Asked by Anonymous
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?
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?
Answers
Answered by
Steve
Since the region is rotated around x = -3, you should have
π((3+√(9-y^2))^2-9)
π((3+√(9-y^2))^2-9)
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