Compute the surface area of the surface generated by revolving the astroid with parametrization $$c(t) = \left(\cos^3 t,\sin^3t\right)$$ about the $x$-axis for $t\in[0,\pi/2]$.
I did not know how to go about answering this question and would really appreciate the help. Thank you in advance.
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We need to integrate $2\pi y\,ds$ over the appropriate interval, where $$ds=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt.$$ The differentiations are straightforward. The $ds$ simplifies to $3|\sin t\cos t|\,dt$, and from $0$ to $\pi/2$ there is no issue of signs. So we want $$\int_0^{\pi/2} 6\pi \sin^4 t\cos t\,dt.$$ Let $u=\sin t$, or just write down an antiderivative. The surface area is $\frac{6\pi}{5}$.
We know $S = \int^{\pi/2}_0 2\pi yds$ where $S$ is the surface area and $ds = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}dt $. (The bounds on the integral come from your bounds).
First you should compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$.
Then compute $ds$ and you could be able to solve it from there.