Converting an integral from Cartesian to Polar coordinates.

Let me assume you are interested in $$ I=\int_0^{2}\int_0^{(5/2)\sqrt{4-x^2}}\left(4-x^2-\frac{4}{25}y^2\right) \mathrm dy \;\mathrm dx $$ A key feature of the integration domain is the upper limit $$y=5\sqrt{1-(x/2)^2}$$ It hints at the fact that a curve of interest is (a part of) the ellipse $$(x/2)^2+(y/5)^2=1$$ which is parametrized by $$x=2\cos(u)\qquad y=5\sin(u)$$

One sees that $0\leqslant y\leqslant5\sqrt{1-(x/2)^2}$ and $0\leqslant x\leqslant2$ can be parametrized by $x=2\cos(u)$ and $y=5r\sin(u)$ with $0\leqslant u\leqslant\pi/2$ and $0\leqslant r\leqslant1$.

The Jacobian of the transformation $(r,u)\mapsto(x,y)=(2\cos(u),5r\sin(u))$ is $10\sin^2(u)$, hence $$ I=\int_0^{1}\int_0^{\pi/2}4(1-r^2)\sin^2(u)\;10\sin^2(u) \mathrm du \;\mathrm dr=5\cdot J\cdot K $$ with $$ J=\int_0^1(1-r^2)\mathrm dr=2/3\qquad K=\int_0^{\pi/2}8\sin^4(u)\mathrm du=3\pi/2 $$ Finally, $$I=5\pi$$