Question: A sphere of radius $R$ has total mass $M$ and density function given by $ρ = kr$, where $r$ is the distance a point lies from the centre of the sphere. Give an expression for the constant $k$ in terms of $M$ and $R$.

My Attempt: $ρ$ is defined as density, meaning $ρ=\frac{M}{Volume}$. Volume of a sphere $= \frac{4}{3}\pi R^3$. Therefore, $ρ=\frac{M}{\frac{4}{3}\pi R^3}=\frac{3M}{4\pi R^3}$. Substituting in $ρ = kr$, I am left with $$kr=\frac{3M}{4\pi R^3}$$ and therefore need to find an expression for $r$ in terms of $k$, $M$ and/or $R$, but am unsure how to continue from here.

Please help!

  • 1
    $\begingroup$ density = mass/volume is only true if density is a constant, which it is not. You are told it is a function. $\endgroup$ May 2, 2020 at 8:45
  • $\begingroup$ The sphere in the question is solid, but think of it as being made up of many hollow concentric spherical shells, each shell being $ \delta r$ thick. Each shell has uniform density and for the shell with radius $ r, 0 \leq r \leq R $ the mass is $ \rho(r) \cdot 4\pi r^2 \delta r $. Use calculus to calculate the total mass in terms of $ \rho$ and equate to $ M $. $\endgroup$
    – WA Don
    May 2, 2020 at 8:50

2 Answers 2


Take a shell of infinitesimal thickness $dr$ at a distance $r$ from the centre. The mass of this shell would be $$dM = 4\pi r^2dr \times \rho(r) = 4 \pi r^2 kr dr$$ Then integrate both sides to find the mass as a function of $k$ $$M = \int_0^R 4 \pi r^2 \cdot kr dr = 4\pi k \frac{R^4}{4}$$ Therefore, the value of k is $$\boxed{k = \frac{M}{\pi R^4}}$$

  • $\begingroup$ I'm a bit confused as to how you got that integral? $\endgroup$
    – Viv4660
    May 2, 2020 at 9:03
  • $\begingroup$ I'll edit to add some verbosity. $\endgroup$ May 2, 2020 at 9:03

If we consider a spherical shell with radius $r$ and thickness $dr$, then

enter image description here

Now, to find the mass of this spherical shell

\begin{align} \text{Volume of spherical shell } &= \text{Surface area} \times \text{Thickness} \\ &= 4\pi r^{2} \cdot dr \end{align}

Since $\rho = kr$,

\begin{align} \text{Mass } &= \text{Volume} \times \text{Density} \\ &= 4 \pi k r^{3}dr \end{align}

Now, to find the mass of the sphere, we integrate the above from $r=0$ to $r=R$

\begin{align} M &= \int_{0}^{R} 4 \pi k r^{3} dr \\ &= 4 \pi k \int_{0}^{R} r^{3} dr \\ &= 4 \pi k \left[\frac{r^4}{4} \right]_{0}^{R} \\ &= \pi k R^{4} \end{align}

Thus, an expression for the constant $k$ in terms of $M$ and $R$ is

$$k = \frac{M}{\pi R^{4}}$$


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