# Volume of bounded by 2 surfaces

1. Sep 19, 2009

### manenbu

1. The problem statement, all variables and given/known data

I need to find the volume of the body bounded by the following surfaces:
z = x2 + y2
z = 1 - x2 - y2

2. Relevant equations

Volume of a body between z=o and the upper surface:
$$\iint_{D} z(x,y) dA$$

3. The attempt at a solution

Ok, this is something I need to do with a double integral, not a triple integral.
So those are 2 symmetric paraboloids, creating in between a body shaped like a lens.
Since I need to find the volume of a body from z=o to the surface, I thought about this.
The 2 surfaces are symmetrical, so if I take their intersection (which is a circle radius 2-1/2 at z=1/2) and shift it down to z=0, I can calculate the volume of the upper half of the solid and then multiply by 2.
I'll need it in polar coordinates of course, so:
$$x=r\cos{\theta}, y=r\sin{\theta}$$
and the function after shifting it is:
z = 1/2 - x2 - y2 = 1/2 - r2(sin2θ + cos2θ) = 1/2 - r2
So the integral is:
$$\int_{0}^{2\pi} \int_{0}^{\frac{1}{\sqrt{2}}} (\frac{1}{2} - r^2) r dr d\theta = \int_{0}^{2\pi} \int_{0}^{\frac{1}{\sqrt{2}}} (\frac{r}{2} - r^3) dr d\theta =$$

$$\frac{1}{4} \int_{0}^{2\pi} (r^2-r^4)_{0}^{\frac{1}{\sqrt{2}}} d\theta = \frac{1}{8} \int_{0}^{2\pi} d\theta = \frac{\pi}{4}$$

And since it was the upper half, I need to multiply by 2 to get the entire volume, it being $\frac{\pi}{2}$.

Was I correct?