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

Apr 23, 2018

$\frac{6 \pi}{5}$

#### Explanation:

We can param this:

Let $x = {\cos}^{3} t$ and $y = {\sin}^{3} t$ so that we do indeed have ${x}^{\frac{2}{3}} + {y}^{\frac{2}{3}} = 1$

Then arc lenth along the curve in Q1 is:

$\mathrm{ds} = \dot{s} \setminus \mathrm{dt} = \sqrt{{\dot{x}}^{2} + {\dot{y}}^{2}} \setminus \mathrm{dt}$

$= \sqrt{{\left(- 3 {\cos}^{2} t \sin t\right)}^{2} + {\left(3 {\sin}^{2} t \cos t\right)}^{2}} \setminus \mathrm{dt}$

$= 3 \cos t \sin t \setminus \mathrm{dt}$

The surface area of the revolution around the y-axis is:

$\mathrm{dS} = 2 \pi x \mathrm{ds}$

$= 6 \pi \setminus {\cos}^{4} t \setminus \sin t \setminus \mathrm{dt}$

And:

$S = 6 \pi \setminus {\int}_{t = 0}^{\frac{\pi}{2}} \setminus \mathrm{dt} q \quad {\cos}^{4} t \setminus \sin t$

$= 6 \pi {\left(- \frac{1}{5} {\cos}^{5} t\right)}_{t = 0}^{\frac{\pi}{2}} = \frac{6 \pi}{5}$