# [SOLVED]Triple integral

#### dwsmith

##### Well-known member
Anyone have any smooth ideas for this triple integral?
$$-\rho Gm\iiint\frac{r'^2\sin\theta}{\sqrt{R^2 + r'^2 - 2Rr'\cos\theta}}d\theta d\rho dr'$$
where $0<r'<a$, $0<\theta<\pi$, and $0<\rho<2\pi$.
The $\rho$ out front is constant.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: triple integral

Anyone have any smooth ideas for this triple integral?
$$-\rho Gm\iiint\frac{r'^2\sin\theta}{\sqrt{R^2 + r'^2 - 2Rr'\cos\theta}}d\theta d\rho dr'$$
where $0<r'<a$, $0<\theta<\pi$, and $0<\rho<2\pi$.
The $\rho$ out front is constant.
Integrating for $\rho$ (shouldn't that be $\phi$?) is simply multiplying by $2\pi$.
Then integrate with respect to $\theta$.
To do so, consider what $\dfrac{\partial}{\partial \theta}\sqrt{R^2 + r'^2 - 2Rr'\cos\theta}$ is.
You'll be left with a pretty simple expression to integrate with respect to r'.

Finally you'll end up with the gravitational energy between a mass $m$ and a sphere (with radius $a$ and density $\rho$) at a distance R.

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#### dwsmith

##### Well-known member
Re: triple integral

Integrating for $\rho$ (shouldn't that be $\phi$?) is simply multiplying by $2\pi$.
Then integrate with respect to $\theta$.
To do so, consider what $\dfrac{\partial}{\partial \theta}\sqrt{R^2 + r'^2 - 2Rr'\cos\theta}$ is.
You'll be left with a pretty simple expression to integrate with respect to r'.

Finally you'll end up with the gravitational energy between a mass $m$ and a sphere (with radius $a$ and density $\rho$) at a distance R.
Yes, it should be $\phi$.

#### dwsmith

##### Well-known member
Re: triple integral

I keep ending up with
\begin{alignat*}{3}
V & = & -2\pi\rho Gm\int_0^a\left[\sqrt{1 +\left(\frac{r'}{R}\right)^2-2\frac{r'}{R}\cos\theta}\right|_0^{\pi}dr\\
& = & -2\pi\rho Gm\int_0^a\left[\sqrt{\frac{(r' + R)^2}{R^2}} - \sqrt{\frac{(r' - R)^2}{R^2}}\right]dr\\
& = & -\frac{2\pi\rho Gm}{R}\int_0^a[r' + R - r' + R]dr\\
& = & -4\pi\rho Gm\int_0^a dr\\
& = & -4\pi\rho Gma
\end{alignat*}
Then $\rho = \frac{M}{V}$ so $V = -\frac{4\pi GMm a}{V}$ but I need to conclude $V = -\frac{GMm}{R}$.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: triple integral

I keep ending up with
\begin{alignat*}{3}
V & = & -2\pi\rho Gm\int_0^a\left[\sqrt{1 +\left(\frac{r'}{R}\right)^2-2\frac{r'}{R}\cos\theta}\right|_0^{\pi}dr\\
& = & -2\pi\rho Gm\int_0^a\left[\sqrt{\frac{(r' + R)^2}{R^2}} - \sqrt{\frac{(r' - R)^2}{R^2}}\right]dr\\
& = & -\frac{2\pi\rho Gm}{R}\int_0^a[r' + R - r' + R]dr\\
& = & -4\pi\rho Gm\int_0^a dr\\
& = & -4\pi\rho Gma
\end{alignat*}
Then $\rho = \frac{M}{V}$ so $V = -\frac{4\pi GMm a}{V}$ but I need to conclude $V = -\frac{GMm}{R}$.
Assuming you did the integral right ,it is impossible to get rid of (a) unless you have a value for it. May I ask what is r' ?

#### dwsmith

##### Well-known member
Re: triple integral

Assuming you did the integral right ,it is impossible to get rid of (a) unless you have a value for it. May I ask what is r' ?
r' is r

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: triple integral

Notice $$\displaystyle \sqrt{(r'-R)^2}\not =(r'-R)$$ unless r'>R

#### dwsmith

##### Well-known member
Re: triple integral

Here is what I am working on. It is on page 1. Maybe there is something you will see that will help.
View attachment 230 Notes.pdf

#### dwsmith

##### Well-known member
Re: triple integral

Notice $$\displaystyle \sqrt{(r'-R)^2}\not =(r'-R)$$ unless r'>R
How can I handle that then?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: triple integral

Can we deduce from the figure that R>r' ?

#### dwsmith

##### Well-known member
Re: triple integral

Can we deduce from the figure that R>r' ?
No because the differential volume can be anywhere. I just drew so that r' > R. As the distance between the planets go to infinity, $r'\to R$ but even doing that only yields:
$$-\frac{Gma\pi(a+2R)\rho}{R}$$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: triple integral

Ok,what do you mean by a ? Is it the radius of M ?

#### dwsmith

##### Well-known member
Re: triple integral

Ok,what do you mean by a ? Is it the radius of M ?
Yes

#### dwsmith

##### Well-known member
Re: triple integral

\begin{alignat*}{3}
V & = & -\frac{2\pi\rho Gm}{R}\int_0^a r'\left[r' + R - \sqrt{(r' - R)^2}\right]dr'\\
& = & \frac{2\pi\rho Gm}{R}\int_0^a 2r'^2dr'\\
& = & -\frac{4Gma^3\rho\pi}{3R}\\
& = & -\frac{GMm}{R}
\end{alignat*}
Since $\rho = \frac{M}{V}$ where $V = \frac{4\pi r^3}{3}$, we have the desired result.

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#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: triple integral

Excellent , so you are choosing that R>r' , but what is the meaning of that ? Are you raking some unnecessary restrictions ?

#### dwsmith

##### Well-known member
Re: triple integral

Excellent , so you are choosing that R>r' , but what is the meaning of that ? Are you raking some unnecessary restrictions ?
If we let $R\to\infty$, then $r'\to R$.
So I took r'=R

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: triple integral

If we let $R\to\infty$, then $r'\to R$.
So I took r'=R
That doesn't look quite correct.
If we let $R\to\infty$, then $r' \le a \ll R$.

But really, there is no need to do this.
Your formula works for any $R$ with $r' \le a \le R$.

Actually, the additionally interesting case is when $R < a$.
You can also calculate what happens then.
It should come out that $V = -\frac 4 3 \pi R^3 \rho \cdot \frac {Gm }{R}$.
That is, the attractive force when inside a solid sphere is completely determined by the subsphere between you and the center.
The forces by the part of the sphere that is on the outside cancel out.

#### dwsmith

##### Well-known member
Re: triple integral

That doesn't look quite correct.
If we let $R\to\infty$, then $r' \le a \ll R$.

But really, there is no need to do this.
Your formula works for any $R$ with $r' \le a \le R$.

Actually, the additionally interesting case is when $R < a$.
You can also calculate what happens then.
It should come out that $V = -\frac 4 3 \pi R^3 \rho \cdot \frac {Gm }{R}$.
That is, the attractive force when inside a solid sphere is completely determined by the subsphere between you and the center.
The forces by the part of the sphere that is on the outside cancel out.
How do I go from the first integral to the second then?
\begin{alignat*}{3}
V & = & -\frac{2\pi\rho Gm}{R}\int_0^a r'\left[r' + R - \sqrt{(r' - R)^2}\right]dr'\\
& = & \frac{2\pi\rho Gm}{R}\int_0^a 2r'^2dr'
\end{alignat*}
Since $0 < r' < a < R$, we can write $\sqrt{(r' - R)^2} = \sqrt{(R - r')^2} = R - r'$??

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: triple integral

How do I go from the first integral to the second then?
\begin{alignat*}{3}
V & = & -\frac{2\pi\rho Gm}{R}\int_0^a r'\left[r' + R - \sqrt{(r' - R)^2}\right]dr'\\
& = & \frac{2\pi\rho Gm}{R}\int_0^a 2r'^2dr'
\end{alignat*}
If $R \ge r'$, then $\sqrt{(r' - R)^2} = |r' - R| = (R - r')$.
Substitute?

Since $0 < r' < a < R$, we can write $\sqrt{(r' - R)^2} = \sqrt{(R - r')^2} = R - r'$??

Yes!