# Triple integral

• May 31st 2010, 10:15 AM
Em Yeu Anh
Triple integral
Evaluate the following using cylindrical coordinates:
$\int\int\int_E(x^3+xy^2)dV$
where E is the solid that lies in the first octant beneath $z=16-x^2-y^2$
• May 31st 2010, 10:39 AM
running-gag
Quote:

Originally Posted by Em Yeu Anh
Evaluate the following using cylindrical coordinates:
$\int\int\int_E(x^3+xy^2)dV$
where E is the solid that lies in the first octant beneath $z=16-x^2-y^2$

Hi

$x^2+y^2 =16-z$ means that z varies from 0 to 16

$x^2+y^2 = r^2 = 16-z$ therefore for any value for z between 0 and 16, r varies from 0 to $\sqrt{16-z}$

The first octant means that $\theta$ varies from 0 to $\frac{\pi}{2}$

Then $x^3+xy^2 = r^3 \cos^3 \theta + r \cos \theta r^2 \sin^2 \theta = r^3 \cos \theta$

and $dx dy$ is to be replaced by $r dr d\theta$

$\int\int\int_E(x^3+xy^2)dV = \int_{0}^{16} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\sqrt{16-z}} r^4 \cos \theta dr d\theta dz$
• May 31st 2010, 12:52 PM
Jester
What I might do is

$
\int_0^{\pi/2} \int_0^4 \int_0^{16-r^2} r^4 \cos \theta dz\,dr\, d \theta
$

it would avoid the $\sqrt{\;\;\;}$.