Question

Determine the area of the surface generated by revolving a given curve about a given axis.

a.) y=x³ on (0,1), about the x-axis.

b. ) y=x² on (0,√2), about the y-axis.

Answer 1

a)

$y=x^3$

$\frac{dy}{dx}=3x^2$

$Surface\ area\ of\ revolution$

$2\pi\int_{0}^{1}{x^3\sqrt{1+\left(3x^2\right)^2}\ dx}$

$let\ u=1+9x^4$

$du=36x^3\ dx$

$2\pi\int{\sqrt u\frac{du}{36}}$

$\frac{\pi}{18}\left(\frac{2}{3}\right)\left(u\right)^\frac{3}{2}$

$\frac{\pi}{27}\left(1+9x^4\right)^\frac{3}{2}\ \ \ \ \ \left(0,\ 1\right)$

$\frac{\pi}{27}\left(\left(10\right)^\frac{3}{2}-1\right)$

$1.13412\pi$

b)

$y=x^2$

$x=\ \sqrt y$

$\frac{dx}{dy}=\frac{1}{2\sqrt y}$

$Surface\ are\ of\ revolution$

$2\pi\int_{0}^{\sqrt2}{\sqrt y\sqrt{1+\left(\frac{1}{2\sqrt y}\right)^2}\ dy}$

$2\pi\int_{0}^{\sqrt2}{\sqrt{y+\frac{1}{4}}\ dy}$

$2\pi\ \left(\frac{2}{3}\right)\left(y+\frac{1}{4}\right)^\frac{3}{2}\ (0,\ \sqrt2)$

$\frac{4\pi\ }{3}\left({\ \left(\sqrt2+\frac{1}{4}\right)}^\frac{3}{2}-\left(\frac{1}{4}\right)^\frac{3}{2}\right)$

2.6958\pi\