- #1

kmr159

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1. Homework Statement

2. Relevant equation

3. The Attempt at a Solution

**∫∫**_{S}xz dS where S is the boundary region enclosed by the cylinder y^{2}+ z^{2}= 9 and the planes x = 0 and x + y = 5.2. Relevant equation

**∫∫**_{S}f(x,y,z)dS = ∫∫_{D}f(**r**(u,v)) * |**r**χ_{u}**r**|dA_{v}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 y

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

r

r

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

r

r

The double integral becomes 3∫

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

r

r

The double integral( without limits) becomes 2

2

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

The cylinder y

^{2}+ z^{2}= 9 that is intersected by the plane x + y = 5 - S2and 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 1So 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 3The double integral becomes 3∫

_{0}^{2∏}∫_{0}^{5 - 3cos(θ)}x(3sin(θ))d_{x}d_{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_{y}d_{z}->2

^{.5}∫∫5z - yz d_{y}d_{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).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**
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