Calculating Volume of Overlapping Regions using Integration

[daivinhtran]

Homework Statement

THe region bounded by y = -x + 3 and y = x^2 - 3x
the region revolve about a, x-axis, and b, y=axis

Homework Equations

V = π∫r^2 dx

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.

 

[SteamKing]

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.

 

[daivinhtran]

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.

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

 

[haruspex]

Treat the overlapping and non-overlapping regions separately.

 

[daivinhtran]

what do you do with the overlapping region?

 

[SammyS]

what do you do with the overlapping region?

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)

 

[daivinhtran]

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)

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

 

[SammyS]

...

Then, pi x ∫ (-x+3)^2 - (-x^2 + 3x)^2 dx for interval [0,1]
...

This one is incorrect.

Added in Edit:

For x > 0, the inner radius is zero. -- This becomes the disc method rather than the washer method for x > 0.

 

[haruspex]

I'll modify my previous suggestion. If you treat overlapping and non-overlapping entirely separately you'll have five separate integrals (2 for overlap, 3 for non).
It's a bit easier to process the whole shape without worrying about the overlap, then subtract the overlap parts on the basis that they have been counted twice. That should reduce it to the three integration steps.