Using double integration to find area of a cylinder

liishii
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Homework Statement


Find the area of that portion of a cylinder x^2+y^2=1 lying above the xy-plane and below the plane z=3y.

Homework Equations


dr=sqrt(1+fx2+fy2[/SUB])dxdy

or

dr=|ru x rv| dudv

The Attempt at a Solution


I attempted to parametrize the function in terms of theta (t) and z. I got r(t,z)=icost+jsint+kz. Then, I found the partials with respect to t and z and got: rt=-sinti+costj and rz=z, where t=0 to 2pi and z=0 to 3y=3sint.
I took the cross product of the two partials to get icost+jsint+0k. I used the cross product to find |rt x rz|=1. Then, plugging this into the second equation above, I got the answer to be 0, which is incorrect. What am I doing wrong?? Thanks for your help!
 
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Your limits for t are wrong. Remember, you only want the part above the xy-plane.
 
oh that makes sense! so the t limits had to be from 0 to pi, not 2pi.
I got the right answer! Thanks so much vela!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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