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# How do you find the area inside the circle ##r=6## and above the line ##r=3cscθ##?

First, you need to determine to which ##theta## the curve ##r=3csc theta## is within the circle ##r=6##.

Since we want ##3csc theta <=r<=6##, we implicitly need that:

##3csc theta<=6##, or ##1/(sin theta) <=2##, that is, ##sin theta >=1/2##

We have, then:

##pi/6<=theta<=5pi/6## or ##7pi/6<=theta<=11pi/6##

The curve is symmetric through the x axis, so we need only to integrate in the first interval.

Using the change of coordinates to polar ordinates, we have that ##dV=r dr d theta##.

Then,

##V=2 int_(pi/6)^(5pi/6) int_(3csc theta)^6 r dr d theta=##

##2 int_(pi/6)^(5pi/6) [(6^2)/2-((3 csc theta)^2)/2] d theta=##

##2 int_(pi/6)^(5pi/6) (18 - 9/2*csc^2 theta) d theta=##

##24pi - 2 int_(pi/6)^(5pi/6) 9/(2*sin^2 theta) d theta=##

##24pi - 9 (ctg(5pi/6)-ctg(pi/6))=24pi-18sqrt(3)##