For more information, please see full course syllabus of Multivariable Calculus

For more information, please see full course syllabus of Multivariable Calculus

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## Practice Questions

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### Surface Area

- The area of the parallelogram formed by two vecotrs →a and →b is the norm of the cross product of the vectors, || →a ×→b ||.
- Hence →a ×→b = det(

) = 0→i + 0→j + 1→k and so →a ×→b = (0,0,1).→i →j →k 1 0 0 0 1 0

- The area of the parallelogram formed by two vecotrs →a and →b is the norm of the cross product of the vectors, || →a ×→b ||.
- Hence →a ×→b = det(

) = 0→i + 12→j + 0→k and so →a ×→b = (0,12,0).→i →j →k − 4 0 2 2 0 − 4

- The area of the parallelogram formed by two vecotrs →a and →b is the norm of the cross product of the vectors, || →a ×→b ||.
- Hence →a ×→b = det(

) = − 30→i − 2→j − 38→k and so →a ×→b = ( − 30, − 2, − 38).→i →j →k − 3 7 2 5 1 − 4

- Hence →a ×→b = det(

) = 0→i + 0→j + 1→k and so →a ×→b = ( [1/(2√3 )] + [1/(√6 )], − [1/(2√3 )] − [1/(√6 )],0 ).→i →j →k 1 \mathord / phantom 1 22 1 \mathord / phantom 1 22 1 \mathord / phantom 1 √2 √2 − 1 \mathord / phantom 1 √3 √3 − 1 \mathord / phantom 1 √3 √3 1 \mathord / phantom 1 √3 √3

- We can find the area of the triangle by taking half the area of the parallelogram formed by the vertices.
- Let →a = ( − 1,2, − 2) − (2,1,1) = ( − 3,1, − 3) and →b = (1, − 1,2) − (2,1,1) = ( − 1, − 2,1) then →a and →b are vectors extending from the same vertex (2,1,1)
- To find the area of the parallelogram formed by →a and →b we compute || →a ×→b ||
- Now →a ×→b = ( − 3,1, − 3) ×( − 1, − 2,1) = det(

) = − 5→i + 6→j + 7→k and so →a ×→b = ( − 5,6,7).→i →j →k − 3 1 − 3 − 1 − 2 1

- The surface area of a sphere is found by computing S
_{A}= 4πR^{2}where R is the radius of the sphere.

_{A}= 4π(5)

^{2}= 100π and so the surface area of a sphere of radius 5 is 100π.

^{2},4tu,2t). i) Find ∫

_{a}

^{b}|| [dC/dt] || dt from 0 ≤ t ≤ 1. Do not integrate.

- We have [dC/dt] = (6t,4u,2) and so || (6t,4u,2) || = √{36t
^{2}+ 16u^{2}+ 4} = 2√{9t^{2}+ 4u^{2}+ 1}

_{a}

^{b}|| [dC/dt] || dt = ∫

_{0}

^{1}2√{9t

^{2}+ 4u

^{2}+ 1} dt.

^{2},4tu,2t). ii) Find ∫

_{a}

^{b}|| [dC/du] || du from 0 ≤ u ≤ 1. Do not integrate.

- We have [dC/du] = (0,4t,0) and so || (0,4t,0) || = √{0
^{2}+ 16t^{2}+ 0^{2}} = 4t

_{a}

^{b}|| [dC/du] || du = ∫

_{0}

^{1}4t du.

^{2},4tu).

iii) Find dtdu from 0 ≤ u ≤ 1, 0 ≤ t ≤ 1. Do not integrate.

- Since [dC/dt] = (6t,4u,2) and [dC/du] = (0,4t,0) then [dC/dt] ×[dC/du] = det(

) = − 8t→i − 0→j + 24t→i →j →k 6t 4u 2 0 4t 0 ^{2}→k and so [dC/dt] ×[dC/du] = ( − 8t,0,24t^{2}) - Then || ( − 8t,0,24t
^{2}) || = √{64t^{2}+ 0^{2}+ 576t^{4}} = 8t√{1 + 9t^{2}} .

_{0}

^{1}∫

_{0}

^{1}8t√{1 + 9t

^{2}} dtdu.

^{2}+ tu + u

^{2}) for − 2 ≤ t ≤ 2 and − 2 ≤ u ≤ 2. Do not integrate.

- The surface area of a solid described parametrically through P(t,u) is given by S
_{A}= dtdu where S is the region where the surface lies. - Now [dP/dt] = (1,0,2t + u) and [dP/du] = (0,1,t + 2u) then [dP/dt] ×[dP/du] = det(

) = − (2t + u)→i − (t + 2u)→j + →k→i →j →k 1 0 2t + u 0 1 t + 2u - So [dP/dt] ×[dP/du] = ( − 2t − u, − t − 2u,1) and || ( − 2t − u, − t − 2u,1) || = √{5t
^{2}+ 8tu + 5u^{2}} .

_{A}= dtdu = ∫

_{ − 2}

^{2}∫

_{ − 2}

^{2}√{5t

^{2}+ 8tu + 5u

^{2}} dtdu.

- The surface area of a solid described parametrically through P(t,u) is given by S
_{A}= dtdu where S is the region where the surface lies. - Now [dP/dt] = (1,1,u) and [dP/du] = (1, − 1,t) then [dP/dt] ×[dP/du] = det(

) = (t + u)→i − (t − u)→j − 2→k→i →j →k 1 1 u 1 − 1 t - So [dP/dt] ×[dP/du] = (t + u, − t + u, − 2) and || (t + u, − t + u, − 2) || = √{2t
^{2}+ 2u^{2}+ 4} .

_{A}= dtdu = ∫

_{0}

^{1}∫

_{0}

^{1}√{2t

^{2}+ 2u

^{2}+ 4} dtdu.

- The surface area of a solid described parametrically through P(f,θ) is given by S
_{A}= dfdθ where S is the region where the surface lies. - Now [dP/df] = ( − cosf,0,0) and [dP/(dθ)] = (0, − sinθ,0) then [dP/df] ×[dP/(dθ)] = det(

) = 0→i − 0→j + cosfsinθ→k→i →j →k − cosf 0 0 0 − sinθ 0 - So [dP/df] ×[dP/(dθ)] = (0,0,cosfsinθ) and || (0,0,cosfsinθ) || = cosfsinθ
- Hence S
_{A}= dfdθ = ∫_{0}^{p}∫_{0}^{p \mathord/ phantom p 2 2}cosfsinθ dfdθ

_{0}

^{p}∫

_{0}

^{p \mathord/ phantom p 2 2}cosfsinθ dfdθ = ∫

_{0}

^{p}sin θdθ = 2

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Surface Area

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Surface Area 0:27
- Introduction to Surface Area
- Given a Surface in 3-space and a Parameterization P
- Defining Surface Area
- Curve Length
- Example 1: Find the Are of a Sphere of Radius R
- Example 2: Find the Area of the Paraboloid z= x² + y² for 0 ≤ z ≤ 5
- Example 2: Writing the Answer in Polar Coordinates

### Multivariable Calculus

### Transcription: Surface Area

*Hello and welcome back to educator.com and multi-variable calculus.*0000

*So, today's lesson we are going to be talking about surface area. How to find the area of the surface given a particular parameterization.*0004

*From there we are going to jump off into subsequent lessons. We are going to be talking about integrals of functions and vector fields over given surfaces, but we want to start with the notion of surface areas.*0013

*So, let us just go ahead and jump in. Okay. So, in the last few lessons, we have introduced some concepts, so let us go ahead and recall real quickly.*0026

*We introduced this concept of a cross product, so given a vector a and a vector b, we can form a cross b and we used it to construct the normal vector to a surface.*0035

*Well, we also said that the magnitude of a cross b, well, was just the norm of a cross b.*0052

*I do not need to write that. The magnitude is equal to the norm of a cross b, so it's length, basically.*0072

*We also introduced another formula... it's the norm(a) × norm(b) × sin(θ), the sine of the angle between them.*0081

*I am not sure if we will have a lot of opportunity to use this particular formula. It may come up, it may not, depending on a particular example that we are doing, but mostly we are concerned with constructing the normal vector and then finding its norm.*0094

*That is what is important. Then, once we have the vector, finding a norm is really, really easy.*0108

*Okay. Now, we also said that a cross b, that this magnitude is the area... and we said it real quickly, but it will be really, really important here -- but we also said that this magnitude is the area of the parallelogram that is spanned by the vectors a and b.*0112

*So, if we are given a vector, given a couple of vectors like this, let us just say this is a and this is b, well, this... when we take those vectors and we close it out and form the parallelogram, well when we take a cross b, we end up with this vector which is perpendicular to both, right?*0160

*That is the a cross b, well this parallelogram that is actually spanned by a and b, the area of that parallelogram happens to be the norm of this vector. That is the thing.*0181

*So, there is an association between the vector itself and the area. So, there is some number we can attach to it and that number is the norm of the cross product.*0195

*Okay. So, now we can go ahead and jump into our notion of surface area.*0207

*So, now, given a surface in 3-space and the notions we are about to develop are complete analogs of the lower dimensional cases.*0213

*We work with curves, we define line integrals, we define the length of a curve, now we are dealing with a surface, we are going to define a surface area which is analogous to the length of a curve.*0228

*It is going to be the exact same thing. You are going to see that the integral is going to look the same, and that is what is nice.*0239

*So, if you ever lose your way, just sort of... if you know the lower dimensional case and feel comfortable with that, you can sort of extract the higher dimensional cases.*0245

*Okay. Given a surface in 3-space and a parameterization, which is the important part... and a parameterization... let us call that p, we found dp dt cross dp du.*0254

*So, now let me draw a surface here. This one... it is very, very important that we draw this one properly, so that is the surface, and we have a point -- I am going to draw one vector like that, and I am going to draw another vector like that.*0287

*Then, I am going to go that way, and I am going to -- oops, close this one out, let me just color this in here... so we have -- okay, so this point is our p(t,u) for some value of t and u.*0312

*This vector right here, let us call it dp du, and let us call this vector right here... that is the dp dt, okay?*0349

*Well dp dt and dp du, they span this particular parallelogram, okay?*0365

*So, dp dt and dp du span a parallelogram -- all of a sudden I do not know how to spell parallel -- parallelogram, whose area, we said, was the norm of that cross product. The area is... the norm of dp dt cross dp du, right? So this parallelogram has the area of a cross product.*0373

*Now, notice that this parallelogram right here, the reason I have drawn this little thing on the surface, what you have is this parallelogram which is the plane, so this parallelogram is part of the plane that is tangent to the surface at this point.*0413

*This colored is the shadow of this parallelogram if you happen to be shining a light from on top. So, we wanted to give the sense that there is this shadow, which is also in the shape of a rectangle on the surface.*0428

*Well, the idea is if you take these parallelograms small enough, what you end up actually getting is... you are adding up all of the surface area elements on the surface itself. That is essentially what we are doing.*0444

*I just wanted you to see this. This is just the projection, the shadow of the parallelogram on the surface.*0458

*Okay. So now we are ready to define our notion of surface area.*0466

*So, define surface area is equal to, we are going to use this symbol dΣ, that is the symbolic representation of that, and the double integral itself is the double integral over the entire surface of dp dt cross dp du with respect to t and with respect to u. This right here.*0475

*So, what we are doing is when we form dp dt and dp du, when we take their cross product we get the vector. When we take the magnitude of that vector, we get the area of the parallelogram.*0515

*Well, what we do is we take the area of the parallelogram and we basically add up all of the areas of the parallelogram, the differential elements and we end up getting the surface area.*0529

*This is what is important here. I am going to talk about this a little bit more, but we just want to be able to take a look at that.*0539

*So, the surface area given a parameterization, it is going to be the double integral of the normal vector, or the norm of dp dt cross dp du, integrated with respect to t and with respect to u. That is it. That is the definition of surface area.*0547

*Okay. Now, let us discuss this a little bit.*0567

*So, symbolically, ds, when we use the symbol ds, that is equal to dp dt cross dp du dt du.*0573

*This is a differential area element. Okay.*0590

*When we integrate, first with respect to t, then with respect to u, we have integrated over the entire surface, right? That is what we are doing.*0608

*When we do an integral, we can only go one variable at a time. We go along t first, then we go along u, and we have covered the entire surface. We have integrated over the entire surface.*0635

*Okay. Now, recall -- oops, let me do this one in blue -- recall the following. We said that curve length, we define curve length as the integral from a to b of the norm of the tangent vector, with respect to t.*0652

*We take the tangent vector and then we integrate it along the curve. Well, this is a complete analog. Except now, we are not dealing with a curve, we are dealing with a surface.*0681

*So, we are going to integrate along one curve, and then we are going to take that curve and we are going to integrate along the other parameter, and we are going to cover the entire surface. That is what we are doing.*0694

*This is the analog for a two-dimensional object, which a surface is, a two-dimensional object.*0705

*Okay. So, this area integral, I hope I am not belaboring too many points, I just want to make sure that there is a good intuitive sense of what is going on.*0720

*I mean, it is true ultimately that we are going to give you the definitions and we are going to lay these integrals out and then just sort of see if we can follow... take a look at what you need in the integrand, then fill in the blanks and integrate, but it is important to know where these integrands come from. Why it is that they take the form that they do.*0732

*So, we want to help develop a sense of intuition. We are still dealing with 2 and 3-space, so we still have the geometric intuition, we might as well use it.*0751

*Okay, so the area integral is a double integral. What we are doing, well not essentially, what we are doing explicitly, what we are doing is finding the length of p(t,u) holding one variable constant, then integrating this curve along the other variable -- I am going to draw this out in just a minute -- until we have covered the surface.*0761

*What we are doing, essentially, is finding the length of a curve and then we are going to integrate that curve, that whole curve. We are going to sweep it out along the other variable until we have covered the surface. What we have is something like this.*0845

*So if this is our surface, what we do when we take the first integral is we find essentially that curve length, and then we integrate in this direction, the other parameter.*0858

*What we do is we swing this, we sweep this curve out until we have covered the entire surface. So we move in that direction.*0870

*That is all that we are doing. It is like two curve integrals, one after the other. That is it.*0878

*Okay. So, let us go ahead and do an example, and again, when you are doing your examples, when you are doing your problems for your homework or whatever it happens to be, it is always best to write out what the integral is... the definition just to get use to writing it out symbolically.*0885

*Very, very important. So, let us do example 1.*0901

*Find the area of a sphere of radius r, find the surface area of a sphere of radius r. Well, you already know what that is from geometry, but the idea is where did that formula come from.*0911

*You know, when you learned it we just sort of dropped it on you, now we are going to show you where the formula actually comes from.*0931

*Okay. So, we have the parameterization, we have p(φ) and θ, so let us use the standard parameterization, and we know that this equals rsin(φ)cos(θ), rsin(φ)sin(θ), oops, that is sin(θ), and then we have rcos(φ), well we already found what the norm was.*0939

*We already, in the previous lesson, we did this, we already found the norm of dp dφ cross dp dθ.*0969

*That was just the norm of the normal vector, and that was r ^{2} × sin(φ).*0984

*So, our ds element, dΣ, that is equal to this thing dφdθ = r ^{2}sin(φ)dφdθ.*0994

*Well, we know that φ, for a complete sphere is going to be > or = 0 and < or = pi, which is 180 degrees.*1018

*θ is going to run all the way from 0 to 2pi, which is 360 degrees, so, and like I just said a minute ago, let me go ahead and write the integral symbolically so we get used to writing it.*1027

*So, we are taking the integral over s of the differential area element. We are adding a differential area element and we are going to integrate, we are going to add all of the differential area elements over the surface.*1042

*Again, that is what integration is. Do not forget that. It is very important to remember that all you are doing is you are adding something. It is an infinite sum. The idea of integration is a particular technique, but it is just addition.*1056

*As it turns out in mathematics, all you can do to something is add things. That is it.*1067

*So, that is equal to the integral over s of -- alright -- dp dφ cross dp dθ, dφ dθ*1073

*Now we are actually going to write out the integral itself, and that is going to equal the integral from 0 to 2pi, so let us go ahead and do θ last.*1090

*It is going to be from 0 to pi, so this is going to be φ, and then we have this thing.*1100

*This thing is that, or ds is this whole thing.*1107

*So, we have r ^{2}, sin(φ), and we are going to be integrating with respect to φ first, then with respect to θ next, and when you go ahead and solve this integral... what you get is 4pi r^{2}, which is exactly the formula you learned in geometry.*1115

*This is where it comes from. It is based on the definition of surface area given a particular parameterization which is given by this. That is the definition of surface area. That is it.*1137

*Okay. Let us do another example here... excuse me.*1151

*Alright, so, example number 2. We want to find the area of the paraboloid z = x ^{2} + y^{2}, so now it is given... this paraboloid is not given to us in parameterized form.*1159

*It is given to us in just a straight Cartesian fashion... z = some function of x and y... for z > or = 0, and < or = 5.*1188

*So, we are trying to find the area of... so we have this paraboloid and z is > or = 0, so obviously it is going to be above the xy plane, and up to a length... up to a height of 5.*1211

*So, we want the surface area of that paraboloid.*1230

*Real quickly before I get into this problem, I want to talk about the notion of length, area, and volume.*1234

*Length is -- you know what length is -- and there is a physical way of actually measuring length.*1243

*If I said what is the length of some curve, well what you can do is you can take a string, run it along a curve, make sure that it is there, and then pull the string taught and you can measure the length of that string.*1250

*There is a physical representation, a physical manifestation of length.*1263

*As far as volume is concerned, if you want me to measure the volume of some strange looking object, what you do is you have a cup of water, you measure the height... you measure how much... you read off the graduation and then what you do is you insert this object into the water until it is just -- you know -- under the surface, and of course the water level is going to rise.*1267

*Well, the difference in water level gives you the volume of the object. Unfortunately for area, for surface area, there is no physical manifestation, so in some sense, area is a very, very odd thing.*1289

*Length we can deal with physically, volume we can deal with physically, area we cannot really deal with physically.*1302

*Area is strictly a mathematical notion. I just wanted you to be aware of that. It is very, very curious that that is the case.*1308

*Okay, so let us go ahead and finish this problem. So we are given our equation that way, so let us go ahead and do our parameterization.*1317

*So, our parameterization, when we are given z = f(x,y), well that is equal to x and y and the function itself x ^{2} + y^{2}, so this is our parameterization.*1326

*Coordinate function 1, number 2, and number 3. Okay.*1339

*Let us find dp dx, so this is going to be a vector. This is going to be (1,0,2x), I am just differentiating this.*1344

*let us go ahead and find dp dy, so dp dy, that is going to equal (0,1,2y), excuse me.*1355

*Now we need to find the cross product of these two vectors, so dp dx cross dp dy, and we are going to go ahead and use our i,j,k, symbolism. We have (1,0,2x), and we have (0,1,2y).*1366

*When you expand this determinant you are going to end up with the following. You are going to end up with -2x,-2y, and 1.*1387

*In this particular case, I ended up choosing the opposite orientation. In this particular case, I ended up choosing such that this normal vector -- this is the normal vector -- is actually pointing into the paraboloid as opposed to out of it.*1396

*It is not really going to be a problem because when we take the norm of this, the norm is the same whether it is pointing in or pointing out, but it just so happens that I noticed that I have the incorrect orientation.*1412

*Later, that might be a problem, but for right now, it is not... and we will deal with it when we get to it.*1424

*Okay. So, this is our vector. Now we need the norm of this vector. So, the norm of dp dx cross dp dy = norm of this.*1430

*Well, it equal 4x ^{2} + 4y^{2} + 1 under the radical.*1448

*So, our differential area element that we are going to integrate, ds is equal to the radical 4x ^{2} + 4y^{2} + 1 dy dx.*1457

*That is what we are going to integrate. Now, for... now for z = x ^{2} + y^{2}, we said that we wanted z < or = 5, so x^{2} + y^{2} < or = 5.*1474

*Well, if that is 5, then the disc that we are looking at, this length right here is actually going to end up being sqrt(5), and this length right here is going to be sqrt(5).*1499

*So, what we are going to be doing is we are going to be integrating... this is x and this is y, the z axis is going like this.*1514

*We are integrating over a region of the xy plane, right? dy dx, so we need to know what x goes from and we need to know what y goes from.*1522

*In this particular case, z = x ^{2} + y^{2} < 5.*1534

*Well, that means this circle is in the xy plane, that is this way, and of course the paraboloid is going up this way.*1543

*Well, if z itself is 5, well then this radius, our circle underneath has a radical of sqrt(5), simply based on the parameterization... x ^{2} + y^{2}.*1552

*I hope that that makes sense. So, in this particular case, our x is going to run from - sqrt(5) all the way to +sqrt(5).*1567

*It is going to go from this point to this point... excuse me... and our y value is going to go from... our y, the height is going to change... so y is going to run from -5 - x ^{2} all the way to sqrt(5) - x^{2}.*1583

*That comes from if I move this x ^{2} over here, and I take this square root, I get y explicitly in terms of x, because remember, again, I am integrating with respect to y and I am integrating with respect to x, so I have to know what x does and what y does.*1606

*So, my area integral ends up being as follows: area = the integral from -sqrt(5) to sqrt(5) -- excuse me -- the integral - 5 - x ^{2} under the radical, to sqrt(5) - x^{2} under the radical.*1627

*Of course I have my function, the norm of the cross product, which was 4x ^{2} + 4y^{2} + 1 under the radical, dy dx.*1653

*Because this is y and this is x, so I am integrating that first.*1667

*When I put this into my mathematical software I get the number 49.864. There we go. That is the surface area of the paraboloid, that has a height of 5.*1672

*Okay. Now we did this in Cartesian coordinates. Now let us go ahead and run this integral in polar coordinates, just to see what it looks like. Just for the sake of some practice.*1688

*Okay, now, in polar coordinates -- forgive me, I seem to have something in my throat here -- in polar coordinates, it looks like this.*1698

*Again, we have a circle and we have this paraboloid on top and of course it is on top of this circle.*1724

*The radius is sqrt(5), so r is going to run from 0 to sqrt(5), because now we are going to be integrating r out this way, and then we are going to sweep this line around... all the way around the circle... to pi.*1741

*So θ is going to run from 0 to 2pi, so these are our lower and upper limits of integration. All I am doing is I am taking this region and I am expressing it in terms of r and θ. I am converting to polar coordinates.*1765

*When I convert to polar coordinates, I have to make the appropriate changes to the integral. Now, this is my -- let me go to red here -- this is my integrand.*1780

*I have to express this integrand in terms of r and θ.*1794

*Well, the conversion, the transformation is x = rcos(θ) for polar coordinates and y = rsin(θ), and when you put rcos(θ), rsin(θ) into this, you end up with the following.*1798

*So, 4x ^{2} + 4y^{2} + 1 in polar form is.. .when you put these into here and solve, what you get is sqrt(4)r^{2} + 1.*1813

*That is our integrand. Okay. Now, dy dx... remember when we convert to polar coordinates, the differential area element is equal to r, dr, dθ, there is this extra factor of r.*1836

*Now let us go ahead and put everything together. Our area in terms of polar coordinates.*1856

*I am going to go ahead and do r last. Go from 0 to 2pi, this thing is that thing, it is 4r ^{2} + 1, and now dy dx is equal to r dr dθ.*1863

*So I write r, dr, dθ. When I go ahead and put this into my mathematical software, I get the number 49.866. That is exactly write. This is .864, this is .866.*1883

*These had to be done numerically, there was no simply symbolic way of doing it to come up with nice closed form expression.*1897

*That is nice. The software will do that for us. That is it.*1904

*All I have done is, I was given a particular surface, I did the Cartesian xy, I found the integral this way, I wanted you to see what it looked like in polar coordinates, it is not necessarily easier, it is just a question of what is easier for you.*1910

*Or perhaps... and again, it is going to depend on the problem.*1928

*We cannot say do it this way or do it this way, but when you do make a conversion from a particular coordinate system, and you solve the integral in another coordinate system, you have to remember that the differential area element, or the differential volume element if you are dealing with triple integrals... it also undergoes a transformation.*1931

*That is it. For every object that you have in one integral, you are going to have a corresponding object in the other integral. Just keep things straight.*1951

*Okay. Well, thank you for joining us here at educator.com to discuss surface area, we will see you next time for a discussion of surface integrals. Take care, bye-bye.*1960

1 answer

Last reply by: Professor Hovasapian

Sat Aug 3, 2013 2:56 PM

Post by michael Boocher on August 3, 2013

Very profound, turns out all you can do is add things. (17:45)

1 answer

Last reply by: Professor Hovasapian

Sun May 26, 2013 4:30 PM

Post by Josh Winfield on May 26, 2013

Thanks raffi, example 2 should end with rd(theta)dr not other way around but the numerical answer is still correct