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 \(\tfrac{1}{3} \pi ab^2k(3-k^2)\).
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.
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.
