Volume under $z=10x^2+5y^2$

Question

I have to find the volume under the surface $$z=10x^2+5y^2$$ and above $z=0$ in the region bounded by the plans $y=x,y=2$ and $x=0$.

At first, I thought the volume was $$\int_{0}^{2}\int_{x}^{2}(10x^2+5y^2)dydx$$ but I plotted the surface in Geogebra, it seems the upper limit of $x$ is lower than $2$. How am I supposed to find the upper limit of $x$?

I also tried using polar coordinates $$x=\frac{r\cos\theta}{\sqrt{10}},\quad y=\frac{r\sin\theta}{\sqrt{5}}$$ which implies $$dxdy=\frac{r}{\sqrt{50}}drd\theta$$ with $$\frac{\pi}{4}\leq\theta\leq\frac{\pi}{2},\quad 0\leq r\leq2$$ But I'm not sure about the interval of $r$.

Answer 1

It should be\begin{align}\int_0^2\int_x^210x^2+5y^2\,\mathrm dy\,\mathrm dx&=\int_0^2-\frac{5}{3} \left(7 x^3-12 x^2-8\right)\,\mathrm dx\\&=33+\frac13.\end{align}