Question

Determine the surface area generated by revolving the curve y=1/3 x³ over the interval (0,2), about the x-axis.

Answer 1

Solution

If the curve y = f(x), a≤x≤b is rotated about the x-axis, then the surface area is given by

$A=2\pi\int_a^bf(x)\sqrt{1+[f'(x)]^2}dx$

In this case f(x) = 1/3 x³, f’(x) = x² ,a = 0, b = 2 

$A=2\pi\int_0^2\frac{1}{3}x^3\sqrt{1+x^4}dx$

Substitution y = x⁴ 

$A=\frac{\pi}{6}\int_0^{16}\sqrt{1+y}dy=\frac{\pi}{9}(1+y)^{3/2}|_0^{16}=\frac{\pi}{9}(17^{3/2}-1)=24.118$

Answer

$A=\frac{\pi}{9}(17^{3/2}-1)=24.118$