - plane
is the square S of vertices (0, 0), (2, 0), (2, 2), (0, 2) in the uv - plane.
24
Solving for x, y, the system of equations
u = x +y
v = x −y
we get:
x =
u +v
2
=
1
2
u +
1
2
v
y =
u −v
2
=
1
2
u −
1
2
v
Consider then the transformation
T(u, v) = (x, y)
where
x =
1
2
u +
1
2
v
y =
1
2
u −
1
2
v
The image of the square S via the transformation T is then the square R (see the picture above).
The Jacobian of this transformations is
∂(x, y)
∂(u, v)
=
¸
¸
¸
¸
1/2 1/2
1/2 −1/2
¸
¸
¸
¸
= −
1
2
Using the change of variables formula for double integrals, we have:
__
R
xy dA =
__
S
(
1
2
u +
1
2
v) · (
1
2
u −
1
2
v) · | −
1
2
| dA =
=
_
2
0
_
2
0
(
1
4
u
2
−
1
4
v
2
) ·
1
2
dudv ==
1
8
_
2
0
[
_
2
0
(u
2
−v
2
) du] dv =
=
1
8
_
2
0
(
u
3
3
−v
2
u)
u=2
u=0
dv =
1
8
_
2
0
(
8
3
−2v
2
) dv =
=
1
8
· (
8
3
v −2
v
3
3
)
v=2
v=0
=
1
8
· (
8
3
· 2 −2 ·
8
3
) = 0
Therefore,
__
R
xy dA = 0
Note: There is a much shorter way of showing that this integral is 0 (using a trick and a change of
variables). Think about it.
Problem 7.4. Find the volume and the area of the sphere of radius 1.
Solution: The area of the sphere of radius 1 is twice the area of the upper hemisphere.
The equation of the sphere of radius 1 is x
2
+y
2
+z
2
= 1, so z
2
= 1 −x
2
−y
2
.
Then the upper hemisphere (pictured above) is the graph of the function f(x, y) =
_
1 −x
2
−y
2
defined on the unit disk D = {(x, y) : x
2
+y
2
≤ 1} pictured below.
It follows that the area of the upper hemisphere is
A =
__
D
_
f
2
x
+f
2
y
+ 1 dA
25
We have: f
x
= −
x
√
1−x
2
−y
2
and f
y
= −
y
√
1−x
2
−y
2
. Then
f
2
x
+f
2
y
+ 1 =
x
2
1 −x
2
−y
2
+
y
2
1 −x
2
−y
2
+ 1 =
=
x
2
+y
2
+ 1 −x
2
−y
2
1 −x
2
−y
2
=
1
1 −x
2
−y
2
Hence
A =
__
D
1
_
1 −x
2
−y
2
dA
The unit disk D is a polar rectangle:
D = {(r, θ) : 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π}
We use polar coordinates: x = r cos θ, y = r sin θ so x
2
+y
2
= r
2
.
Then
A =
__
D
1
_
1 −x
2
−y
2
dA =
_
2π
0
_
1
0
1
√
1 −r
2
· r dr dθ = 2π ·
_
1
0
1
√
1 −r
2
· r dr
To compute the integral above we use the substitution
s = 1 −r
2
ds = −2rdr
rdr = −
1
2
ds
Then
A = 2π ·
_
1
0
1
√
1 −r
2
· r dr = 2π ·
1
2
_
1
0
1
√
s
ds =
= π
_
1
0
s
−1/2
ds = 2π
We conclude that the area of the (whole) unit sphere is 4π.
Similarly, the volume of the unit sphere is twice the volume of the upper hemisphere, which is the
solid above the unit disk D and below the graph of the function z =
_
1 −x
2
−y
2
.
Therefore, the volume of the upper hemisphere is:
V =
__
D
_
1 −x
2
−y
2
dA
Using polar coordinates again, we get:
V =
_
2π
0
_
1
0
_
1 −r
2
· r dr dθ = 2π
_
1
0
_
1 −r
2
· r dr
Using the same substitution s = 1 −r
2
, the integrals becomes:
V = 2π
_
1
0
_
1 −r
2
· r dr = 2π ·
1
2
·
_
1
0
s
1/2
ds = π ·
2
3
We conclude that the volume of the (whole) unit sphere is
4π
3
.
26
8. Line integrals of functions and of vector fields, Green’s theorem
Problem 8.1. Let f(x, y) = x
3
and let C be the curve that consists of the arc C
1
of the parabola
y = x
2
from (0, 0) to (1, 1), followed by the straight line C
2
from (1, 1) to (4, 1).
Sketch the curve C. Then find the area of the surface between the curve C and the graph of f(x, y)
along C. That is, find the area of the ”fence” whose base is C and whose height above (x, y) is
f(x, y).
Solution: Here is the sketch of the curve C.
Using the geometric interpretation of the line integral of a scalar function with respect to arc length,
the area of the surface between the curve C and the graph of f along C is (see the picture below):
A =
_
C
f(x, y) ds =
_
C
1
f(x, y) ds +
_
C
2
f(x, y) ds
We compute each of the two integrals separately.
→ The integral along C
1
: x = t, y = t
2
, where 0 ≤ t ≤ 1.
ds =
_
(
dx
dt
)
2
+ (
dx
dt
)
2
dt =
_
1 + (2t)
2
dt =
√
1 + 4t
2
dt.
Then
_
C
1
f(x, y) ds =
_
1
0
t
3
_
1 + 4t
2
dt
Use the substitution u = 1 + 4t
2
, so du = 8t dt.
Solving for t
2
, we have t
2
=
u−1
4
=
1
4
(u −1).
Then
_
1
0
t
3
_
1 + 4t
2
dt =
1
8
_
1
0
t
2
_
1 + 4t
2
8 t dt =
1
8
_
5
1
1
4
(u −1)
√
u du =
=
1
32
_
5
1
(u −1) u
1/2
du =
1
32
_
5
1
(u
3/2
−u
1/2
) du =
=
1
32
(
2
5
· u
5/2
5
1
−
2
3
· u
3/2
5
1
) =
1
32
[
2
5
· (5
5/2
−1) −
2
3
· (5
3/2
−1)]
27
→ The integral along C
2
: x = t, y = 1, where 1 ≤ t ≤ 4.
Then ds =
_
(
dx
dt
)
2
+ (
dx
dt
)
2
dt =
√
1 + 0 dt = dt, so
_
C
2
f(x, y) ds =
_
4
1
t
3
dt =
t
4
4
4
1
= 64 −
1
4
We conclude:
A =
1
32
[
2
5
· (5
5/2
−1) −
2
3
· (5
3/2
−1)] + 64 −
1
4
,
which looks like the right answer for that area :)
Problem 8.2. Let f(x, y) = ln(4x
2
+y
2
+ 1).
a) Find the gradient vector field ∇f.
b) Sketch the gradient vector field ∇f together with a contour map of f. Explain how they are
related to each other.
c) Compute
_
C
∇f · dr
where C is a piecewise - smooth curve from (0, 0) to (1, 2).
Solution: a)
∇f(x, y) =<
∂f
∂x
,
∂f
∂y
>=<
8x
4x
2
+y
2
+ 1
,
2y
4x
2
+y
2
+ 1
>
b) The contour map of f is a picture showing a few level curves of f.
The level curve of f at level k is the curve of equation f(x, y) = k.
That is, ln(4x
2
+y
2
+ 1) = k. This becomes 4x
2
+y
2
+ 1 = e
k
or 4x
2
+y
2
= e
k
−1
If k < 0 there is no level curve because 4x
2
+y
2
≥ 0, while e
k
−1 < e
0
−1 = 0.
If k = 0, then 4x
2
+y
2
= e
0
−1 = 0 so the level curve at level 0 is just the point (0, 0).
If k > 0, the equation 4x
2
+y
2
= e
k
−1 is the equation of an ellipse with major axis y.
Therefore, the contour map consists of ellipses with major axis y at every level k (see below).
We know that ∇f(x, y) ⊥ level curve of f through (x, y).
Here is the sketch of the gradient vector field ∇f.
Contour map Contour map with gradient vector
c) Using the FTC for line integrals, we have:
_
C
∇f · dr = f(1, 2) −f(0, 0) = ln(4 + 4 + 1) −ln(0 + 0 + 1) = ln 9 −ln 1 = ln 9
28
Problem 8.3. Find the work done by the force field
F(x, y, z) = zi +xj +yk
in moving a particle from the point (3, 0, 0) to the point (0,
π
2
, 3) along:
a) A straight line.
b) The helix x = 3 cos t, y = t, z = 3 sin t. It would be nice if you also drew the helix.
Solution: We can write F(x, y, z) =< z, x, y >.
The work done by the force vector field F in moving a particle along a curve C is
W =
_
C
F · dr
a) The vector equation of the straight line (or line segment) C
1
from r
1
=< 3, 0, 0 > to r
2
=<
0,
π
2
, 3 > is
r(t) = (1 −t) · r
1
+t · r
2
=
= (1 −t)· < 3, 0, 0 > +t · < 0,
π
2
, 3 >=< 3 −3t,
π
2
t, 3t >
where the parameter t is between 0 and 1. Hence the vector equation of the line C
1
is r(t) =<
3 − 3t,
π
2
t, 3t >, where 0 ≤ t ≤ 1 (or if you prefer parametric equations, they are : x = 3 − 3t,
y =
π
2
t, z = 3t). Moreover, r

(t) =< −3,
π
2
, 3 >.
Then we have:
W =
_
C
1
F · dr =
_
1
0
F(r(t)) · r

(t)dt =
=
_
1
0
< 3t, 3 −3t,
π
2
t > · < −3,
π
2
, 3 > dt =
_
1
0
(−9t + (3 −3t)
π
2
+
π
2
t · 3) dt =
=
_
1
0
(−9t +
3π
2
) dt = −9
t
2
2
1
0
+
3π
2
= −
9
2
+
3π
2
b) The parametric equations of the given helix (let us call it C
2
) are x = 3 cos t, y = t, z = 3 sin t,
so the vector equation of the helix is r(t) =< 3 cos t, t, 3 sin t >.
Then r

(t) =< −3 sin t, 1, 3 cos t >.
We have to determine the interval [a, b] where the parameter t lies. We know that this helix starts
at (3, 0, 0) and ends at (0,
π
2
, 3). Therefore, r(a) =< 3, 0, 0 > and r(b) = (0,
π
2
, 3).
Then a = 0 and b =
π
2
, so 0 ≤ t ≤
π
2
.
Below is the sketch of the helix.
29
We have:
W =
_
C
2
F · dr =
_
1
0
F(r(t)) · r

(t) dt =
=
_ π
2
0
< 3 sin t, 3 cos t, t > · < −3 sin t, 1, 3 cos t > dt =
=
_ π
2
0
(−9 sin
2
t + 3 cos t + 3t cos t) dt = −9
_ π
2
0
sin
2
t dt + 3
_ π
2
0
cos t dt + 3
_ π
2
0
t cos t dt
_ π
2
0
sin
2
t dt =
_ π
2
0
1 −cos 2t
2
dt =
1
2
·
π
2
−
1
2
sin 2t

π
2
0
=
π
4
_ π
2
0
cos t dt = sin t

π
2
0
= 1
To compute
_
π
2
0
t cos t dt we need to use integration by parts:
_
b
a
udv = uv
b
a
−
_
b
a
v du
Let u = t and v = sin t. Then dv = cos t dt and du = 1 · dt = dt.
Then
_ π
2
0
t cos t dt = t sin t

π
2
0
−
_ π
2
0
sin t dt =
π
2
+ cos t

π
2
0
=
π
2
−1
After doing the calculations we get:
W = −9 ·
π
4
+ 3 · 1 + 3 · (
π
2
−1) = −
3π
4
Problem 8.4. Let F(x, y) =< 3x
2
+ ln y,
x
y
+e
7y
> be a vector field.
a) Show that F is a conservative vector field. Indicate the domain of F.
b) Find a function f(x, y) such that F = ∇f.
c) Compute
_
C
F · dr
where C is a curve from (0, 1) to (1, e).
Solution: F(x, y) =< 3x
2
+ ln y,
x
y
+e
7y
>.
a) Let P(x, y) = 3x
2
+ ln y and Q(x, y) =
x
y
+e
7y
.
Then
∂P
∂y
=
1
y
and
∂Q
∂x
=
1
y
, so
∂P
∂y
=
∂Q
∂x
The domain of F is the set of points (x, y) for which both 3x
2
+ ln y and
x
y
+e
7y
are well defined.
For ln y to be well defined we need y > 0, in which case
x
y
+e
7y
is well defined too. Therefore, the
domain of F is {(x, y) : y > 0}, which is the upper half plane (pictured below) without the x axis
(the boundary).
This is clearly an open, simply-connected domain. It follows that F is conservative.
b) F = ∇f =<
∂f
∂x
,
∂f
∂y
>.
30
We need to find f(x, y) such that
∂f
∂x
= P(x, y) = 3x
2
+ ln y
∂f
∂y
= Q(x, y) =
x
y
+e
7y
From the first equation we get
f(x, y) = x
3
+xln y +g(y)
Then
∂f
∂y
=
x
y
+g

(y)
But
∂f
∂y
=
x
y
+e
7y
so g

(y) = e
7y
, hence g(y) =
1
7
e
7y
+C, where C is any constant.
We conclude:
f(x, y) = x
3
+xln y +
1
7
e
7y
+C
c) Using the FTC for line integrals, we have:
_
C
F · dr =
_
C
∇f · dr = f(1, e) −f(0, 1) =
= (1 + 1 · ln e +
1
7
e
7e
+C) −(0 + 0 · ln 1 +
1
7
e
7
+C) = 2 +
1
7
e
7e
−
1
7
e
7
Problem 8.5. Compute
_
C
x
2
y dx −xy
2
dy
where C is the circle x
2
+y
2
= 4 with counterclockwise orientation.
Solution: The circle C is a closed (and also simple, piecewise smooth) curve, therefore Green’s
theorem is applicable. Let D be the domain enclosed by the curve C. The positive orientation of
the curve C (relative to the domain D) is counterclockwise.
Let P = x
2
y and Q = −xy
2
. Using Green’s theorem we have:
_
C
x
2
y dx −xy
2
dy =
_
C
Pdx +Qdy =
__
D
(
∂Q
∂x
−
∂P
∂y
) dA =
=
__
D
(−y
2
−x
2
) dA = −
__
D
(y
2
+x
2
) dA
Since D is the disk of radius 2, we use polar coordinates to compute the above double integral.
We have: x = r cos θ, y = r sin θ, where 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 2.
Then y
2
+x
2
= r
2
and the Jacobian is r. Therefore,
__
D
(y
2
+x
2
) dA =
_
2π
0
_
2
0
r
2
· r dr dθ = 2π
_
2
0
r
3
dr = 2π
r
4
4
2
0
= 8π
We conclude: _
C
x
2
y dx −xy
2
dy = −8π
31
Problem 8.6. Compute
_
C
e
x
2
dx + (x
3
+ sin y) dy
where C is the boundary of the region between the circles x
2
+y
2
= 1 and x
2
+y
2
= 4 in the first
quadrant.
Solution: Let P(x, y) = e
x
2
, Q(x, y) = x
3
+ sin y.
Using Green’s theorem we have:
_
C
e
x
2
dx + (x
3
+ sin y) dy =
_
C
Pdx +Qdy =
=
__
D
(
∂Q
∂x
−
∂P
∂y
)dA =
__
D
(3x
2
−0)dA =
__
D
3x
2
dA
To compute the above double integral we use polar coordinates, since D is a polar rectangle.
D = {(r, θ) : 1 ≤ r ≤ 2, 0 ≤ θ ≤
π
2
}
x = r cos θ, y = r sin θ
Then
__
D
3x
2
dA =
_
2
1
_ π
2
0
3r
2
cos
2
θ · r dr dθ =
=
_
2
1
_ π
2
0
3r
3
cos
2
θ dr dθ = (
_
2
1
3r
3
dr) · (
_ π
2
0
cos
2
θ dθ)
_
2
1
3r
3
dr = 3
r
4
4
2
1
=
45
4
_ π
2
0
cos
2
θ dθ =
_ π
2
0
1 + cos 2θ
2
dθ =
π
4
Then _
C
e
x
2
dx + (x
3
+ sin y) dy =
__
D
3x
2
dA =
45
4
·
π
4
=
45π
16
32