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anonymous
 5 years ago
volume of a region bounded by x=y^2 +4 and x=0; revolved about x=1
anonymous
 5 years ago
volume of a region bounded by x=y^2 +4 and x=0; revolved about x=1

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01. 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Did that still got it wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It looks like you are doing washer method, which does not look appropriate for this case.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Correction washer merit is appropriate

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Bounds of integration 0 to 2. Integrate [(y^2+4)^2 pi  1^2 pi]
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