A uniform solid right circular cylinder of radius \(a\) and height \(2a\) stands on a horizontal plane. Rectangular co-ordinate axes \(Ox\), \(Oy\) are taken in the plane of the base, with \(O\) at the centre of the base. The cylinder is cut into two equal parts by the plane which passes through the line \(x = -a\) and makes an angle of \(45^\circ\) with the horizontal. Justify the formula \[\begin{equation*} \int_{-a}^a 2(x+a) \sqrt{a^2 - x^2} \:dx \end{equation*}\]

for the volume of the lower half…

The volume of the lower half can be thought of as the sum of the volumes of thin vertical slices parallel to the \(y\)-axis.

Each slice has a shape approximating a cuboid, with thickness \(\delta x\).

To find the height of a slice we look at the front view of the cylinder. The slice at distance \(x\) from the origin has height \(z=x+a\).

For the width of a slice, we look at the plan view. The base is a circle of radius \(a\) and Pythagoras’ theorem tells us that \(x^2+y^2=a^2\). The width of the slice is twice the \(y\) value, namely \(2\sqrt{a^2 - x^2}\).

The volume of each slice, then, is \[\begin{equation*} 2\sqrt{a^2 - x^2} (x+a)\delta x, \end{equation*}\] and so the total volume of the bottom half (as in the limit, the sum becomes an integral) is \[\begin{equation*} \int_{-a}^a 2\sqrt{a^2 - x^2} (x+a) \:dx, \end{equation*}\]

as required.

…and verify by calculation of the integral that the volume is \(\pi a^3\).

Take \(x = a \sin \theta\), so that \(dx = a \cos \theta\: d\theta\). Then \[\begin{align*} V = \int_{-a}^a 2\sqrt{a^2 - x^2} (x+a) \:dx &= \int_{-\pi/2}^{\pi/2} 2 \sqrt{a^2 - a^2 \sin^2 \theta} (a \sin \theta + a) a \cos \theta \:d\theta \\ &= 2a^3\int_{-\pi/2}^{\pi/2} \cos \theta (\sin \theta + 1) \cos \theta \:d\theta \\ &= 2a^3\int_{-\pi/2}^{\pi/2} \cos^2 \theta \sin \theta + \cos^2 \theta \:d\theta . \end{align*}\] The first term can be integrated by substituting \(u=\cos\theta\), \(du=-\sin\theta \:d\theta\). The second term can be expanded using a trig identity. \[\begin{align*} V &= 2a^3\left\{\int_{0}^{0} -u^2 \: du + \int_{-\pi/2}^{\pi/2} \dfrac{\cos 2\theta+1}{2} \:d\theta \right\} \\ &= 2a^3\left\{ 0 + \left[\dfrac{\sin 2\theta}{4} +\dfrac{\theta}{2}\right]_{-\pi/2}^{\pi/2} \right\}\\ &= 2a^3\left\{ 0+0+\dfrac{\pi}{2} \right\}\\ &= \pi a^3 \end{align*}\]

Prove that the centre of mass of this lower half is vertically above the point \((\tfrac{1}{4}a,0)\).

Since the object has reflective symmetry in the \(x-z\) plane, the centre of mass must lie in this plane, i.e. the plane \(y = 0\).

The \(x\)-coordinate, \(x_0\), of the centre of mass must be such that the mass would balance around it. Taking moments we have that \[\begin{equation*} \int_{-a}^a x \times 2\sqrt{a^2 - x^2} (x+a) \:dx = \pi a^3 \times x_0. \end{equation*}\]

To find this integral we’ll use the substitution \(x = a \sin \theta\) again and trig identities for \(\sin 2\theta\) and \(\cos 2\theta\). We have

\[\begin{align*} \int_{-a}^a x \times 2 \sqrt{a^2 - x^2} (x+a) \:dx &= \int_{-\pi/2}^{\pi/2} a\sin \theta \times 2 \sqrt{a^2 - a^2 \sin^2 \theta} (a \sin \theta + a) a \cos \theta \:d\theta \\ &= 2a^4\int_{-\pi/2}^{\pi/2} \sin \theta \cos \theta ( \sin \theta + 1) \cos \theta \:d\theta \\ &= 2a^4\int_{-\pi/2}^{\pi/2} \cos^2 \theta \sin^2 \theta + \cos^2 \theta\sin \theta \:d\theta \\ &= 2a^4\int_{-\pi/2}^{\pi/2} \dfrac{\sin^2 2\theta}{4} + \cos^2 \theta\sin \theta \:d\theta \\ &= 2a^4\int_{-\pi/2}^{\pi/2} \dfrac{1-\cos 4\theta}{8} + \cos^2 \theta\sin \theta \:d\theta \\ &= 2a^4\left[\dfrac{\theta}{8}-\dfrac {\sin 4\theta}{32} - \dfrac{\cos^3 \theta}{3}\right]_{-\pi/2}^{\pi/2} \\ &= \dfrac{\pi a^4}{4}. \end{align*}\]

Thus \(\dfrac{\pi a^4}{4} = \pi a^3 x_0 \implies x_0 = \dfrac{a}{4},\) as required, and so the centre of mass of the lower half lies directly above the point \((\frac{1}{4}a,0)\).