Double integral to find volume between two surfaces

[toothpaste666]

Homework Statement

set up and evaluate a double integral to find the volume of the solid bounded by the graphs of the equations
y = 4 - x^2
z= 4 - x^2
first octant

The Attempt at a Solution

I am fairly confident in my ability to evaluate double integrals , but I am having a problem figuring out how to set this one up. In the example in my book they give two equations for z and equate them to find the region in the xy plane from this region they find the limits of integration.

I think for this one I have to rewrite the first equation in terms of z?
so z = f(x,y) = y + x^2 - 4 = 0
and then I set this equal to the other equation for z

y + x^2 -4 = 4 - x^2
y -4 = 4 - 2x^2
y = -2x^2 + 8

I am not sure if anything up to here is correct and I don't know where to go from here. can someone give me a hint to help me get started ?

 

[LCKurtz]

No, you are making it too complicated. In the xy plane plot the first quadrant portion of $y=4-x^2$. That region is the base of your volume and you can find the xy limits there. The height of the solid is $z=4-x^2$. Just do a double integral in xy.

 

[toothpaste666]

Oh I think I get it now. This is the integral I came up with:

$\int_0^4 \int_0^{(4-y)^\frac{1}{2}} (4-x^2) dxdy$