# Homework Help: Find Volumes by integration.

1. Feb 3, 2013

### daivinhtran

1. The problem statement, all variables and given/known data
THe region bounded by y = -x + 3 and y = x^2 - 3x
the region revolve about a, x-axis, and b, y=axis

2. Relevant equations
V = π∫r^2 dx

3. The attempt at a solution
I have no clue to solve it since the volume overlap. I try to ignore the overlapped region but didn't get the right answer.

Last edited: Feb 3, 2013
2. Feb 3, 2013

### SteamKing

Staff Emeritus
Your problem statement as worded will not result in a volume, but an area.

Have you left something out, like the region is to be rotated about a certain axis?

In any event, you must determine the limits where the overlap takes place, in order to calculate the proper area.

3. Feb 3, 2013

### daivinhtran

oh yes...I actually left something out....the region revolve about a, x-axis, and b, y=axis..

4. Feb 3, 2013

### haruspex

Treat the overlapping and non-overlapping regions separately.

5. Feb 3, 2013

### daivinhtran

what do you do with the overlapping region?

6. Feb 3, 2013

### SammyS

Staff Emeritus
The following is a graph by WolframAlpha which may help for the rotation about the x axis.

It's a graph of 4 curves:

$y = -x + 3$

$y = -(-x + 3)$

$y = x^2 - 3x$

$y = -(x^2 - 3x)$

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7. Feb 3, 2013

### daivinhtran

I did try this way....I find the region at the interval [-1, 0] by take the integrate of pi ∫(-x+3)^2 - (x^2 - 3x)^2 dx........
Then, pi x ∫ (-x+3)^2 - (-x^2 + 3x)^2 dx for interval [0,1]
Then, pi x ∫ (3x-x^2)^2 for the interval [1,3]
and take the sum of all...and get 56pi/3

but the answer is not that

8. Feb 3, 2013

### SammyS

Staff Emeritus
This one is incorrect.