Calculate the volume between the surfaces f(x,y)=3-x^2-y^2 and g(x,y)=2x^2+2y^2.

The volume between two surfaces is
V=int int_D (f_(t o p)(x,y)-f_(b o t t o m)(x,y))dA
Where D is the region between the two surfaces projected on the xy-plane.
f(x,y) is a parabolic surface that starts at f(0,0)=3 and then opens downward. g(x,y) is a parabolic surface that starts at g(0,0)=0 and opens upward. Therefore, f(x,y) is above g(x,y) and these surfaces must then cross. We need to find the region that these surfaces enclose and project that region onto the xy-plane to find the region to integrate over. This can be done by solving f(x,y)=g(x,y) .

3-x^2-y^2=2x^2+2y^2
3=3x^2+3y^2
1=x^2+y^2
This is the unit circle. The domain to integrate over is then D: x^2+y^2lt=1 .
Now solve the integral.
V=int int_D (f_(t o p)(x,y)-f_(b o t t o m)(x,y))dA
V=int int_D [(3-x^2-y^2)-(2x^2+2y^2)]dxdy
V=int int_D (3-3x^2-3y^2)dxdy
V=3int int_D [1-(x^2+y^2)]dxdy
Notice that the domain is rather messy to solve in cartesian coordinates, so switch this integral over to polar coordinates. Where r^2=x^2+y^2 , and D: 0lt=rlt=1, 0lt=thetalt=2pi .
V=3int_0^2pi int_0^1 (1-r^2)r dr d(theta)
V=3int_0^(2pi) int_0^1 (r-r^3)dr d(theta)
V=3int_0^(2pi) (1/2r^2-1/4r^4)|_0^1 d(theta)
V=3int_0^(2pi) (1/2-1/4) d(theta)
V=3(2pi)(1/2-1/4)=6pi(1/4)
V=(3pi)/2
http://www.vias.org/calculus/12_multiple_integrals_03_04.html