# How do you find the surface area of the solid obtained by rotating about the y-axis the region bounded by y=1-x^2 on the interval 0<=x<=1 ?

Oct 9, 2014

The surface area A of the solid obtained by rotating about the $y$-axis the region under the graph of $y = f \left(x\right)$ from $x = a$ to $b$ can be found by

$A = 2 \pi {\int}_{a}^{b} x \sqrt{1 + {\left[f \left(x\right)\right]}^{2}} \mathrm{dx}$.

Let us now look at the posted question.

By the formula above,

$A = 2 \pi {\int}_{0}^{1} x \sqrt{1 + 4 {x}^{2}} \mathrm{dx}$

by rewriting a bit,

$= \frac{\pi}{4} {\int}_{0}^{1} {\left(1 + 4 {x}^{2}\right)}^{\frac{1}{2}} \cdot 8 x \mathrm{dx}$

by General Power Rule,

$= \frac{\pi}{4} {\left[\frac{2}{3} {\left(1 + 4 {x}^{2}\right)}^{\frac{3}{2}}\right]}_{0}^{1} = \frac{\pi}{6} \left({5}^{\frac{3}{2}} - 1\right)$

I hope that this was helpful.