Surface Integrals: What am I doing wrong?

[math_maj0r]

Problem: evaluate the double integral of yz over that part of the plane z=y+3 which is inside the cylinder x²+y²=1

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

I got (y³+9y+6y²)/2. Then I integrated this over x²+y²=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 [r³sin³(theta) +9sin(theta)+3r²sin²(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 2^1/2pi/4

What did I do wrong?

 

[mighty2000]

Hi there,

Thank you for sharing your work and your question. It looks like you have set up the integral correctly and used the appropriate change of variables to switch to polar coordinates. However, there may be a mistake in your integration process.

I would suggest double-checking your integration steps and making sure that you are correctly integrating each term in the integrand. It is possible that there was an error in one of your calculations or in setting up the limits of integration.

Another possibility is that there may be a typo in the original problem or in the provided correct answer. I would recommend double-checking the problem statement and the correct answer to make sure they match up with your work.

If you are still having trouble, I would suggest consulting with a colleague or your professor to get a second opinion on your work. They may be able to spot the error or provide additional guidance.

I hope this helps. Good luck!