triple integral

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle \iiint\limits_{Q} \cos(x^2 + y^2 + z^2)^{3/2} \ dV\)

where \(\displaystyle Q\) is the unit ball.

 

[khansaheb]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle \iiint\limits_{Q} \cos(x^2 + y^2 + z^2)^{3/2} \ dV\)

where \(\displaystyle Q\) is the unit ball.

Please show us what you have tried and exactly where you are stuck.

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[logistic_guy]

Let us write the integral again.

\(\displaystyle \iiint\limits_{Q} \cos(x^2 + y^2 + z^2)^{3/2} \ dV\)

The first thing that I will do is to change \(\displaystyle dV\) to the spherical coordinate.

\(\displaystyle \iiint\limits_{Q} \cos(x^2 + y^2 + z^2)^{3/2} \ \rho^2\sin \phi \ d\rho \ d\phi \ d\theta\)

A spherical ball is:

\(\displaystyle x^2 + y^2 + z^2 = \rho^2\)

Then,

\(\displaystyle \iiint\limits_{Q} \cos(\rho^3) \ \rho^2\sin \phi \ d\rho \ d\phi \ d\theta\)

Or

\(\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1} \cos(\rho^3) \ \rho^2\sin \phi \ d\rho \ d\phi \ d\theta\)

\(\displaystyle u = \rho^3\)
\(\displaystyle du = 3\rho^2 \ d\rho\)

\(\displaystyle \frac{1}{3}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{1} \cos u \sin \phi \ du \ d\phi \ d\theta\)


\(\displaystyle \frac{1}{3}\sin 1\int_{0}^{2\pi}\int_{0}^{\pi} \sin \phi \ d\phi \ d\theta\)


\(\displaystyle \frac{2}{3}\sin 1\int_{0}^{2\pi} d\theta = \frac{4\pi}{3}\sin 1 = \textcolor{blue}{3.5247}\)