for the volume of the lower half…
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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 \(\pi a^3\).
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 \((\frac{1}{4}a,0)\).