Answer to Question #346595 in Calculus for Likhu

Question

Find the volume of the solid by revolving the astroid x⅔+y⅔=a⅔ along x axis

Answer 1

Due to symmetry, we can consider the region lying in the first quadrant and then multiply the volume of the region by two.

Let's solve the astroid equation for $y^2$

$y^{2/3}=a^{2/3}-x^{2/3}$

$y^2=(a^{2/3}-x^{2/3})^3$

$y^2=a^2-3a^{4/3}x^{2/3}+3a^{2/3}x^{4/3}-x^2$

Hence, the total volume of the solid bounded by the astroid is given by

$V=2\pi\displaystyle\int_{0}^ay^2dx$

$=2\pi\displaystyle\int_{0}^a(a^2-3a^{4/3}x^{2/3}+3a^{2/3}x^{4/3}-x^2)dx$

$=2\pi[a^2x-\dfrac{9}{5}a^{4/3}x^{5/3}+\dfrac{9}{7}a^{2/3}x^{7/3}-\dfrac{1}{3}x^3]\begin{matrix} a \\ 0 \end{matrix}$

$=2\pi(a^3-\dfrac{9}{5}a^{4/3}a^{5/3}+\dfrac{9}{7}a^{2/3}a^{7/3}-\dfrac{1}{3}a^3)$

$=\dfrac{32\pi a}{105} ({units}^3)$