Efficiently Solve a Challenging Rotation Problem | Homework with Shell Method

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The discussion focuses on solving a rotation problem using the shell method, specifically addressing the setup of the integral. The integral proposed is 2π ∫₀¹ (4-y)(√sin(y)) dy, but there is confusion regarding the limits of integration. Participants clarify that the integration should be with respect to y, and the correct limits should reflect the graph of x = √sin(y) from 0 to π. Visualizing the radius in these problems is highlighted as a common challenge. Overall, the conversation emphasizes the importance of correctly interpreting the functions and limits in the context of the shell method.
iRaid
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Homework Statement


serocp.png



Homework Equations


Shell method


The Attempt at a Solution


Not sure if this is right, but the integral I set up is:
2\pi \int_0^1 (4-y)(\sqrt{siny})dy


Finding the radius is my big problem with these problems, I can't visualize it very well.

Any help is appreciated.
 
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iRaid said:

Homework Statement


serocp.png


Homework Equations


Shell method

The Attempt at a Solution


Not sure if this is right, but the integral I set up is:
2\pi \int_0^1 (4-y)(\sqrt{siny})dy
Finding the radius is my big problem with these problems, I can't visualize it very well.

Any help is appreciated.
That looks fine except for the limits of integration.

How did you get those limits?
 
graphing \sqrt{siny} from 0 to pi gives me a curve that goes to 1 and back to 0. Since I'm integrating wrt y, I figured it's 0 to 1?
 
Last edited:
iRaid said:
graphing \sqrt{siny} from 0 to pi gives me a curve that goes to 1 and back to 0. Since I'm integrating wrt y, I figured it's 0 to 1?
You don't have \ \ y=\sqrt{\sin(x)}\ \ with x going from 0 to π.

You have \ \ x=\sqrt{\sin(y)}\ \ with y going from 0 to π --- and your integration is w.r.t. y.
 
SammyS said:
You don't have \ \ y=\sqrt{\sin(x)}\ \ with x going from 0 to π.

You have \ \ x=\sqrt{\sin(y)}\ \ with y going from 0 to π --- and your integration is w.r.t. y.


Thanks, probably shouldn't agree with wolfram straight after I graph...
 
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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