Hello Everyone, I\'m having some difficulty setting up this problem since the re

Question

Hello Everyone,

I'm having some difficulty setting up this problem since the region is not centered. Any help would be appreciated.

Problem- Find the surface area of that portion of the sphere x^2+y^2+z^2=9 that lies inside the cylinder x^2+y^2=3y.

So far I have converted to polar but cannot find the region for dr.

Explanation / Answer

Use polar coordinates. Since R(r, ?) = represents the top half of the region, we may use symmetry and double the surface area of this region. As for the bounds, note that x^2 + y^2 = 3y ==> r^2 = 3r sin ? ==> r = 3 sin ?. So, the bounds are 0 = r = 3 sin ? with ? in [0, p]. Next, R_r x R_? = ==> ||R_r x R_?|| = r sqrt[r^2 /(9 - r^2) + 1] = 3r / sqrt(9 - r^2). So, the surface area equals 2 ?(? = 0 to p) ?(r = 0 to 3 sin ?) [ 3r / sqrt(9 - r^2)] dr d? = 2 ?(? = 0 to p) -3(9 - r^2)^(1/2) {for r = 0 to 3 sin ?} d? = 2 ?(? = 0 to p) 9(1 - cos ?) d? = 18(? - sin ?) {for ? = 0 to p} = 18p. I hope this helps! please do rate my answer

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