Multivariable Calculus - Surface integrals

[kmr159]

1. Homework Statement ∫∫_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)) * |r_u χ r_v|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,
r_y(the partial derivative of r with respect to y) = <0,1,0>
r_x = <0,0,1>

r_x X(cross) r_y = <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
r_θ = <0,-3sin(θ),3cos(θ)>
r_x = <1,0,0>

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

The double integral becomes 3∫₀^2∏∫₀^{5 - 3cos(θ)} x(3sin(θ))d_xd_θ

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>

r_y = <-1,1,0>
r_x = <0,0,1>

r_y χ r_x = <1,1,0>. The magnitude is 2^.5

The double integral( without limits) becomes 2^.5∫∫xz d_yd_z ->

2^.5∫∫5z - yz d_yd_z

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

 

[member 428835]

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

 

[member 428835]

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

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.

 

[member 428835]

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

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.

 

[member 428835]

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