Volume between two paraboloids.

[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.

 

[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.

 

[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?*

 

[HallsofIvy]

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
$$\int_{\theta= 0}^{2\pi}\int_{r= 0}^2 [(10- r^2)- (2r^2- 2)]drd\theta$$
$$= \int_{\theta= 0}^{2\pi}\int_{r= 0}^2 (12- r^2)r drd\theta$$
$$= 12\int_{\theta= 0}^{2\pi}\int_{r= 0}^2 (r- r^2) drd\theta$$

 

[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!)

 

[Dick]

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?*

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