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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.

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  • $\begingroup$ So, did you compute the integral? $\endgroup$ – Rodrigo de Azevedo Apr 10 at 20:14
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Hint: Use the substitution $u=\sin{(t)}$. The bounds would be from $0$ to $\pi$.

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