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Calculate the area of the bounded surface enclosed by the curve $(x+y)^4 = x^2y$ with the help of the coordinate transformation $x = r\cos^2 t, y = r\sin^2 t$.

As I see it the area is unbounded, so this exercise is not solvable, or am I wrong?

Here you can see a plot of the curve.

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Note that $y\geq0$, while $x$ could be also negative.
For $x\geq0$ the graph is given by the following image:

enter image description here

The relation between $x$ and $y$, with the proposed change of coordinates, becomes $$ r(t)=\cos^4t\sin^2t $$ (but note that $r$ and $t$ do not correspond to radius and argument in ordinary polar coordinates).
The area is given, using Gauss-Green formula, by $$ \int_D dxdy=\int_{\partial D}x dy=-\int_{\partial D}y dx $$

Taking \begin{align} x(t)&=r(t)\cos^2t=\cos^6t\sin^2t\\ y(t)&=r(t)\sin^2t=\cos^4t\sin^4t \end{align} and $0\leq t\leq\pi/2$, substituting into the line integral, we get, after some tedious calculation, to $$ A=\frac{1}{210} $$

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