MHB How do you evaluate the spherical coordinate integral at 244.15.7.24?

[karush]

$\tiny{244 .15.7.24}$
$\textsf{Evaluate the spherical coordinate integral}\\
\begin{align}\displaystyle
DV_{24}&=\int_{0}^{3\pi/4}
\int_{0}^{\pi}
\int_{0}^{1}
\, 5\rho^3 \sin^3 \phi \, d\rho \, d\phi \, d\theta \\
&=\int_{4}^{3\pi/4}
\int_{0}^{\pi} 5\biggr[\frac{p^4}{4}\biggr]_0^1
\sin^3\phi \, d\phi \, d\theta\\
% &=\color{red}{123}
\end{align} $

just seeing if I got this first step ok
no book answer

 

[Ackbach]

Good so far!

 

[karush]

$\displaystyle\int_{0}^{3 \pi/4} \int_{0}^{\pi}
\frac{5}{4}\sin^3 \phi \, d \theta$
so then $u=\cos \phi \therefore du =\sin \phi \, d\theta$
$\displaystyle\int_{0}^{3 \pi/4}\frac{5}{4} \int_{0}^{\pi}
\left(1-u^2\right) \, du \, d\theta$

 

[tkhunny]

Evaluate the spherical coordinate integral

This is a very strange command. Why is it important that the integral came from a 'spherical coordinate' exercise? It's just an integral.

Just one tiny thing. Why did rho turn into p? :-) No harm.

 

[karush]

This is a very strange command. Why is it important that the integral came from a 'spherical coordinate' exercise? It's just an integral.

Just one tiny thing. Why did rho turn into p? :-) No harm.

\rho=$\rho$

it was the books suggestion

 

[Ackbach]

$\displaystyle\int_{0}^{3 \pi/4} \int_{0}^{\pi}
\frac{5}{4}\sin^3 \phi \, d \theta$
so then $u=\cos \phi \therefore du =\sin \phi \, d\theta$
$\displaystyle\int_{0}^{3 \pi/4}\frac{5}{4} \int_{0}^{\pi}
\left(1-u^2\right) \, du \, d\theta$

You should either write
$$\int_{1}^{-1}(1-u^2) \, du \quad \text{or} \quad \int_{u(0)}^{u(\pi)}(1-u^2) \, du.$$
That way, you're less likely to make the category error of plugging in $0$ and $\pi$ as $u$-values.

 

[MarkFL]

...so then $u=\cos \phi \therefore du =\sin \phi \, d\theta$...

If:

$$u=\cos(\phi)$$

then:

$$du=-\sin(\phi)\,d\phi$$

which would result in reversing the limits of integration. :)

 

[tkhunny]

it was the books suggestion

I think you did not understand my question.

Not my question:
Why did you tackle this problem using spherical coordinates? It was the book's suggestion.

My question:
After setting up the integral, why does it matter that it can from a spherical coordinate system? It's just an integral.

 

[MarkFL]

ok I presume we are here
these next steps are !

$\displaystyle\int_{4}^{3\pi/4} \frac{5}{4} \biggr[ u-\frac{u^3}{3}\biggr]_1^{-1} d\theta\\$

how do you plug $$\cos(\phi)$$ after this?

You want to reverse the limits of integration, because of the negative sign in the differential resulting from your $u$-substitution (I don't know why you changed the lower limit on the outermost integral):

$$I=\frac{5}{4}\int_{0}^{\Large\frac{3\pi}{4}}\left[ u-\frac{u^3}{3}\right]_{-1}^{1} d\theta$$

You are done with $\phi$ because of the substitution you made.

 

[karush]

You want to reverse the limits of integration, because of the negative sign in the differential resulting from your $u$-substitution (I don't know why you changed the lower limit on the outermost integral):

$$I=\frac{5}{4}\int_{0}^{\Large\frac{3\pi}{4}}\left[ u-\frac{u^3}{3}\right]_{-1}^{1} d\theta$$

You are done with $\phi$ because of the substitution you made.

So we are down to this? hopefully

$\biggr[\frac{5}{4}\cos\theta\biggr]_{0}^{3\pi/4}$

 

[MarkFL]

So we are down to this? hopefully

$\biggr[\frac{5}{4}\cos\theta\biggr]_{0}^{3\pi/4}$

Where did the $\cos(\theta)$ come from?

 

[karush]

Where did the $\cos(\theta)$ come from?

so is it

$$\biggr[\frac{5}{4}\cos\phi\biggr]_{0}^{3\pi/4}$$

well wasn't there a $d\theta$ ?

 

[MarkFL]

so is it

$$\biggr[\frac{5}{4}\cos\phi\biggr]_{0}^{3\pi/4}$$

well wasn't there a $d\theta$ ?

I'm going to walk through this problem, using your substitution...let's begin with:

$$I=\int_{0}^{\Large\frac{3\pi}{4}}\int_{0}^{\pi}\int_{0}^{1} 5\rho^3\sin^3(\phi)\,d\rho\,d\phi\,d\theta$$

My first step would be to move everything as far to the left as possible:

$$I=5\int_{0}^{\Large\frac{3\pi}{4}}\int_{0}^{\pi} \sin^3(\phi)\int_{0}^{1} \rho^3\,d\rho\,d\phi\,d\theta$$

Now, let's consider the innermost integral:

$$I=5\int_{0}^{\Large\frac{3\pi}{4}}\int_{0}^{\pi} \sin^3(\phi)\left(\int_{0}^{1} \rho^3\,d\rho\right)\,d\phi\,d\theta$$

$$\int_{0}^{1} \rho^3\,d\rho=\frac{1}{4}\left[\rho^4\right]_0^1=\frac{1}{4}\left(1^4-0^4\right)=\frac{1}{4}$$

And so we have:

$$I=5\int_{0}^{\Large\frac{3\pi}{4}}\int_{0}^{\pi} \sin^3(\phi)\left(\frac{1}{4}\right)\,d\phi\,d\theta$$

Move the constant to the far left:

$$I=\frac{5}{4}\int_{0}^{\Large\frac{3\pi}{4}}\int_{0}^{\pi} \sin^3(\phi)\,d\phi\,d\theta$$

Again, let's consider the innermost integral:

$$I=\frac{5}{4}\int_{0}^{\Large\frac{3\pi}{4}}\left(\int_{0}^{\pi} \sin^3(\phi)\,d\phi\right)\,d\theta$$

$$\int_{0}^{\pi} \sin^3(\phi)\,d\phi=\int_{0}^{\pi} \sin^2(\phi)\,\sin(\phi)\,d\phi=\int_{0}^{\pi} \left(1-\cos^2(\phi)\right)\,\sin(\phi)\,d\phi$$

Let:

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

So we have:

$$-\int_{1}^{-1} \left(1-u^2\right)\,du=2\int_{0}^{1} 1-u^2\,du=\frac{2}{3}\left[3u-u^3\right]_0^1=\frac{4}{3}$$

And our integral is now:

$$I=\frac{5}{4}\int_{0}^{\Large\frac{3\pi}{4}}\left(\frac{4}{3}\right)\,d\theta$$

Move the constant to the far left:

$$I=\frac{5}{3}\int_{0}^{\Large\frac{3\pi}{4}}\,d\theta=\frac{5}{3}\left(\frac{3\pi}{4}-0\right)=\frac{5\pi}{4}$$

 

[karush]

how do you get the $$\frac{5}{3}$$ ?

\begin{align*}\displaystyle
&=\frac{5}{4}\int_{4}^{3\pi/4} \int_{1}^{-1}(1-u^2) \, du \, d\theta\\
&=\frac{5}{4}\int_{0}^{3\pi/4} \biggr[ u-\frac{u^3}{3}\biggr]_1^{-1} d\theta\\
&=\frac{5}{4}\int_{0}^{3\pi/4}\frac{4}{3} d\theta\\
&=\frac{5}{4}\biggr[\frac{4}{3}\theta\biggr]_{0}^{3\pi/4}\\
&=\frac{5}{4}\cdot\frac{4}{3}\cdot\frac{3\pi}{4}\\
&=\color{red}{\frac{5\pi}{4}}
\end{align*}

hopefully

 

[MarkFL]

how do you get the $$\frac{5}{3}$$ ?

$$\frac{5}{4}\cdot\frac{4}{3}=\frac{5}{3}$$ :)