Calculating area by using integral

Question

I need to calculate the area that confined by:

$$|x|^{\frac{2}{3}}+|y|^{\frac{2}{3}}=a^{\frac{2}{3}}\ ,$$

$$a\gt 0\ .$$

I need to do this with some integral.

Answer 1

Hint

From the equation $$x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$$ you can extract $$y=\left(a^{2/3}-x^{2/3}\right)^{3/2}$$ and you need to integrate from $x=0$ to $x=a$. A first change of variable $x=at$ makes the formula simpler $$y=a^2\left(1-t^{2/3}\right)^{3/2}$$ to be integrated from $t=0$ to $t=1$ but I must confess that the antiderivative is not very pleasant to establish (even if doable). Using now $t^{1/3}=\sin(z)$ simplifies a lot since $$y=3 \sin ^2(z) \cos ^4(z)$$ to be integrated from $z=0$ to $z=\frac {\pi}{2}$.

I am sure that you can take from here (just remember the relation between $\cos^n (z)$ and the cosine of multiple angles).

Answer 2

$A = \displaystyle \int_{-a}^a \displaystyle \int_{-\left(a^{\frac{2}{3}} - y^{\frac{2}{3}}\right)^{\frac{3}{2}}}^{\left(a^{\frac{2}{3}} - y^{\frac{2}{3}}\right)^{\frac{3}{2}}} 1dxdy$

Answer 3

Parametrize the part of $A$ lying in the first quadrant by means of $$(r,\phi)\mapsto\left\{\eqalign{x&:=(r\cos\phi)^3\cr y&:=(r\sin\phi)^3\cr}\right.\qquad(0\leq r\leq a^{1/3}, \ \ 0\leq\phi\leq{\pi\over2})\ .$$ The Jacobian computes to $$J(r,\phi)=9r^5\sin^2\phi\>\cos^2\phi={9\over4} r^5\sin^2(2\phi)\ .$$ Therefore we obtain $${\rm area}(A)=4 \int_0^{a^{1/3}} \int_0^{\pi/2}{9\over4} r^5\sin^2(2\phi)\ d\phi\ dr={3\pi\over 8}\>a^2\ .$$