Review question

# Can we find the centre of mass of a solid hemisphere? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5584

## Solution

The region in the first quadrant enclosed by the curve $y = \dfrac{b(a^2 - x^2)^{1/2}}{a}$ and the lines $y = 0$, $x = 0$, $x = ka$ $(0 < k \le 1)$, is rotated through four right angles about the $x$-axis to form a solid of revolution. Show that the volume of the solid is $\tfrac{1}{3} \pi ab^2k(3-k^2)$.

The volume of revolution of a curve around the $x$-axis is given by the expression $\begin{equation*} \int \pi y^2 dx. \end{equation*}$ So the volume we need is \begin{align*} \int_0^{ka} \pi b^2 \frac{a^2 - x^2}{a^2} \:dx = \frac{\pi b^2}{a^2} \left[ a^2x - \frac{x^3}{3} \right]_0^{ka} &= \frac{\pi b^2}{a^2} \left( ka^3 - \frac{k^3a^3}{3} \right) \\ &= \pi b^2 ka \left( 1 - \frac{k^2}{3} \right) \\ &= \frac{1}{3} \pi ab^2k (3-k^2). \end{align*}

If this solid is of uniform density find the coordinates of its centre of mass.

As this solid has rotational symmetry around the $x$-axis, the centre of mass lies on the $x$-axis, let’s say at $(\bar{x},0)$.

The moment of our solid about an axis through $O$ (setting its density to $1$) is therefore $\frac{1}{3}\pi ab^2k (3-k^2) \times \bar{x}.$

This must equal the sum of the moments of the infinitesimal cylinders we added in the first part of the question.

This is (in the limit) \begin{align*} \int_0^{ka} \pi y^2x \:dx &= \int_0^{ka} \pi b^2 \frac{a^2x - x^3}{a^2} \:dx\\ &= \frac{\pi b^2}{a^2} \left[ \dfrac{a^2x^2}{2} - \frac{x^4}{4} \right]_0^{ka}\\ &= \frac{\pi b^2}{a^2} \left( \dfrac{a^4k^2}{2} - \frac{a^4k^4}{4}\right)\\ &= \frac{\pi a^2b^2k^2}{4}(2 -k^2). \end{align*} Thus \begin{align*} \frac{\pi a^2b^2k^2}{4}(2 -k^2)&=\frac{1}{3} \pi ab^2k (3-k^2)\bar{x}\\ \implies \frac{ak}{4}(2 -k^2)&=\frac{1}{3}(3-k^2)\bar{x}\\ \implies 3ak(2 -k^2)&=4(3-k^2)\bar{x}\\ \implies \bar{x} &= \dfrac{3ak(2 -k^2)}{4(3-k^2)}. \end{align*}

By considering the case $b = a$, $k = 1$, show that the centre of mass of a uniform solid hemisphere of radius $a$ is at a distance $\tfrac{3}{8}a$ from the centre.

By taking $b = a$ and $k = 1$, we have that $\begin{equation*} y = \sqrt{a^2 - x^2} \end{equation*}$ and so $\begin{equation*} x^2 + y^2 = a^2. \end{equation*}$

This is the equation of a circle, centred at the origin and with radius $a$.

Our area will rotate to a perfect hemisphere, since our integral’s limits are $0$ and $a$, and so $x=ka$ is tangent to the circle. Moreover, $\begin{equation*} \bar{x} = \frac{3ak(2 - k^2)}{4(3-k^2)} = \frac{3a(2 - 1^2)}{4(3 - 1^2)} = \frac{3a}{8}, \end{equation*}$

and so the centre of mass of a uniform solid hemisphere of radius $a$ is at a distance $\dfrac{3}{8}a$ from the centre.