Using polar coordinates find area of $(x^2+y^2)^3=x^4+y^4$

Question

Using polar coordinates find area of figure bounded by given curves. $(x^2+y^2)^3=x^4+y^4$

Using polar coordinates I get $r^2=\cos^4\phi+\sin^4\phi=\frac{3+\cos4\phi}{4}$

Now $0\leq\phi\leq2\pi$ but how get bounds for $r$?

Answer 1

Polar coordinates give you $r^{6}=r^{4}(\cos^{4}(\theta)+\sin^{4}(\theta))$

This gives $r^{2}=(\cos^{4}(\theta)+\sin^{4}(\theta))$

Here's a nice picture

Now the area is given by

$$\int_{0}^{2\pi}\int_{0}^{\sqrt{(\cos^{4}(\theta)+\sin^{4}(\theta))}}r\,dr\,d\theta=\frac{1}{2}\int_{0}^{2\pi}((\cos^{4}(\theta)+\sin^{4}(\theta))\,d\theta=\int_{0}^{2\pi}\sin^{4}(\theta)\,d\theta $$ . I'll leave the rest to you.

Answer 2

Integrating in polar coordinates, we get $$ \begin{align} \text{Area} &=\int_0^{2\pi}\int_0^{r(\theta)}r\,\mathrm{d}r\,\mathrm{d}\theta\\ &=\int_0^{2\pi}\frac12r(\theta)^2\,\mathrm{d}\theta \end{align} $$ The last equality is valid when $r$ can be expressed as a function of $\theta$ (e.g. when the curve circles the origin).

As has been mentioned, $$ r^2=\frac{3+\cos(4\theta)}4 $$ so we can compute the area as $$ \begin{align} \int_0^{2\pi}\frac12r^2\,\mathrm{d}\theta &=\int_0^{2\pi}\frac{3+\cos(4\theta)}8\,\mathrm{d}\theta\\ &=\frac{3\pi}4 \end{align} $$