volume of a region bounded by x=-y^2 +4 and x=0; revolved about x=-1

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volume of a region bounded by x=-y^2 +4 and x=0; revolved about x=-1

Mathematics
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1. solve for y in terms of x. this is R(x) 1a. since 0 is 1 step away from x = 1 call this r 2. take the integral of this R^2 - r^2 dx 3. multiply the result by pi. 4. find where R and r interect. these are your bounds for the integral
Did that still got it wrong
my bad it looks like your rotating the other way inegrate this: (-y^2 + 4 ) ^2 - 1 dy your bounds of integration will be the values of y where x = 0 and the parabola intersect. then mutliply by pi.

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Other answers:

It looks like you are doing washer method, which does not look appropriate for this case.
Correction washer merit is appropriate
Bounds of integration 0 to 2. Integrate [(-y^2+4)^2 pi - 1^2 pi]

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