# Surface area of ellipsoid created by rotation of parametric curve

I have a parametric curve (elipse) defined as follows

\begin{aligned} x(t) &= \cos(t)\\ y(t) &= 2 \sin(t)\end{aligned}

and need to calculate the surface area of the ellipsoid produced by rotating this curve around the $$x$$-axis. I know the formula

$$S=2\pi \int_{\alpha}^{\beta}{[y(t)]\sqrt{x(t)'^2 +y(t)'^2}} \, \mathrm d t$$

I've managed to calculate the squares of derivatives:

$$x(t)'^2 = \sin^2(t)$$

$$y(t)'^2 = 4\cos^2(t)$$

Then, I've noticed that

$$\sqrt{\sin^2(t)+4\cos^2(t)}=\sqrt{1+3\cos^2(t)}$$

Therefore, I have to calculate definite integral

$$\int_{\alpha}^{\beta}{2\sin(t)\sqrt{1+3\cos^2(t)}} \mathrm d t$$

I have no idea how to solve this integral, and even what the boundaries should be in this case. I need help here.

• So, did you compute the integral? – Rodrigo de Azevedo Apr 10 at 20:14

Hint: Use the substitution $$u=\sin{(t)}$$. The bounds would be from $$0$$ to $$\pi$$.