Evaluate the following using cylindrical coordinates:

$\displaystyle \int\int\int_E(x^3+xy^2)dV$

where E is the solid that lies in the first octant beneath $\displaystyle z=16-x^2-y^2$

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- May 31st 2010, 10:15 AMEm Yeu AnhTriple integral
Evaluate the following using cylindrical coordinates:

$\displaystyle \int\int\int_E(x^3+xy^2)dV$

where E is the solid that lies in the first octant beneath $\displaystyle z=16-x^2-y^2$ - May 31st 2010, 10:39 AMrunning-gag
Hi

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

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

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

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

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

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

$\displaystyle

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

$

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