How do I find the surface area of the solid defined by revolving $r = 3sin(theta)$ about the polar axis?

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Answer 1 · 2015-02-24T00:17:56

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$r = 3 sin(theta)$

$r = 3y/r$ because $y=r sin(t)$

$r^2 = 3y$

$x^2 + y^2 = 3y$ because $r = sqrt(x^2+y^2)$ .

$x^2 + (y-3/2)^2 = 9/4$

You recognize a circle of radius $3/2$ . The area is $pi (3/2)^2 = 9 pi/ 4$ .

How do I find the surface area of the solid defined by revolving r = 3sin(theta)r = 3sin(theta) about the polar axis?