A shell method (calculus) question?

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SUMMARY

The discussion focuses on using the shell method to calculate the volume of the solid formed by rotating the curve defined by the equation y=3+2x-x^2 around the line x=4. The correct formula for volume is V=2π (integral from 0 to 3) (radius)*(height) dx, where the radius is (4-x) and the height is given by the function. The user initially calculated an approximate volume of 155.51, but the exact volume is 99/2π, confirming that the original calculation was indeed correct despite doubts about its magnitude.

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jcfor3ver
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Homework Statement



Heres the question.
Use shell method to find volume:

y=3+2x-x^2 with boundaries: y=0, x=0, intercept x=3 and rotate around x=4



Homework Equations



I know that V=2pi (integral from 0 to 3) (radius)*(height) dx



The Attempt at a Solution



My radius was (4-x) and my ht was y=3+2x-x^2

now my answer was aprox= 155.51, and i triple checked it. What the hell am I doing wrong because it seems like that volume is much too large for an answer. my exact answer was 99/2pi
 
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You didn't do anything wrong. Your answer is correct. Your intuition is off.

To get a really rough estimate of what the volume should be, try approximating the shape as a flat disk of radius 4 with a hole of radius 1 in the middle. If you plot y vs. x, you can see 3 would be a decent approximation for the height of this flat disk. If you calculate the volume of the disk, it comes out to 47π, which is pretty close to the exact volume you found.
 

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