91Ó°ÊÓ

The given information is $x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$.

The Green's theorem connects a line integral around a simple closed curve C to a double integral over the plane region D circumscribed by C in vector calculus. Stokes' theorem has a two-dimensional special case.

Write the given equation with respect to .

$$\begin{gathered} x={\left(4-y^{\frac{2}{3}}\right)}^{\frac{3}{2}} \\ dx=\frac{3}{2}{\left(4-y^{\frac{2}{3}}\right)}^{\frac{1}{2}}\left(-\frac{2}{3}{y}^{\frac{-1}{3}}\right)dy \\ ={\left(4-y^{\frac{2}{3}}\right)}^{\frac{3}{2}{y}^{\frac{-1}{3}}dy} \\ A=\frac{1}{2}âˆ\circledR \left[{\left(4-y^{\frac{2}{3}}\right)}^{{}^{\frac{3}{2}}}+{\left(4-y^{\frac{2}{3}}\right)}^{{}^{\frac{1}{2}}}{y}^{\frac{2}{3}}\right]dy \end{gathered}$$

Use parameterization to solve the integral.

Write the values of and.

$$\begin{gathered} x=aco{s}^3\mathrm{θ} \\ y=asi{n}^3\mathrm{θ} \\ 0<\mathrm{θ}<2\mathrm{π} \end{gathered}$$

Find the differentiation with respect to $\mathrm{θ}$.

$$\begin{gathered} dx=-3a\sin \mathrm{θ}{\cos }^2\mathrm{θ}d\mathrm{θ} \\ dy=a\cos \mathrm{θ}{\sin }^2\mathrm{θ}d\mathrm{θ} \\ a=4^{\frac{3}{2}} \end{gathered}$$

Find the area.

role="math" localid="1659174879691" $$\begin{gathered} A=\frac{1}{2}{∫}_0^{2\mathrm{π}}\left[3a^2{\cos }^4\mathrm{θ}{\sin }^2\mathrm{θ}+3a^2{\sin }^4\mathrm{θ}{\cos }^2\mathrm{θ}\right]d\mathrm{θ} \\ =\frac{3a^2}{2}{∫}_0^{2\mathrm{π}}\left[{\cos }^4\mathrm{θ}{\sin }^2\mathrm{θ}+{\sin }^4\mathrm{θ}{\cos }^2\mathrm{θ}\right]d\mathrm{θ} \\ =\frac{3a^2}{2}{∫}_0^{2\mathrm{π}}\left[{\sin }^2\mathrm{θ}{\cos }^2\mathrm{θ}\right]d\mathrm{θ} \\ A=\frac{3a^2}{2}{∫}_0^{2\mathrm{π}}\left[{\sin }^2\mathrm{θ}-{\sin }^4\mathrm{θ}\right]d\mathrm{θ} \\ =\frac{3a^2}{2}\frac{4t-\sin 4t}{32} \\ =24\mathrm{π} \end{gathered}$$

Hence, the solution to this problem is $24\mathrm{Ï€}.$