Área encerrada por x3+y3=x2+y2x3+y3=x2+y2x^3+y^3=x^2+y^2 y los ejes x,yx,yx,y

En coordenadas polares la curva se convierte en

$$r(\theta)=\frac{\sin^2(\theta)+\cos^2(\theta)}{\sin^3(\theta)+\cos^3(\theta)}=\frac{1}{(\sin(\theta)+\cos(\theta))(\sin^2(\theta)-\cos(\theta)\sin(\theta)+\cos^2(\theta))}$$ $$=\frac{1}{\sqrt{2}\sin\left(\theta+\frac{\pi}{4}\right)(1-\frac{1}{2}\sin(2\theta))}=\frac{\sqrt{2}}{\sin(\theta+\frac{\pi}{4})(2-\sin(2\theta))}$$

El área entre la curva y el eje de coordenadas viene dada por

$$A=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}r^2(\theta)d\theta=\int_{0}^{\frac{\pi}{2}}\frac{1}{\sin^2(\theta+\frac{\pi}{4})(2-\sin(2\theta))^2}d\theta$$

y sustituyendo $u=\theta+\frac{\pi}{4}$tenemos

$$A=\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\frac{1}{\sin^2(u)(2+\cos(2u))^2}du=\frac{1}{27}\left(9+4\pi\sqrt{3}\right)$$

según WA .

Pista:

Usa coordenadas polares para obtener

$$r=\frac{1}{\sin^3 t+\cos^3 t}$$ El área requerida es $$A=\frac{1}{2}\int_{0}^{\pi/2} r^2 dt=\frac{1}{2}\int_{0}^{\pi/2} \frac{dt}{(\sin^3 t+ \cos^3 t)^2}=\frac{1}{2}\int_{0}^{\pi/2}\frac{\sec^6 t dt}{(1+\tan^3 t)^2}$$ deja $\tan t= u$, entonces $$A=\frac{1}{2}\int_{0}^{\infty} \frac{(1+u^2)^2}{(1+u^3)^2}du$$ $$2A{=}\int_{0}^{\infty} \left( \frac{2u^2}{(1+u^3)^2}{+}\frac{2}{9(1+u)^2}{+}\frac{7}{9(1-u+u^2)}{+}\frac{2u-1}{6(1-u+u^2)^2}+\frac{1}{6(1-u+u^2)^2}\right) du$$