Question

The semicircular region bounded by the curve x=sqrt{9-y^2} and the y-axis is revolved about the line x=-3.

The integral that represents its volume is
V= ∫ [a^b] f(y) dy

What is f(y)?



I've gotten f(y) to:

(1+(sqrt(9-y^2)))^2-9

But it's not being accepted. What am I doing wrong?
Since the region is rotated around x = -3, you should have

π((3+√(9-y^2))^2-9)

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