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

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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)` .

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`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`