What is the average density of this spherical planet?

What is the average density of this spherical planet?

Question

A spherical planet of radius $a$ has a variable density $f(r)$ which depends only on the distance $r$ from the planet’s centre. Show that the average density of the planet is $3 \int_0^1 t^2 f(at) \:dt.$

Find the average density correct to two significant figures in each of the three cases:

1. $f(r) = \exp\left[\left(-\dfrac{r}{a}\right)^3\right]$, where $\exp(x)$ denotes $e^x$,

---

1. $f(r) = \exp\left(-\dfrac{r}{a}\right)$,

---

1. $f(r) = \dfrac{a^2r}{(a+r)^3}$.

Suggestion

The tricky part is going to be finding the total mass of the planet. Since the density depends only on $r$ we should think about the mass of a thin hollow spherical shell.

We will need to use the result from the first part.

We will need to find a tool to help us evaluate the integral we get this time.

This time the integral involves an awkward looking fraction. Again, we’ll need to find an appropriate tool to help.

Solution

Let’s consider the hollow sphere of radius $r$, where $r < a$, and thickness $\delta r$, as shown below.

The volume of this hollow sphere is $\delta V = \dfrac{4\pi}{3}(r + \delta r)^3 - \dfrac{4\pi}{3}r^3= 4\pi r^2 \delta r$ (neglecting higher powers of $\delta r$).

Alternatively, $\delta V =$ surface area of the hollow sphere$\times$ its thickness.

We know that the density of this hollow sphere is $f(r)$, so the mass of the hollow sphere is $4\pi r^2 f(r) \delta r$.

Taking the limit, the total mass of the whole sphere is therefore

$$\int_0^a 4\pi r^2 f(r) \:dr.$$

We know that the volume of the whole sphere is $\dfrac{4\pi}{3}a^3$, so the average density of the whole sphere is $\dfrac{\int_0^a 4\pi r^2 f(r) \:dr}{\frac{4\pi}{3}a^3}$ which is $\int_0^a \dfrac{3 r^2 f(r)}{a^3} \:dr.$

If we now put $r = at$, we have $dr = a\: dt$, and substituting in, we find that the average density is $3 \int_0^1 t^2 f(at) \:dt,$ as required.

Our work from the first part tells us that we need to find the following integral, which we can do by inspection or by substitution.

$$\begin{align*} \int_0^1 3t^2 e^{-t^3} \:dt &=\left[-e^{-t^3}\right]_0^1\\ &=1-e^{-1}\\ &=0.63 \quad (2\text{sf}). \end{align*}$$

Using the first part again tells us that we need to find the following. This time we use integration by parts twice.

$$\begin{align*} \int_0^1 3t^2 e^{-t} \:dt &= 3\Big[-t^2e^{-t}\Big]_0^1 - 3\int_0^1 -e^{-t}2t \:dt\\ &=-3e^{-1} +6\int_0^1te^{-t}\:dt \\ &=-3e^{-1} +6\left\{ \Big[-te^{-t}\Big]_0^1 - \int_0^1-e^{-t}\:dt \right\}\\ &=-9e^{-1} -6\Big[e^{-t}\Big]_0^1 \\ &=-15e^{-1} + 6\\ &= 0.48 \quad (2\text{sf}). \end{align*}$$

Once again we can use the first part so that we need to find $\int_0^1 \dfrac{3t^3}{(1+t)^3} \:dt.$

Let’s use

partial fractions

to write

$\dfrac{t^3}{(1+t)^3}$

as

$A + \dfrac{B}{1+t} + \dfrac{C}{(1+t)^2} + \dfrac{D}{(1+t)^3}$

. Putting all this over a common denominator gives

$$\begin{align*} \dfrac{t^3}{(1+t)^3} &= \dfrac{A(1+t)^3+B(1+t)^2+C(1+t)+D}{(1+t)^3} \\ &= \dfrac{At^3+(3A+B)t^2+(3A+2B+C)t+(A+B+C+D)}{(1+t)^3}, \end{align*}$$

which gives us on comparing coefficients

$A = 1$

,

$B = -3$

,

$C = 3$

,

$D = -1$

. So our integral becomes

$$\begin{align*} &3\int_0^1 1 -\dfrac{3}{1+t} + \dfrac{3}{(1+t)^2}-\dfrac{1}{(1+t)^3} \:dt\\ &= 3\left[t-3\ln(1+t)-3\dfrac{1}{1+t} +\dfrac{1}{2(1+t)^2}\right]_0^1\\ &= 3\left(1-3\ln2-\dfrac{3}{2} + \dfrac{1}{8}\right)-3\left(-3+\dfrac{1}{2}\right)\\ &= 3(2.125-3\ln2) \\ &= 0.14 \quad (2\text{sf}). \end{align*}$$

Alternatively, rather than using partial fractions, we could write $u=1+t$ so that our integral becomes $3\int_1^2 \frac{(u-1)^3}{u^3}\:du$ which we can expand, cancel and integrate as powers of $u$ with the same result as above.