# How do you find the area of the surface generated by rotating the curve about the y-axis y=1/3(x^2+2)^(3/2), 1<=x<=2?

Apr 18, 2017

The surface area of a curve $y$ rotated around the $x$-axis on $a < = x < = b$ is given by:

$S = 2 \pi {\int}_{a}^{b} y \sqrt{1 + {\left(y '\right)}^{2}} \mathrm{dx}$

Where $y = \frac{1}{3} {\left({x}^{2} + 2\right)}^{\frac{3}{2}}$ we see that $y ' = \frac{1}{3} \left(\frac{3}{2} {\left({x}^{2} + 2\right)}^{\frac{1}{2}}\right) \left(2 x\right) = x \sqrt{{x}^{2} + 2}$.

Then the surface area is:

$S = 2 \pi {\int}_{1}^{2} \frac{1}{3} {\left({x}^{2} + 2\right)}^{\frac{3}{2}} \sqrt{1 + {\left(x \sqrt{{x}^{2} + 2}\right)}^{2}} \mathrm{dx}$

$S = \frac{2 \pi}{3} {\int}_{1}^{2} {\left({x}^{2} + 2\right)}^{\frac{3}{2}} \sqrt{1 + {x}^{2} \left({x}^{2} + 2\right)} \mathrm{dx}$

Note that $\sqrt{1 + {x}^{2} \left({x}^{2} + 2\right)} = \sqrt{{x}^{2} + 2 {x}^{2} + 1} = \sqrt{{\left({x}^{2} + 1\right)}^{2}} = {x}^{2} + 1$:

$S = \frac{2 \pi}{3} {\int}_{1}^{2} {\left({x}^{2} + 2\right)}^{\frac{3}{2}} \left({x}^{2} + 1\right) \mathrm{dx} \approx 69.9055$