Volume of bounded by 2 surfaces

[manenbu]

Homework Statement

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

Homework Equations

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

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:
[tex]x=r\cos{\theta}, y=r\sin{\theta}[/tex]
and the function after shifting it is:
z = 1/2 - x² - y² = 1/2 - r²(sin²θ + cos²θ) = 1/2 - r²
So the integral is:
[tex]
\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 = [/tex]

[tex]\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}[/tex]

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

Was I correct?

 

[mighty2000]

Hello! Yes, your approach and solution are correct. By finding the intersection of the two surfaces and shifting it down to z=0, you have essentially divided the body into two symmetric parts and only need to calculate the volume of one part and multiply by 2. Your use of polar coordinates is also a good choice for this problem. Keep up the good work!