Find the area of the surface generated by revolving the
astroid about the y-axis
an astroid is defined implicitly by:
$$x^{2/3}+y^{2/3}=a^{2/3}$$
now:
$$y = (a^{2/3}-x^{2/3})^{3/2}$$
$$y' = -\frac{\sqrt{a^{2/3}-x^{2/3}}}{x^{1/3}}$$
by arc length and surface area of revolution definition:
$$SA =2\pi \int_0^a x\sqrt{1+\frac{a^{2/3}-x^{2/3}}{x^{2/3}}} $$
did I get the surface area generated by the astroid right and how would you approach that integral
Best Answer
You are almost done, indeed note that
$$\sqrt{1+\frac{a^{2/3}-x^{2/3}}{x^{2/3}}}=\sqrt{\frac{a^{2/3}}{x^{2/3}}}$$
therefore
$$2\pi \int_0^a x\sqrt{1+\frac{a^{2/3}-x^{2/3}}{x^{2/3}}}dx =2\pi \int_0^a x^{\frac23}a^{1/3}dx=2\pi\left[\frac35x^\frac 53a^\frac13\right]_0^a=\frac 65\pi a^2$$