Evaluate the spherical coordinate integrals

[karush]

$\textsf{Evaluate the spherical coordinate integrals}$
\begin{align*}\displaystyle
DV_{22}&=\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{2}
\, (\rho \cos \phi) \rho^2 \sin \phi
\, d\rho \, d\phi \, d\theta \\
%&=\color{red}{abc}
\end{align*}
so then next ?
\begin{align*}\displaystyle
DV_{22}&=2\pi\int_{0}^{\pi/4}\int_{0}^{2}
\, (\rho \cos \phi) \rho^2 \sin \phi
\, d\rho \, d\phi \,
\end{align*}

no book answer

 

[HOI]

\(\displaystyle DV_{22}= 2\pi \left(\int_0^{\pi/4} sin(\phi)cos(\phi) d\phi\right)\left(\int_0^2 \rho^3 d\rho\right)\)

Both of those integrals should be easy.

 

[karush]

\(\displaystyle DV_{22}= 2\pi \left(\int_0^{\pi/4} sin(\phi)cos(\phi) d\phi\right)\left(\int_0^2 \rho^3 d\rho\right)\)

Both of those integrals should be easy.

are you suggesting multiplication

well here is some of it\begin{align*}
&\displaystyle\int_{0}^{2}\rho^3 \, dp
=\left[\frac{\rho^4}{4} \right]_{0}^{2}
=4
\end{align*}

\begin{align*}\displaystyle
&\int_0^{\pi/4} \sin(\phi)\cos(\phi) \, d\phi\\
=&\frac{1}{2}\int_0^{\pi/4}
2\sin(\phi)\cos(\phi)\, d\phi\\
=&\frac{1}{2}\int_0^{\pi/4}
\sin(2\phi)\, d\phi\\
=&\frac{1}{2}\biggr[2\cos(2\phi) \biggr]_0^{\pi/4}\\
=&4
\end{align*}

 

[MarkFL]

$\textsf{Evaluate the spherical coordinate integrals}$
\begin{align*}\displaystyle
DV_{22}&=\int_{0}^{2\pi}\int_{0}^{\pi/4}\int_{0}^{2}
\, (\rho \cos \phi) \rho^2 \sin \phi
\, d\rho \, d\phi \, d\theta \\
%&=\color{red}{abc}
\end{align*}
so then next ?
\begin{align*}\displaystyle
DV_{22}&=2\pi\int_{0}^{\pi/4}\int_{0}^{2}
\, (\rho \cos \phi) \rho^2 \sin \phi
\, d\rho \, d\phi \,
\end{align*}

no book answer

I would begin by writing:

\(\displaystyle I=\int_0^{2\pi}\int_0^{\Large\frac{\pi}{4}} \sin(\phi)\cos(\phi)\int_0^2 \rho^3\,d\rho\,d\phi\,d\theta\)

Next, work your way from the inside out:

\(\displaystyle I=4\int_0^{2\pi}\int_0^{\Large\frac{\pi}{4}} \sin(\phi)\cos(\phi)\,d\phi\,d\theta=4\int_0^{2\pi}\int_0^{\Large\frac{1}{\sqrt{2}}} u\,du\,d\theta\)

\(\displaystyle I=\int_0^{2\pi}\,d\theta=2\pi\)

 

[karush]

where does the \(\displaystyle 4\) come from?what is u ?

 

[MarkFL]

where does the \(\displaystyle 4\) come from?

It came from the evaluation of the innermost integral.

what is u ?

\(\displaystyle u=\sin(\phi)\implies du=\cos(\phi)\,d\phi\)

 

[karush]

It came from the evaluation of the innermost integral.
\(\displaystyle u=\sin(\phi)\implies du=\cos(\phi)\,d\phi\)

so then we have

$$8\pi\int_{0}^{\sqrt{2}/2} u\,du$$

 

[MarkFL]

so then we have

$$8\pi\int_{0}^{\sqrt{2}/2} u\,du$$

You could look at it that way, but I prefer to work from the inside out. :)

 

[karush]

like this?

\(\displaystyle

\begin{align*}\displaystyle &4\int_0^{2\pi}\int_0^{\Large\frac{1}{\sqrt{2}}} u\,du\,d\theta\\
=&4\int_0^{2\pi}
\left[\frac{u^2}{2} \right]_0^{\sqrt{2}/2}
d\theta
\end{align*}\)