Answer to Question #170294 in Calculus for Kermen Kaur Parihar

Question

Find the area of the surface generated by revolving the curve

y

=

5

√

x

,

y=5x,

       2

≤

x

≤

4

2≤x≤4

around the x

x

-axis.

Answer 1

$y\ =\ 5\sqrt x\$

$\frac{dy}{dx}=\frac{5}{2\sqrt x}$

$\left(\frac{dy}{dx}\right)^2=\frac{25}{4x}$

$\int_{2}^{4}{2\pi y\ ds}=\ \int_{2}^{4}{2\pi5\sqrt x\sqrt{1+\ \frac{25}{4x}}dx}$

$10\pi\int_{2}^{4}{\sqrt{x+\ \frac{25}{4}}dx}$

$10\pi\ \left(x+\frac{25}{4}\right)^\frac{3}{2}\ast\frac{2}{3}$

$\frac{20\pi}{3}\left\{\left(\frac{41}{4}\right)^\frac{3}{2}-\left(\frac{33}{4}\right)^\frac{3}{2}\right\}$

$60.79794\pi$