# Volume between two paraboloids.

1. Sep 13, 2009

### forty

Find the volume of the region between the two paraboloids z1=2x2+2y2-2 and z2=10-x2-y2 using Cartesian coordinates.

I let z1 = z2 and solved this to get the intersection of the two paraboloids which gave y2+x2=4 (Which I can also use as my domain for integration?)

So the volume of the area between them would be the double integral of z2-z1 dA (where dA = dxdy)

x goes from -(4-y2)1/2 to (4-y2)1/2 and y goes from -2 to 2.

so integrating z2-z1 with respect to x first and plugging in the terminals (after some algebra which I hope I've done right) condenses to 4(4-y2)3/2 now I need to integrate this with respect to y from -2 to 2.

I don't know how to solve that integral, I've tried parts and looking up tables. If I've stuffed up somewhere or have done this completely wrong any help would be greatly appreciated.

2. Sep 13, 2009

### Dick

To work out the integral start with a trig substitution, like y=2*sin(t). Then use a double angle formula to deal with things like cos^2.

3. Sep 13, 2009

### forty

I end up with 64 times the integral from -(pi/2) to (pi/2) of (cos(t))^4 dt.

Is this right?

*I found the solution to this using trig identities (24pi), but have I ended up with the right integral?*

Last edited: Sep 13, 2009
4. Sep 14, 2009

### HallsofIvy

Staff Emeritus
Or use cylindrical coordinates. $z= 2x^2+ 2y^2- 2$ becomes $z= 2r^2- 2$ and $z= 10- x^2- y^2$ becomes $z= 10- r^2$. Of course, the "differential of area" for the xy-plane is $r dr d\theta$ in cylindrical coordinates so the integral is
$$\int_{\theta= 0}^{2\pi}\int_{r= 0}^2 [(10- r^2)- (2r^2- 2)]drd\theta$$
$$= \int_{\theta= 0}^{2\pi}\int_{r= 0}^2 (12- r^2)r drd\theta$$
$$= 12\int_{\theta= 0}^{2\pi}\int_{r= 0}^2 (r- r^2) drd\theta$$

5. Sep 14, 2009

### forty

I would use cylindrical coordinates usually but the question says explicitly to use Cartesian. And also doing the integration you suggested I get -16pi. How can the volume be negative? (I always feel so bad questioning you!)

Last edited: Sep 14, 2009
6. Sep 14, 2009

### Dick

That looks right to me. Hall's integrand should have been 12-3*r^2. It works out to be the same thing.

Last edited: Sep 14, 2009
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