# Thread: Triple integral

1. ## Triple integral

Please, I need some help to evaluate integral:
$\displaystyle \int \frac {z^2}{(x^2+y^2+z^2)^\frac{3}{2}} dxdydz$
in area that is bounded with
$\displaystyle 4z \geq x^2+y^2+z^2+(8/3)$
$\displaystyle z\geq\sqrt{3x^2+3y^2}$
Thanks in advance.

2. Originally Posted by TCKony
Please, I need some help to evaluate integral:
$\displaystyle \int \frac {z^2}{(x^2+y^2+z^2)^\frac{3}{2}} dxdydz$
in area that is bounded with
$\displaystyle 4z \geq x^2+y^2+z^2+(8/3)$
$\displaystyle z\geq\sqrt{3x^2+3y^2}$
Thanks in advance.
Have you drawn the region?

3. $\displaystyle 4z= x^2+ y^2+ z^2+ (8/3)$ is the same as $\displaystyle x^2+ y^2+ z^2- 4z+ 4- 4+ 8/3= x^2+ y^2+ (z- 2)^2- (4/3)= 0$, a sphere with center at (0, 0, 2) and radius $\displaystyle 2/\sqrt{3}$. $\displaystyle 4zk\ge x^2+ y^2+ z^2+ (8/3)$ is the same as $\displaystyle x^2+ y^2+ z^2- 4z+ 4- 4+ 8/3= x^2+ y^2+ (z- 2)^2\le (4/3)$, the region on and inside that sphere. $\displaystyle z= \sqrt{3x^2+ 3y^2}= \sqrt{3}\sqrt{3}$ is the same as $\displaystyle z= \sqrt{3}r$ where r is the radius is the xy-plane in any direction. In other words, it is the upper part of a cone- the line $\displaystyle z= \sqrt{3}x$ rotated around the z- axis. $\displaystyle z\ge \sqrt{3x^2+3y^2}$ is the region above (interior to) that cone.

4. I have drawn that, but I need the result to check my solution, that is $\displaystyle \frac {\pi*373}{270\sqrt{3}}$.
Thanks.