Problem: evaluate the double integral of yz over that part of the plane z=y+3 which is inside the cylinder x(adsbygoogle = window.adsbygoogle || []).push({}); ^{2}+y^{2}=1

I evaluated with respect to z, from z=0 to z=y+3

I got (y^{3}+9y+6y^{2})/2. Then I integrated this over x^{2}+y^{2}=1. To do that, I switched to polar coordinates, letting x=rcos(theta), y=2sin(theta)

To set up the integral, I used the formula: double integral of [f(x,y,z) dS] = double integral of [f(R(u,v))(multitude of cross product of derivative of R with respect to u, and derivative of R with respect to v) dA] where R is a vector

In my case, R = rcos(theta)i + rsin(theta)j

and u=r, and v=theta

I got (multitude of cross product of derivative of R with respect to u, and derivative of R with respect to v) = r

Then I integrated [r^{3}sin^{3}(theta) +9sin(theta)+3r^{2}sin^{2}(theta)]r r dr d(theta) over r from 0 to 1, and theta from 0 to 2pi

My final answer was 3pi/5. The correct answer is 2^{1/2}pi/4

What did I do wrong?

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# Homework Help: Surface Integrals: What am I doing wrong?

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