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## Multivariable calculus

### Course: Multivariable calculus > Unit 4

Lesson 11: Surface integrals- Introduction to the surface integral
- Find area elements
- Example of calculating a surface integral part 1
- Example of calculating a surface integral part 2
- Example of calculating a surface integral part 3
- Surface integrals to find surface area
- Surface integral example, part 1
- Surface integral example part 2
- Surface integral example part 3: The home stretch
- Surface integral ex2 part 1
- Surface integral ex2 part 2
- Surface integral ex3 part 1
- Surface integral ex3 part 2
- Surface integral ex3 part 3
- Surface integral ex3 part 4

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# Surface integral ex3 part 4

Evaluating the third surface integral and coming to the final answer. Created by Sal Khan.

## Want to join the conversation?

- Why don't you add another r before you start integrating? Isn't the Jacobian of polar coordinates r? Or is that just when you switch from rectangular to cylindrical, where here you're never started in rectangular?(9 votes)
- You add r only if you are in polar coordinates. In this video Sal is not using polar coordinates but he is using variables r and theta. You could name those variables s and t and nothing would change.(5 votes)

- after watching all the videos for this example, although we omit the bottom surface because of the parameter 'z', how do we incorporate the surface area for it? It seems as though we could use algebra for the majority of this surface, except for s2.(2 votes)
- If the objective of the problem was to obtain the surface area of the figure, then yes, you could probably have found the area of the upper ellipse using algebra. But the problem wasn't to find the surface area of the figure, but to find how the scalar field
`f(x,y,z) = z`

behaved over the surface of the figure, and I doubt there is a way to do that without using integration.(3 votes)

- When evaluating the last integral at time2:46, shouldn't it be positive (1/3)sin(theta) because the negatives cancel out? I know that this doesn't change the answer because either way that term is zero but just checking.(1 vote)
- Nope! You're integrating cos and not differentiating cos. d(sin(ø))/dø = cos(ø). d(cos(ø))/dø = -sin(ø). So there is no negative from the integration of cosine :)(2 votes)

- The surface s3 is making a projection on xy plane which is a circle of area pi. This surface s3 is making an angle of 45 degree to xy plane.

So, area of circle= area of s3 x cos 45

=> area of s3 = area of circle x sec 45=pi x root2.

Is this approach also correct? Thank You.(1 vote) - What is the point of making several shorter videos? Why not just combine them into one...?(0 votes)

## Video transcript

So I stopped abruptly 12
minutes into the last video because 12 minutes was too long. I needed a break. But where we left off, we
were in the home stretch. We were evaluating the
third surface integral. And we were setting it up as
a double integral with respect to r and theta. And we just had to
set up the bounds. And we know that r takes
on values between 0 and 1. And theta takes on values
between 0 and 2 pi. So r, we're going to integrate
with respect to r first. r takes on values
between 0 and 1. And theta takes on values
between 0 and 2 pi. And so now we are
ready to integrate. So let's do the first part. Let's do this inside
part right over here. And I'm just going to
rewrite the outside part. So this is going
to be equal to-- we have our square root of 2
times the integral from 0 to 2 pi of-- and we have
d theta right over here. So that's the outside. This inside part
right over here, we can rewrite it as
if we distribute the r. It's r minus r squared
cosine of theta. Now we're going to
integrate with respect to r. So when we integrate with
respect to r, cosine of theta is just a constant. So if you integrate
this with respect to r, you get-- and I'll do
it in that pink color-- if you integrate with respect
to r, the antiderivative of r is r squared over 2. And the antiderivative
of r squared is r cubed over 3 minus r
cubed over 3 cosine theta. This is just a constant. And we're going to
evaluate that from 0 to 1. So when you would
evaluate it at 1, you get 1/2 minus
1/3 cosine theta. So you get 1/2-- I'll just
do it right over here-- 1/2 minus 1/3 cosine theta. And then minus both of
these evaluated at 0, well those are
just going to be 0. 0 squared minus 0
squared times whatever. It's all going to be 0. So this business
right over here just evaluates to 1/2 minus
1/3 cosine of theta. And so we get the integral. This is all going
to be equal now to the square root of 2 times
the integral from theta 0 to 2 pi of 1/2 minus 1/3
cosine of theta d theta. And this is equal to
square root of 2 times the antiderivative
of 1/2 is 1/2 theta. And the antiderivative of
cosine theta is sine theta. So minus 1/3 sine theta. And we're evaluating
it from 0 to 2 pi. When you evaluate
these at 2 pi, you have-- let me just write it all
out-- this is the home stretch. I don't want to make
a careless mistake. We have square root of 2 times--
let's evaluate it at 2 pi-- 1/2 times 2 pi is pi minus
1/3 times sine of 2 pi-- well that's just going to be 0. And when you evaluate it
at 0, 1/2 times 0 is 0. Sine of 0 is 0. So that all comes out to 0. So all of this business
simplifies to pi. And we are done. We've evaluated surface three. It is square root of 2, or
the third surface integral, or the surface integral
over surface three. It is square root of 2 pi. And we are done. So this part right over
here is square root 2, or root 2, times pi. So this entire surface
integral that we started like 80 videos ago, evaluates
to-- the orange part evaluates to 0. The blue part, 3 pi over 2. And then the magenta part
is square root of 2 pi. So we're left with 3 pi over
2 plus the square root of 2 pi is the value of our
original surface integral.