<s>[INST] Find the area bounded by the curve $\left(\frac{x}a+\frac{y}b\right)^5=\frac{x^2y^2}{c^4}$ Find the area bounded by the curve $\left(\dfrac{x}a+\dfrac{y}b\right)^5=\dfrac{x^2y^2}{c^4}$.
Let $x=ar\cos\varphi$, $y=br\sin\varphi$, then 
$$r=\dfrac{a^2b^2\cos^2\varphi\sin^2\varphi}{c^4(\cos\varphi+\sin\varphi)^5}.$$
But how to find the polar angle?
 [/INST] It's probably better to use the substitution 
$$\begin{cases} x = ar\cos^2\theta \\ y = br\sin^2\theta \\ \end{cases} $$
This has a Jacobian of $abr\sin(2\theta)$. From looking at the equation, it only encloses a loop in the first quadrant. Then plugging in we get that
$$ r^5 = \frac{a^2b^2}{c^4}r^4\sin^4\theta\cos^4\theta \implies r = \frac{a^2b^2}{16c^4}\sin^42\theta$$
Now we can set up our integral
$$\int_0^{\frac{\pi}{2}} \int_0^{\frac{a^2b^2}{16c^4}\sin^42\theta} abr\sin 2\theta \:dr \: d\theta = \frac{a^5b^5}{512c^8} \int_0^{\frac{\pi}{2}} \sin^92\theta \: d\theta = \frac{a^5b^5}{1024c^8} \int_{-1}^1 (1-x^2)^4dx$$
The last integral evaluates to $\frac{256}{315}$, making the value of the area
$$A = \iint_{\text{Loop}} 1 \: dA = \frac{a^5b^5}{1260c^8}$$
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