Volume of Region Rotated about X-Axis Using Shell Method

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Homework Help Overview

The discussion revolves around using the shell method to calculate the volume of a region rotated about the x-axis, specifically involving the equations y=3x+10 and y=x^2.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the integral for the shell method and question the correctness of the initial attempts. There is a focus on whether a single integral is sufficient for the problem, particularly given the changing boundary function.

Discussion Status

Some participants are exploring the setup of the integral and questioning its correctness, while others suggest that a visual representation of the region might clarify the problem. There is acknowledgment of differing results from calculations, indicating ongoing exploration of the problem.

Contextual Notes

Participants note the importance of understanding the boundaries of the region, particularly at the point (-2, 4), which may affect the integral setup.

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Homework Statement


use the shell method to find the volume of the region rotated about the x-axis.

Homework Equations


y=3x+10
y=x^2

The Attempt at a Solution


2∏ (integral) (0 to 25) [ (y) (sqrt(y) - (y-10)/3) dy ]
 
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whatlifeforme said:

Homework Statement


use the shell method to find the volume of the region rotated about the x-axis.


Homework Equations


y=3x+10
y=x^2


The Attempt at a Solution


2∏ (integral) (0 to 25) [ (y) (sqrt(y) - (y-10)/3) dy ]

What's the question?
 
is my attempt correct so far? just setting up the integral? because i keep getting the wrong answer.

my answer: 1111 (pi); correct answer: 5488pi/5
 
First off, did you draw a picture of the region, and did you draw a sketch of the solid of revolution?

If so, you should have noticed that one integral isn't going to work for this problem. The boundary function on the left changes at the point (-2, 4).
 

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