MHB Evaluate the spherical coordinate integrals

karush
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$\textsf{Evaluate the spherical coordinate integrals}$
\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:mad:
 
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[math]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)[/math]

Both of those integrals should be easy.
 
Country Boy said:
[math]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)[/math]

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*}
 
Last edited:
karush said:
$\textsf{Evaluate the spherical coordinate integrals}$
\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:mad:

I would begin by writing:

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

Next, work your way from the inside out:

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

$$I=\int_0^{2\pi}\,d\theta=2\pi$$
 
where does the $$4$$ come from?what is u ?
 
karush said:
where does the $$4$$ come from?

It came from the evaluation of the innermost integral.

karush said:
what is u ?

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

so then we have

$$8\pi\int_{0}^{\sqrt{2}/2} u\,du$$
 
karush said:
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. :)
 
like this?

$$

\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*}$$
 
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