Question

• Find the volume of the solid generated by revolving the area enclosed between the evolute, 27ay^2 = 4(x-2a) ^3 and the parabola, y^2 = 4ax about x-axis.

Answer 1

The given evolute is:

$$27ay^2=4(x-2a)^3 \qquad \cdots(i)$$

And the given parabola is:

$$y^2=4ax \qquad \cdots(ii)$$

To obtain the enclosed area, substitute (ii) in (i):

$27a(4ax)=4(x-2a)^3\\ 4(27a^2x)=4(x-2a)^3\\ 27a^2x=x^3-6ax^2+12a^2x-8a^3\\ x^3-6ax^2-15a^2x-8a^3=0\\ (x-8a)(x+a)^2=0\\$

Thus

$$x=8a, x=-a$$

Therefore, about x-axis, the solid revolved from -a to 8a.

The required volume is thus:

$V=\pi\int_{-a}^{8a}\Big(4ax-\frac{4(x-2a)^3}{27a}\Big)dx\\ \quad = \pi\int_{-a}^{8a}\Big(\frac{108a^2x-4(x-2a)^3}{27a}\Big)dx\\ \quad = \frac{4\pi}{27a}\int_{-a}^{8a}\Big(27a^2x-(x-2a)^3\Big)dx\\ \quad = \frac{4\pi}{27a}\int_{-a}^{8a}\Big(-x^3+6ax^2+15a^2x+8a^3\Big)dx\\ \quad = \frac{4\pi}{27a}\Big[-\frac{x^4}{4}+\frac{6ax^3}{3}+\frac{15a^2x^2}{2}+8a^3x\Big]_{-a}^{8a}\\ \quad = \frac{4\pi}{27a}\Big[(-\frac{4096a^4}{4}+1024a^4+960a^4+64a^3)-(-\frac{a^4}{4}-2a^4+\frac{15a^4}{2}-8a^4)\Big]\\ \quad = \frac{4\pi}{27a}\Big(1024a^4-(-\frac{11a^4}{4})\Big)\\ \quad = \frac{4\pi}{27a}\Big(1024a^4+\frac{11a^4}{4})\Big)\\ \quad = \frac{4\pi}{27a}\Big(\frac{4107a^4}{4}\Big)\\ \quad = \frac{1369a^3}{9}\pi \quad cubic \,units\\ V= 152.1a^3\pi \quad cubic \,units\\$

Hence the required volume