# Surface Integrals: What am I doing wrong?

1. Jan 26, 2010

### 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