Use Problem 6 to find the area inside the curve${\mathit{x}}^{\frac{\mathbf{2}}{\mathbf{3}}}\mathbf{+}{\mathit{y}}^{\frac{\mathbf{2}}{\mathbf{3}}}\mathbf{=}\mathbf{4}$.

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 .

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

Use parameterization to solve the integral.

Write the values of and.

$$\begin{gathered} x=aco{s}^3\theta \\ y=asi{n}^3\theta \\ 0<\theta <2\pi \end{gathered}$$

Find the differentiation with respect to $\theta$.

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

Find the area.

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

Hence, the solution to this problem is $24\pi .$