Answer in Calculus for KABIR #206141
- 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.
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\n4(27a^2x)=4(x-2a)^3\n27a^2x=x^3-6ax^2+12a^2x-8a^3\nx^3-6ax^2-15a^2x-8a^3=0\n(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=piint_{-a}^{8a}Big(4ax-frac{4(x-2a)^3}{27a}Big)dx\nquad = piint_{-a}^{8a}Big(frac{108a^2x-4(x-2a)^3}{27a}Big)dx\nquad = frac{4pi}{27a}int_{-a}^{8a}Big(27a^2x-(x-2a)^3Big)dx\nquad = frac{4pi}{27a}int_{-a}^{8a}Big(-x^3+6ax^2+15a^2x+8a^3Big)dx\nquad = frac{4pi}{27a}Big[-frac{x^4}{4}+frac{6ax^3}{3}+frac{15a^2x^2}{2}+8a^3xBig]_{-a}^{8a}\nquad = frac{4pi}{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]\nquad = frac{4pi}{27a}Big(1024a^4-(-frac{11a^4}{4})Big)\nquad = frac{4pi}{27a}Big(1024a^4+frac{11a^4}{4})Big)\nquad = frac{4pi}{27a}Big(frac{4107a^4}{4}Big)\nquad = frac{1369a^3}{9}pi quad cubic ,units\nV= 152.1a^3pi quad cubic ,units\”
Hence the required volume
