# 2nd semester calc question : calculating volume of solid of revolution

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• #1
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## Homework Statement

The solid formed when the region bounded by y = x^2 and y = 2 - x^2 is revolved about the x-axis

## Homework Equations

disc method with respect to x-axis

the integral of : (pi * (f(x)^2 - g(x)^2))

## The Attempt at a Solution

When I square each function and integrate the resulting difference I get :

4x - 4x^3/3 with limits of integration from 0 to 1 (x = 0 and x = 1)

My final answer is 8pi/3 (eight pi thirds)

My solutions manual gives me an answer of 16pi/3. However, I noticed that they have a constant factored out of their integral of 2pi instead of just pi. You are supposed to use pi instead of 2pi for the disc method, hence their answer being exactly double what mine is.

Either this is a mistake on my book's part or I am missing something :(

## The Attempt at a Solution

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• #2
Dick
Homework Helper
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Check your limits again. You want the region bounded between the two curves. It doesn't say the y-axis is a boundary. The '2' is there for a different reason than you think.

• #3
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Dick, thanks for the response.

However, I did leave out the boundary was supposed to be between x^2, 2 -x^2, and the x axis.

I'm not sure if that makes a difference.

• #4
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I think it does. Would my new limits of integration be from 0 to where (2-x^2) hits the x axis?? (+ and - root (2))

• #5
Dick
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I think it does. Would my new limits of integration be from 0 to where (2-x^2) hits the x axis?? (+ and - root (2))
Where do the two curves intersect? They don't intersect at x=0.

• #6
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LOL

Ok, they intersect at 1 and -1. They evaluated the integral from 0 to 1 and multiplied it by two since these functions are even and have symmetry.

So, with my constant of pi I should have evaluated it from -1 to 1.

If that isn't it I think I may just cry.

• #7
Dick
Homework Helper
26,258
618
Don't cry. That's it. They probably shouldn't have used symmetry without saying so. It's doesn't really make it that much easier.

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