Answer to Question #326261 in Calculus for Kishore

Question

find the surface area of the object obtained by rotating y=4+3x^2, 1<=x<=2 about the y axis

Answer 1

If the continuous curve is described by the function y = f(x), a ≤ x ≤ b, then the area is equal to integral

$A=2\pi\int_{a}^{b}{x\sqrt{1+\left(\frac{dy}{dx}\right)^2}dx}$

For given function

$A=2\pi\int_{1}^{2}{x\sqrt{1+\left(6x\right)^2}dx}=\frac{\pi}{36}\int_{1}^{2}{2\left(6x\right)\sqrt{1+\left(6x\right)^2}d\left(6x\right)}$

Substitution t=6x:

$A=\frac{\pi}{36}\int_{6}^{12}{2t\sqrt{1+t^2}dt}$

Substitution  u=t²:

$A=\frac{\pi}{36}\int_{36}^{144}{\sqrt{1+u}du}=\frac{\pi}{36}\frac{2}{3}\left(\sqrt{1+u}\right)^3\left|\begin{matrix}144\\36\\\end{matrix}=\frac{\pi}{54}\left(145\sqrt{145}-37\sqrt{37}\right)\right.=88.486$

Answer

A = 88.486