# Homework Help: Multivariable Calculus - Surface integrals

1. Jan 5, 2014

### kmr159

1. The problem statement, all variables and given/known data ∫∫S xz dS where S is the boundary region enclosed by the cylinder y2 + z2 = 9 and the planes x = 0 and x + y = 5.

2. Relevant equation∫∫Sf(x,y,z)dS = ∫∫Df(r(u,v)) * |ru χ rv|dA

3. The attempt at a solution
I think I have broken this up into 3 surfaces. The bottom circle in the y-z plane. - S1

The cylinder y2 + z2 = 9 that is intersected by the plane x + y = 5 - S2

and the slanted ellipsoid resulting from that intersection - S3

Calculations for S1

The equation for r for the S1 is r = <0,y,z> - I think,
ry(the partial derivative of r with respect to y) = <0,1,0>
rx = <0,0,1>

rx X(cross) ry = <1,0,0>. The magnitude is 1

So we integrate the double integral of x * z dy dz = 0 & since x = 0

Calculations for S2

The equation for r for the S2 is r = <x,3cos(θ),3sin(θ)> - I think
rθ = <0,-3sin(θ),3cos(θ)>
rx = <1,0,0>

rθ X(cross) rx = <0,3cos(θ),3sin(θ)>. The magnitude is 3

The double integral becomes 3∫02∏05 - 3cos(θ) x(3sin(θ))dxdθ

I have a question - do I need to apply a jacobian or something similar because I changed variables from z to θ?

When I solved this double integral I got a value of 0.

Calculations for S3

I am not sure what r is supposed to be but I guessed r = <5-y,y,z>

ry = <-1,1,0>
rx = <0,0,1>

ry χ rx = <1,1,0>. The magnitude is 2.5

The double integral( without limits) becomes 2.5∫∫xz dydz ->

2.5∫∫5z - yz dydz

I am not sure where to go from here. If I convert to polar coordinates I only have one variable that is changing - θ

Additionally I am not sure how jacobians figure into surface integrals.

I am asking for help understanding this problem - validating that I am approaching it correctly. I also don't know how to parametrize and solve S3 (not that I am sure of my parametrizations for S1 or S2).

Thanks

Last edited: Jan 5, 2014
2. Jan 6, 2014

### joshmccraney

why are you not using polar coordinates for the entire problem?

Last edited: Jan 6, 2014
3. Jan 6, 2014

### joshmccraney

nope. you are thinking of $dA$, in which case you would need the $r$ as a jacobian. however, your formula directs you to use $dS$, a parametized length, so don't worry about the jacobian.

4. Jan 6, 2014

### joshmccraney

well, it seems ideal to use polar coordinates, right??? thus if the "top" plane is $x + y = 5$, should we use coordinates, say $x=r\cos\theta, y=r\sin\theta$? only remember, our coordinate system isn't around the z axis, its around the x-axis, so some slight modification should be used, which it seems you are aware of.

so now, we should have $$\vec{r}=f(x,y,z)\hat{i}+r\cos\theta\hat{j} +r\sin\theta\hat{k}$$ and i think ${f}=x=5-y=5-r\cos\theta$ now perhaps integrate over this flat surface through $r$ and $\theta$ (after taking that messy cross product)

you of course know when i use $r$ i speak of radius and $\vec{r}$ is in reference to your adopted r notation.

5. Jan 6, 2014

### joshmccraney

let me know if this doesn't make sense, but i think it's clear.