Area and Volume Revisited — Question 1

Question 1 – Solution

Strategy. The cubic substitution converts the astroid into a disk, after which polar coordinates handle the transformed integral.

Step 1: Transform the region

Because the real cube function is one-to-one, $$x^{2/3}=u^2,\qquad y^{2/3}=v^2.$$ Thus the interior maps one-to-one to $u^2+v^2\le a^{2/3}$, a disk of radius $a^{1/3}$.

Step 2: Jacobian and integral The absolute Jacobian is $$\left|\frac{\partial(x,y)}{\partial(u,v)}\right| =\left|(3u^2)(3v^2)\right|=9u^2v^2.$$ Writing $u=r\cos\theta$, $v=r\sin\theta$ gives $$\begin{align*} A&=9\int_0^{2\pi}\!\int_0^{a^{1/3}} r^5\cos^2\theta\sin^2\theta\,dr\,d\theta\\ &=9\left(\frac{a^2}{6}\right)\left(\frac{\pi}{4}\right) =\boxed{\frac{3\pi a^2}{8}}. \end{align*}$$

Verification Replacing $a$ by $ka$ enlarges every length by $k$, so area must scale by $k^2$. The result is proportional to $a^2$, as required.