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Workshop 1

Math 1920

Spring 2012

In this workshop you will work on problems that use vectors and dot products to represent the fundamental engineering and physics concepts of work and circulation.

Problem 1 You do work by exerting a force on an object as it undergoes a displacement from one place to another. If a constant force F acts in the same direction as the displacement s, the work done is W = |F ||s|. If the force is constant, but not in the direction of the displacement, the work done is W = F ·s. In the figure below the force F is constant on each of the segments of the path connecting the four points, but as you can see it is not in the direction of the displacement.

Figure 1

a) Can you determine from the information given whether the work done by F in moving an object once around the path in a clockwise direction is positive, negative, or zero? Explain.

1

b) A force is called conservative if the work done in moving an object from one point to another is the same for all possible paths. Is F conservative? Explain. For instance, what if, instead of following a clockwise trajectory, we had taken a counterclockwise path along the same segments in part (a)?

Note: Conservative forces have a special place in physics, with the most prominant examples of conservative forces being gravity and Coulomb (electrostatic) forces.

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Problem 2 You might wonder, what makes an airplane fly, or a sail boat sail? The answer is lift. A full discussion of lift is well beyond the scope of Math 1920 but the exercise below will introduce you to a mathematical concept, called circulation that is used by aeronautical engineers to describe and study lift. A simple but powerful tool for calculating lift,the Kutta-Zhukovsky Theorem, says Flift = Sair Γ ρair d or Lift = airspeed × circulation × density of the air × span of the wing But what is circulation? Mathematically, circulation is computed in the same way we compute work around a loop.

θ

Figures 2a, 2b The bold line in Figure 2b is a simplified sideview of the left wing of the plane in Figure 2a. The simplified wing cross-section has length L. Arrows show the direction of air flowing above, under and behind the wing; while the dotted lines show an imaginary path around the wing. Go around the path clockwise, and take the dot product of the wind velocity vector and the vector of each path segment. The sum of these dot products is the circulation. Let’s assume that the velocity of the air above and well behind the wing (i.e. on the vertical and horizontal parts of the loop) is V = 5i m/sec. In real life the air velocity close to the wing is very complex. But for simplicity we will assume that the air speed along a path parallel to the underside of the wing is V = 0.5(cos(θ)i − sin(θ)j) m/sec. 3

a) Find the circulation of the velocity vector V along the loop shown in Figure 2 if the angle of attack θ is 30 degrees. (Leave your answer in terms of L and don’t forget the units!)

b) Assume that the density of the air is 1.229 kg/m3 , that the length L of the wing from leading to receding edge is 4 m, and that the span d of the wing is 20m. Compute the lift. What are the units of lift? Would this lift you? (1 N = 0.225 lb)

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c) Place a sheet of paper on top of two erasers or books spaced well apart. If you blow air horizontally under the paper will the paper lift up or down? Use the Kutta-Zhukovsky Theorem to predict what should happen. Carry out the experiment!!! (If you want to be sure you are directing the air under the paper you can construct a paper straw to blow under the sheet.)

Note: It might appear that the loop we chose was quite arbitrary. We will see later as a consequence of Stokes’ Theorem that if we can assume the the flow of the air is irrotational, then the circulation (and hence the lift) is essentially independent of the size and shape of the loop used to compute the circulation.

5

Math 1920 Problem 1

Workshop 1: Lift Solutions

Spring 2012

a) We know that the dot product of two vectors is the product of their magnitudes and the cosine of the angle between them. Hence if the angle is acute the cosine will be positive and so the dot product would be positive too, because magnitudes are always positive. In reverse if the angle is obtuse then the cosine and thus the dot product would be negative. So we should look at the angles for each segment. We can start at the top left corner of the path, moving clockwise. It may help if you draw arrows on the path to show the vectors of the path segments. For the first and last segments, the angle between the vector of the path and the force is acute; while for the middle two segments the angle is 90 degrees, which gives a dot product of zero. So overall the sum of the dot products (the work done in this case) is positive. b) F is not conservative. For instance, if instead of moving clockwise, we moved counterclockwise around the path - all the acute angles would become obtuse, which now results in the overall sum of dot products (work done) being negative! Problem 2 a) We take the slanted side to have length L and then sum up the dot products of the wind velocity and the vectors making up the sides of the triangle, moving in a clockwise direction. circulation = 5i· L cos θi +5i· −L sin θj +0.5 cos θi − sin θj ·L − cos θi + sin θj . √ θ = 30◦ , so the circulation is L 5 3 − 1 m2 s−1 . 2 b) Air density is 1.229 kg/m3 , L is 4 m, and the span of the wing is 20m. Using the Kutta-Zhukovsky Theorem we get a lift of 1880 N, or 424 lb. c) The paper moves down! Take the paper to be the airplane wing in figure 2, with θ = 0. The air velocity on the top and side will be zero, while the velocity on the bottom is some positive number (since you are blowing out and not sucking in). Replacing the relevant numbers in part a), you will find that the circulation is negative. Now, since all the other terms in the Kutta-Zhukovsky Theorem are positive, that means that lift will be negative, and hence the paper should move downwards.

Page 1 of 1

Workshop 2

Math 1920

Spring 2012

In this workshop you will explore the idea of flux across a surface.

Problem 1

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Figure 1

a) Fluid is flowing down a pipe at 5 m/sec. The cross sectional area of the pipe is 3 square meters. What volume of fluid flows out of the pipe each second ?

b) Suppose the pipe is cut on the diagonal as shown in Figure 1. Is the rate at which the fluid flows out of the pipe greater than, less than or equal to your answer to part a)? Give convincing reasons for your answer.

1

c) The flux of vector field v, constant over a flat surface S and outward across S is A(v · n), where A is the area of S, and n is the unit outward pointing normal to S. Use this definition of flux to revisit question b) above, by comparing the flux of the fluid across the vertical crosssectional area versus the flux across the area of the pipe cut on an angle.

2

Problem 2: A fun question to ponder

Imagine that you are on your way home when suddenly it rains - and you don’t have an umbrella! For simplicity assume that the top of your head and shoulders are horizontal planes, and your front is a vertical plane (see Figure 2).

Figure 2 Figures 3a and 3b show how the rain falls when you are stationary, and when you are moving forward.

Figure 3a 3

Figure 3b If you move forward with velocity w and the rain falls vertically with velocity v, then the relative velocity of the rain is v − w (see Figure 4).
v−w

v

−w

Figure 4 Home is a distance s away when the rain starts falling. If you want to minimize how wet you get, is better to walk or run through the rain?

S

Figure 5 Hint: It will be easier to consider the horizontal and vertical planes separately. Also make sure your directions are consistent - you want the overall amount of rain going towards both planes, or away from both planes (not one towards and one away!).

4

Math 1920 Problem 1

Workshop 2: Flux Solutions

Spring 2012

a) Assuming that the fluid velocity is the same over the entire cross-section, the volumetric flow rate is then 5 ms−1 × 3 m2 = 15 m3 s−1 . b) The volumetric flow rate is the same as in a). Since the amount of fluid coming in from the left side of the pipe remains the same, the amount that comes out from the right side should also be unchanged regardless of the shape of the cut. (Conservation of mass, with constant density) c) We can think of the volumetric flow rate as the volume of a slant-heighted column of water, with the slanted cross-section at an angle θ to the vertical. This volume is then (area of the cut A)× (cos θ) ×(magnitude of v). (Similar to the triple scalar product, volume of a parallelepiped.) A cos θ |v| = A · v, where A is the ‘area vector’ with magnitude being the area of the slanted cut, and acting in the direction normal to the cut. Compare this to the expression given for flux - this shows that the vertical cross-sectional area can be thought of as the product of cos θ and the area A of the slanted cut, OR that the component of v across the slanting cut is |v| cos θ. Now since the area of the vertical cut A cos θ and |v| are both constants, that means the volumetric flow rate A cos θ |v| is also constant, regardless of the value of θ.

Page 1 of 2

Math 1920

Workshop 2: Flux Solutions

Spring 2012

Problem 2

We can solve this problem in a similar way to problem 1, i.e. the volumetric flow rate of the rain = flux = product of the surface area and the dot product of the rain velocity and unit normal to the surface. First, let the area of the top of your head and shoulders be A1 , and the area of your front be A2 . For the unit normals we have to have both pointing outwards, or both inwards. Let’s take them both inwards, so that the flux would be the volumetric flow rate of rain going towards you. Then the unit inward normal to the top of your head and shoulders is n1 = −j, and the unit inward normal to your front is n2 = −i. The rain velocity relative to you walking is v − w = −wi − v j. The volumetric flow rates of the rain through your top and your front are then respectively: Q1 = A1 −wi − v j · −j = A1 v Q2 = A2 −wi − v j · −i = A2 w

But we’re not interested in the volumetric flow rate, we want the actual volume of rain (the larger the volume, the wetter you get). So we need to multiply by the amount of time spent in the rain which is simply the distance s divided by your walking speed w. This gives us a volume of A1 vs towards your head and shoulders, and A2 s towards your front. w Since w is in the denominator for the volume of rain coming at you from on top, a larger w would mean a smaller volume. However the volume of rain coming towards your front is independent of w, which means that your front is going to be just as wet no matter how fast you run. So overall, you should run. :)

Page 2 of 2

Workshop 3

Math 1920

Spring 2012

In your engineering careers you will discover that almost all of electromagnetics and consequently all the electronic gadgets one may study - can be nicely summed up by just four fundamental equations, known as the Maxwell equations. In this workshop we will explore one of these, Ampere’s Law, which turns out to be nicely expressed in the mathematical formalism of line integrals. Ampere’s Law is a statement about the line integral around a closed loop of the magnetic field caused by an electric current in a long straight wire. For our purposes, there are two important pieces: i) When electric current is carried by a long straight thin wire it induces a magnetic field, B. The magnitude of B is given by µ0 I 2πr where I is the current, r is the distance from the wire, and µ0 is a constant. B = ii) Since it’s hard to draw the induced magnetic field vectors at every point in space, the magnetic field is often represented as a set of field lines to illustrate the direction of the magnetic field (as shown in Figure 1a). These magnetic field lines give the direction of B, at a point r units from the wire, to be tangential to the circle of radius r. When the current is directed out of the page, as shown in Figure 1b, the magnetic field vectors have a counterclockwise direction (and vice versa, when the current is directed into the page, the magnetic field vectors have a clockwise direction).

B(r)

Current out of the page

Figure 1a/b 1

With this in mind, and using what you know about line integrals, a) Find the integral of B around a circle of radius r, centered on a wire carrying current out of the page, in the counter clockwise direction.
B

r

Current out of the page Path of integration

b) Find the integral of B around a circle of radius r, centered on a wire carrying current out of the page, in the clockwise direction.

B

r

Current out of the page Path of integration

2

c) Find the integral of B around the path shown below. The path has four segments, two are circular arcs of radius r1 and r2 centered on the wire, and the others are segments of radial lines.

B

r1

r2

Current out of the page Path of integration

d) Optional. A coaxial cable is a special type of cable composed of an inner conducting wire, a separating insulating layer that does not transfer current, and an outer conductor (all wrapped in a protective cladding). These cables are important in many electronics applications where you do not want fields radiating out from your wires (for instance, they carry cable television signals, preventing leeching of the signal without getting into the cable itself). In other words, they have the special property that outside the coaxial cable, the net magnetic field is zero. Using what you have discovered about the relationship between the line integral of B around a single current carrying wire, postulate how this could be so.

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Math 1920

Workshop 3: Line Integral and Ampere’s Law

Spring 2012

a) Here B is parallel to the unit tangent vector of the circle, so B · dr =
C C

B 1 ds = B
C

ds = 2πr B = µ0 I.

b) B is still parallel to the unit tangent vector, but now in the opposite direction. So the dot product is now negative: B · dr = −
C C

B 1 ds = −2πr B = −µ0 I.

c) Let the inner arc be C1 , the radial line from r1 to r2 be C2 , the outer arc be C3 , and the last radial segment be C4 . The integral of B around the whole path is the sum of the integrals of B along each of the 4 segments. Along C2 and C4 , B is perpendicular to their tangential vectors. Therefore, the dot products will be zero along these two segments. Along C1 and C3 , B is parallel to their tangential vectors, like in b) and a) respectively. Instead of a circle we now have arcs spanning an angle θ. So B · dr = −
C1 C

B 1 ds = −θr1 B = −

µ0 Iθ 2π

and B · dr =
C3 C

B 1 ds = θr2 B =

µ0 Iθ . 2π

Thus overall we have B · dr = −
C1 ∪C2 ∪C3 ∪C4

µ0 Iθ µ0 Iθ +0+ + 0 = 0. 2π 2π

d) From the previous problems, we can restate Ampere’s Law as B · dr = µ0 Ienclosed

C

When the closed path contains a current, the path integral is equal to a constant times the current. This is true even if we hadn’t chosen such a nicely symmetric path (it’s just harder to calculate)! When the closed path does not enclose a net current (as in part (c)) the integral along the path is zero. Therefore, in order for a coaxial cable to have a net magnetic field of zero outside of the wire, its inner and outer conductors must carry uniform currents with equal magnitude and opposite direction.

Page 1 of 1

Workshop 4 Problem 1

Math 1920

Spring 2012

The Matterhorn is a mountain in the Alps with a peak height of 4,478m. Due to it’s often near vertical cliffs, it is one of the most common mountain peaks for prospective climbers in the world (and consequently sees several fatalities every year). Beyond it’s renown as a climbers mountain, the Matterhorn’s massive slopes give us a perfect opportunity to explore the concepts of the gradient and directional derivative.

To begin, there are multiple ways we could map the Matterhorn’s surface. For our purposes, we’ll model the Matterhorn as the graph of a function of two variables, F (x, y) which we’ll take to be (in units of km): F (x, y) = 5 · e−
x2 +y 2 25

+ 2.5 · e−

(x−10)2 +(y−5)2 25

1

The level curves or elevation map of the model function is illustrated below. A basecamp for prospective climbers is located at (10, −5).

a) Find the direction of steepest ascent at the basecamp.

2

b) Find the magnitude of the rate of change in elevation at the basecamp, when you move due north (in the +j direction).

c) Find the magnitude of the rate of change in elevation at the basecamp in the direction of the summit, located in our model at (0, 0). Feel free to use a calculator.

d) After a great deal of effort, you have arrived at the summit. However, as in any climbing movie ever made, a very bad storm is about to set in. Assuming you have to initially leave the summit heading due east (in the +i direction), sketch the quickest path down the Matterhorn on the elevation level map. Briefly justify your answer.

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Problem 2 In 1979 topical storm Claudette produced torrential rains when it made landfall in Texas. The highest one-day total was reported in Alvin, Texas where 42 inches of rain fell in a single day. This remains the 24 hour record for any location in the United States. A rectangular region R of a National Weather Service isohyet map has been subdivided into grid areas, each 5 miles by 5 miles. The isohyets show levels of rainfall in inches within the 3 day period July 24-27, 1979.

a) Use the map to find upper and lower estimates for the total amount of rain that fell over the region during the three days. (1 mi = 63360 in)

4

b) Find upper and lower estimates for the average rainfall over the grid region during the three days.

c) What would happen to your upper and lower estimates if you used smaller grid sizes? Would they increase or decrease? Explain.

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Math 1920 Problem 1

Workshop 4: Gradient

Spring 2012

a) The direction of greatest increase in elevation at the basecamp is given by evaluated at (10, −5). Which means we need the partial derivatives: 2x − x2 +y2 ∂F (x, y) −2(x − 10) − (x−10)2 +(y−5)2 25 =5 − e 25 + 2.5 e ∂x 25 25 ∂F (x, y) ∂x =5 −
x=10,y=−5

F (x, y)

4 20 − 125 e 25 + 0 = − 5 25 e

∂F (x, y) −2(y − 5) − (x−10)2 +(y−5)2 2y − x2 +y2 25 =5 − e 25 + 2.5 e ∂y 25 25 ∂F (x, y) ∂y =5 −
x=10,y=−5

20 − (x−10)2 +(y−5)2 10 − 125 25 e 25 + 2.5 e 25 25 = 2 2 + 4 5 e e

∂F (x, y) ∂y Putting it all together: F (x, y)

x=10,y=−5

=
x=10,y=−5



4 2 , e5 e5

+

2 e4

b) Since the direction of increase is +j, the magnitude of the rate of change of elevation is simply
∂F (x,y) ∂y x=10,y=−5

=

2 e5

+

2 . e4

c) Now we want to find the magnitude of the rate of change in elevation at (10,-5) in the direction towards the point (0,0). This direction is given by the normalized unit vector u= by
1 √ 5

− 2, 1 . The magnitude of the rate of change in this direction is then given

F · u. F ·u= − 4 2 2 , + 4 5 e5 e e · 2 1 −√ ,√ 5 5 =√ 8 2 + √ (1 + e) ≈ 0.0465 5 5e 5e5

Page 1 of 3

Math 1920

Workshop 4: Gradient

Spring 2012

d) Sketched below.

Problem 2 a) The total amount of rainfall is (level of rainfall) (area of grid region) , so to obtain upper and lower estimates we need to find a maximum and minimum for each grid region. As such your answers for the upper and lower estimates may be different from another person’s, because their ‘maximum’ and ‘minimum’ for each grid region may be different. One possible set of maxima and minima for the region is shown in the grid below: 25 15 25 20 25 20 35 20 40 30 35 25 35 20 35 25 35 25 30 20 30 20 30 20 30 20 25 20 25 15

Since the grid regions all have the same area, our computation simplifies to level of rainfall (area of grid region) , where the grid region area is 25 sq.mi. The upper estimate is then 460 in × 25 sq.mi = 4.62 × 1013 in3 = 7.57 × 108 m3 ; and the lower estimate is 315 in × 25 sq.mi = 3.16 × 1013 in3 = 5.18 × 108 m3 . Page 2 of 3

Math 1920

Workshop 4: Gradient

Spring 2012

b) The upper and lower estimates for the average rainfall over the grid region is found by diving by the total area, which is 15×25 sq.mi. This gives us an upper estimate of 30.6 in, and a lower estimate of 21 in.

c) If you used smaller grid sizes, then you would expect your upper and lower estimates to get closer to the actual amount of rainfall. Hence your upper estimate would decrease, while your lower estimate would increase. (If you’re not convinced then find an upper and lower estimate but this time use the entire 3 by 15 grid as one region. What do you find?)

Page 3 of 3

Workshop 5

Math 1920

Spring 2012

In 1928, C. W. Cobb and P.M. Douglas developed a model for the gross output, Q, of a nation or company given by the function Q(K, L) = AK α L1−α where K represents the amount of capital, L the amount of labor, and A and a are positive constants with 0 < a < 1. When one attempts to maximize the output given by the Cobb-Douglas Production Function, it is often done with a budgetary constraint. Each unit of labor and capital will cost a certain amount, and there is a maximum amount of money, M, budgeted to invest. If the price per unit capital is p and the price per unit labor is w, the budget constraint B(K, L) is B(K, L) = pK + wL = M The figure shown below plots several level curves of Q(K,L).

L (units of labor) L(1000 units)

Q(K,L

)= $5

00 m

il

Q(K,L)

= $400

mil

Q(K,L)= $3

00 mil

0 mil Q(K,L)= $100 mil

Q(K,L)= $20

K (units of capital)

1

These level curves are curves of constant output (economists fancily call them isoquants). The output increases for level curves farther from the origin. Here we have taken the constants A and a such that, Q(K, L) = 160, 000K 0.4 L0.6 a) Now imagine you are a financial executive in a board meeting, and an assistant hands you the previous plot detailing the company’s theorized output curves. You know that the budget for your company for the next fiscal year is B(K, L) = 800K + (40, 000)L = $40 × 106 Without making any explicit calculations, and using only a straight edge and the figure, explain how you could find the critical point, that is the optimum allocation of capital and labor to maximize your companies output given the budgetary constraint? (Hint: consider the meaning of f = λ g).

b) Use your proposal to estimate the optimum allocation of capital and labor given by the figure.

2

c) Having estimated the critical point, you now want to solve for it explicitly. First consider the general case: Q(K, L) = AK α L1−α B(K, L) = pK + wL = M Using what you know about Lagrange Multipliers, solve for the critical point in terms of A, a, p, w, M .

d) Using the solution from part (c), check your estimate. How well did you do?

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Math 1920

Workshop 5: Lagrange Multipliers

Spring 2012

a) The Lagrange vector equation states, f = λ g. In other words, the point that maximizes the function f with constraint g is the point where the normals to the constraint and level curves of f are parallel. Equivalently, this is the point where the tangent of a level curve of f is parallel with the tangent of the constraint. Here, we are considering Q(K, L) = λ B(K, L), where B(K, L) is a straight line. The critical point that optimizes this can therefore be found by sketching the budget on the graph, and looking for the point where the tangent of the constraint line is parallel with the tangent of a level curve of Q. b) The budget constraint B(K, L) = 800K + (40, 000)L = 40 × 106 is a straight line, sketched in bold in the figure below. The critical point can then be estimated as (K, L) = (20 × 103 , 600)

L (units of labor)

Q(K,L

)= $5

00 m il

Q(K,L)

= $400

mil

Q(K,L)= $3

0 mil Q(K,L)= $100 mil

Q(K,L)= $20

00 mil

K (units of capital)
c) To solve for the critical point explicitly, we take Q = λ B. In the general case, we have Q(K, L) = AK α L1−α and B(K, L) = pK + wL = M . This gives us the system of equations: AaK a−1 L1−a = λp A(1 − a)K a L−a = λw pK + wL = M Page 1 of 2

Math 1920

Workshop 5: Lagrange Multipliers

Spring 2012

Solving for p and w in the first two equations, p= w= Aa a−1 1−a K L λ

A(1 − a) a −a K L λ Plugging in for p and w into the third equation, Aa a 1−a A(1 − a) a 1−a K L + K L =M λ λ A a 1−a K L =M λ Rearranging for λ, A a 1−a K L M Finally, solving for K and L we find the only critical point is λ= (K, L) =
M a M (1−a) , w p

d) In the earlier example, M = 40 × 106 , a = 0.4, p = 800, and w = 40, 000. Plugging in the relevant numbers we arive at the critical point (K, L) = (20 × 103 , 600) Turns out that in this case our graphical estimate was spot on.

Page 2 of 2

Workshop 6

Math 1920

Spring 2012

The natural coordinate system for a problem is not always rectangular coordinates. As one series of examples, in this workshop we look at common foods and consider the coordinate systems more natural to them. a) Suppose you had to integrate a function δ over half a section of an orange (the orange had 8 sections originally), shown in the figure. What coordinate system would you use, and why? Where would you put the axes? Why? Set up an integral to integrate δ over the section.

b) Set up an integral to integrate a function f over the region consisting of the rind of a grapefruit (the rind has a thickness, it is not just a surface). Let the grapefruit have a radius R and rind thickness t.

1

c) Suppose that your friend likes only the pizza crust, while you like only the interior. If the thickness of toppings varies inversely with the distance from the center of the pizza, set up an integral for the volume of your part of a 16 inch diameter pizza with a 2 inch wide crust.

Note: The toppings’ distribution looks really unrealistic, coming from a ‘reasonable’ mathematical model. But let’s suspend the disbelief for now.

2

d) Set up an integral for the mass of a slice that is 1/12th of an angel food cake with a flat bottom, and variable height h(r) where r is the radial distance from the center of the cake.

e) If you had to integrate over a pineapple which coordinate system would you use and why? Where would you put the coordinate axes? How about a pear? What would the difference in integrals be?

3

f) Suppose you buy a cantaloupe that has an 8 inch diameter, and that after scooping out the seeds the fruit and rind is 2 inches thick (yum). Slice the melon into 12 equal slices. i) Use spherical coordinates and inequalities to describe the first slice. (Let the origin be the center of the melon.)

ii) Now cut a melon slice into 8 melon chunks by subdividing the interval of φ evenly (ie for each chunk the change in φ is the same). Do the 8 chunks have the same volume? Why not? What spherical dimensions are the same, and which are different for each of the chunks?

4

Workshop 6 Solutions

Math 1920

Spring 2012

a) Spherical coordinates, because the orange section is approximately a sector of a sphere.

An integral for δ over this orange section would be
π/4 0 0 π/2 0 R

δ ρ2 sin φ dρdφdθ.

b) Easiest to use spherical coordinates here. Since we only want the rind, the limits of ρ go from R − t to R:
2π 0 0 π R

f ρ2 sin φ dρdφdθ.
R−t

c) Cylindrical coordinates would be easiest here - looking from the top the pizza would have a constant radius and goes full circle from 0 to 2π; the toppings’ height z would have the equation z = h(r), where h(r) is some function that varies inversely with r. Then the volume of your share (pizza minus crust) would be:
2π 0 0 D/2 −t 0 h(r)

1 dz rdrdθ, where D/2 −t is half the pizza diameter minus the crust width = 6 inches. 1

d) Cylindrical coordinates would again be best here. However since we only want 1/12th of the whole cake, then 0 ≤ θ ≤ π/6. This gives us a mass of
π/6 R h(r)

δ dz rdrdθ.
0 0 0

e) For both the pineapple and pear cylindrical coordinates would be best because they both have a long axis and mostly circular cross-sections along that axis.

The pineapple has circular cross-sections of mostly constant radius, while the pear’s circular cross-sections have radii that change along the long axis. This means that for the pineapple, the limits of integrations would all be constants while for the pear, the upper limit of integration for the r variable would be a function of z. f) Assuiming that we’ve cut down the vertical axis, and since we’ve scoopd out the center of the melon, the ranges for the different spherical coordinates for the first slice are given by the inequalities: 6in ≤ r ≤ 8in 0◦ ≤ φ ≤ 180◦ 2

0◦ ≤ θ ≤ 30◦ Now the question is what happens when we divide this first slice into 8 equally divided φ chunks. The first chunk will be given by 0 ≤ φ ≤ π , 8 the second by π ≤ φ ≤ π and so on. Recall that the differential volume 8 4 in spherical coordinates is: dV = ρ2 sin φ dρdφdθ Consequently, the total volume for each chunk won’t be the same. The same conclusion can be reached graphically.

3

Workshop 7 Problem 1

Math 1920

Spring 2012

An important result about electric fields in a conductor is that in static equilibrium, when all charges in the conductor are at rest, the surface is an equipotential surface. Under this condition, no work is done by the field in moving a particle from point a to point b along the equipotential surface. Assume that 1. E is a conservative field, and 2. within the conductor the electric field E is 0 (this is necessary for the charges to be at rest). a) Use assumptions 1. and 2. to show that the tangential component of E at any point a on the surface of the conductor MUST be 0.
E E

E a conductor

vacuum

Hint: Take b to be a point a short distance on the conductor surface from point a. Consider two paths between a and b, one path tangential to the conductor surface, and the other path going through the interior of the conductor.

1

b) Use your result from part a) to conclude that the work in moving a test charge from one point to another along the surface is 0, and that hence the surface of the conductor is an equipotential surface of E.

Problem 2 Suppose you know that the x, y, z components of a field F are the partial derivatives of a scalar field f . Must F be conservative (i.e. independent of path)? Why or why not?

2

Workshop 7 Solutions

Math 1920

Spring 2012

1. (a) Suppose that at some point a, on the surface, the tangential component of E is NOT 0. Let C1 be a path starting from point a, going a very short distance in the direction of the tangential component to point b. The work done in moving along C1 is positive. Let C2 be another path from point b to point a, but going through the conductor. Because E = 0 within the conductor, the work done moving along C2 is 0. The total work done moving a charge along the loop C1 ∪ C2 is then positive. But E was assumed to be conservative, which should have given us zero work done → contradiction! Therefore the assumption that the tangential component of E is not 0, is wrong. (b) Let C be any curve on the surface. From part a) we know that the tangential component of E at each point on C is 0, and hence the work done C E · T ds = 0. 2. Let C be any smooth curve from point A to point B, parametrized by r(t). By the chain rule the work done by F in moving from A to B along C is:
tA

F · dr =
C tB tA

F (t) ·

dr dt dt dr dt dt

=
tB tA

fx (t) , fy (t) , fz (t) · fx (t)
tB tA

= =

dy dz dx + fy (t) + fz (t) dt dt dt dt

df dt tB dt = f (tB ) − f (tA ) Hence F is path-independent. This is essentially the Fundamental Theorem of line integrals. 1

Workshop 8

Math 1920

Spring 2012

Sometimes a physical law will be expressed in a differential form (one quantity is a derivative of another). Other times, it is more natural and convenient to express the law in an integral form (one quantity is expressed as the integral of another). In this workshop we explore how the Divergence theorem can be used to show that each form implies the other. Problem 1 Gauss’s Law expresses the relationship between electric charge and electric field. Let E be an electrostatic field, which arises from a distribution of charges in space. The differential form of Gauss’s Law states that the divergence of the electric field at each point, · E, is proportional to the charge density there, ρ, divided by 0 , a physical constant called the electric constant: ρ ·E =
0

The integral form of Gauss’s Law states that the total electric flux through any closed surface is proportional to the total (net) electric charge, Q, enclosed inside the surface:
§¤ ¦ ¥E
S

· n dσ =

Q
0

where Q is the total charge enclosed. a) Use the Divergence theorem to show that the differential form of Gauss’s Law implies the integral form of the law.

1

b) Showing that the integral form of the law implies the differential form requires a more subtle argument. We will get you started. 1) Suppose that ¦ ¥E · ndσ =
S

§¤

Q
0

for every surface, but that

·E =

ρ
0

.

In fact suppose that · E − ρ > 0 at some point P0 . Then by continuity 0 there must be a small sphere, S0 centered at P0 where · E − ρ > 0 as 0 well. Use the Divergence theorem to obtain the contradiction
§¤ ¦¥
S0

E · n dσ >

Q0
0

where Q0 is the total charge inside S0 . 2) Make a similar argument for the case argument. ·E−
ρ
0

< 0, and finish the

2

Optional Challenge Problem Now that we’ve used the Divergence theorem to prove the equality of both forms of Gauss’s law, one might ask: why bother with multiple forms of the same law? Well, one of the main reasons is that it can save us a lot of work if we pick the right one! We’ll explore this through a classic problem. To begin, we need one other bit of information: a positive electric point charge emits a radially symmetric and outward electric field. Using that knowledge, and the two formulations of Gauss’s law that we’ve explored above, consider the following. A charge of +Q sits at the corner of a square box of length L, as shown. Calculate the flux of the electric field E, due to the point charge +Q, through the shaded sides. Hint: This problem is very simple if you use the above information, select a useful version of Gauss’s law, and think a little outside the box!

L +Q L L

3

Workshop 8 Solutions

Math 1920

Spring 2012

1a) Integrate the differential form over a volume D which is enclosed by surface S: · E dV
D

=
D

ρ
0

dV ρ dV

E · n dσ =
S

1
0 D

E · n dσ =
S

Q
0

1b) Start with the flux of E through the surface of the small sphere S0 . E · n dσ =
S0 D

· E dV ρ
D 0

> = ⇒
S0

dV

Q0
0

E · ndσ >

Q0
0

,
S

which contradicts the starting supposition that every surface.

E · ndσ =

Q
0

for

Similarly if we made the assumption that · E − ρ < 0 at some point 0 P0 , leading to the small sphere, S0 centered at P0 where · E − ρ < 0, 0 this would also give us a contradiction. Therefore the only way not to contradict our initial assumption is if · E − ρ = 0 everywhere. 0 Challenge) First, we need to pick the relevant expression of Gauss’s Law. Since the problem is interested in flux through a closed surface, we choose the integral formulation. (In general, the integral formulation tends to be more useful for macroscopic systems and symmetric problems, where as 1

the differential form tends to be useful for microscopic interpretations, among other things.) Immediately we realize that the surface asked for will not work, at least not as is. Gauss’s Law (and the divergence theorem) require a closed surface. Instead of the surface as given, we choose to consider the surface of a cube of length 2L, who has the charge +Q located at it’s center (see figure below). Now, since the cube of length 2L is a closed surface, Gauss’s law and the divergence theorem apply. The integral that we are interested in, due to symmetry, is just 1/8 of the flux through the entire closed surface. In short,
§¤ ¦¥
2LSquare

E · n dσ =

Q
0 Q 8 0

E · n dσ =
Sides

1 §¤ E · n dσ = ¦¥ 8 2LSquare

+Q 2L 2L 2L

2

Workshop 9

Math 1920

Spring 2012

We have now explored three different theorems that relate an integral of a function taken over a region to an integral taken over the boundary of the region. This workshop explores Stokes’ Theorem. Problem 1 Stokes’ Theorem and Green’s Theorem have an important relationship: Green’s Theorem is actually a special case of Stokes’ Theorem. a) State each theorem in your own words, and provide a verbal argument justifying the above claim.

b) Now make your case mathematically. Show that Green’s Theorem is a special case of Stokes’ Theorem.

1

Optional Challenge Problem As we have previously seen with Gauss’s Law and the Divergence Theorem, the fundamental laws of electromagnetism (the Maxwell Equations) can be stated cleanly in both an integral form and a differential form. Moreover, these two formulations can be shown to imply one another. Now we consider another of these fundamental laws: Faraday’s Law. The differential version of Faraday’s law states that the curl of the electric field about a point in space is equal to the negative rate of change of the magnetic field. The negative sign is important; it is necessary to get the correct direction of the induced electric field. ∂B ∂t The integral version of Faraday’s Law states that the change in magnetic flux across a surface equals the electromagnetic force around the boundary of the surface (this version makes it more clear that this is, in fact, the principle behind the electric generator!). If we write the magnetic flux over the surface as Φ(t) = S B · n dσ, and note that the electromagnetic force around the boundary is C E · dr, then ×E =− E · dr = −
C

dΦ dt

Use Stokes’ Theorem to show that the differential form of Faraday’s Law implies the integral form of the law. Assume that the only thing that changes with time is the B-field (that is we don’t allow the surface shape to change with time).

2

Workshop 9 Solutions

Math 1920

Spring 2012

1)

a) Stokes’ Theorem states that the total net curl (aka ”infinitesimal rotation”) of a vector field F over a surface is equal to the circulation of F along the boundary of the surface. Green’s Theorem states that the circulation of a vector field F around a closed curve in the plane is equal to the total net circulation density of F (expressed as ∂N − ∂M ) over the region enclosed by the curve. ∂x ∂y Green’s Theorem is Stokes’ Theorem under the condition that the surface of interest andF are restricted to the plane (that is, they are independent of z and the k-component of F is zero). b) Now we make our argument from part (a) explicit. Stokes’ Theorem states: F · dr =
C S

× F · n dσ

Following our argument in (a), we assume F = M (x, y)i+N (x, y)j+ 0k. We then calculate the curl of F :  i j k 0  = ∂N ∂M − k ∂x ∂y
∂ ∂ × F =  ∂x ∂y M (x, y) N (x, y) ∂  ∂z

Since the surface is restricted to a region in the plane, n = k and dσ = dx dy. Putting everything together: F · dr =
C S

× F · n dσ ∂M ∂N − k · k dx dy ∂x ∂y ∂N ∂M − dx dy ∂x ∂y


R

F · dr =
C R

We’ve arrived at Green’s Theorem (the tangential form) as expected. 1

Optional Challenge Problem We start with ×E =− . Taking the component parallel to the surface normal integrated over an arbitrary surface of interest, × E · n dσ = −
S S

∂B ∂t

∂B · n dσ ∂t

Since we are told the surface doesn’t change with time, we can rewrite the right hand side of the equation: d ∂B · n dσ = ∂t dt B · n dσ =
S

S

dΦ dt

By Stokes’ Theorem, we have: × E · n dσ =
S C

E · dr

Putting it all together, E · dr = −
C

dΦ dt

2

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