Help with solid of revolution volume question

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Homework Help Overview

The discussion revolves around a volume calculation for a solid of revolution using the shell method. Participants are examining the setup of the integral and the limits of integration based on the given problem context.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the integral using shell's method, questioning the limits of integration and the region being considered. There is an exploration of how to determine appropriate limits through inspection versus calculation.

Discussion Status

Some participants have provided guidance on correcting the limits of integration, suggesting that the original poster was integrating over an incorrect region. There is acknowledgment of a successful resolution for one participant after adjusting the limits, but further questions about general strategies for determining limits remain open.

Contextual Notes

Participants are navigating the constraints of the problem as presented, including the need to accurately identify the region enclosed by the curves and the implications of integrating over incorrect limits.

student93
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Homework Statement



The problem is attached in this post.

Homework Equations



The problem is attached in this post.

The Attempt at a Solution



I used shell's method and set up my integral as 2π∫(4-x)(x^2)dx, from -2 to 2 and got an answer of 128π/3 which is incorrect. The actual answer is 40π/3.

I set my radius as 4-x and I set my height as x^2.

To calculate the limits of integration with respect to x (since I used shell's method), I set x^2=2 and got -2 and 2.
 

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Your setup is correct, but your limits are not. You have to integrate over the region with respect to x, and stay inside of that region too. Going outside would mean you're creating shells that aren't part of the volume you're calculating. It's usually easiest to find your limits by inspection rather than calculation. In this case, the leftmost side of the region is the origin, and then the rightmost side (as given in the problem) is x = 2. Try those limits instead.
 
student93 said:

Homework Statement



The problem is attached in this post.


Homework Equations



The problem is attached in this post.


The Attempt at a Solution



I used shell's method and set up my integral as 2π∫(4-x)(x^2)dx, from -2 to 2 and got an answer of 128π/3 which is incorrect. The actual answer is 40π/3.

I set my radius as 4-x and I set my height as x^2.

To calculate the limits of integration with respect to x (since I used shell's method), I set x^2=2 and got -2 and 2.

You are integrating over a larger region than you should. The part of your region between -2 and 0 isn't really 'enclosed' by your curves.
 
Thanks, I ended up getting the correct answer after using the right limits. Also, do you have any suggestions as to how to make sure that I've used the right limits of integration in other similar problems? How do you know when to calculate the limits of integration or when to inspect the graphs for the limits?
 
student93 said:
Thanks, I ended up getting the correct answer after using the right limits. Also, do you have any suggestions as to how to make sure that I've used the right limits of integration in other similar problems? How do you know when to calculate the limits of integration or when to inspect the graphs for the limits?

Make graphs and inspect the graphs for the limits. Always. Don't just blindly solve an equation without looking at what it means for the graphs.
 

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