A shell method (calculus) question?

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The discussion centers on using the shell method to calculate the volume of the solid formed by rotating the curve y=3+2x-x^2 around the line x=4. The user initially calculated a volume of approximately 155.51 but later found the exact answer to be 99/2pi. It was clarified that the initial intuition about the volume being too large was incorrect, as the calculated value is indeed accurate. A rough estimate using a flat disk model suggested a volume of about 47π, which aligns closely with the exact result. The conversation emphasizes the importance of understanding the geometric implications of the shell method.
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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