Compute $$\iiint_V (z+2)\, dx\, dy\, dz$$ where $V$ is the region bounded by the planes $z=1$, $z=3$ and the paraboloid $\frac{x^2}{25}+\frac{y^2}{9}=z$.
I think that the projection of $V$ on the $xy$ plane is $D=\{(x, y)\in \mathbb{R}^2| \frac{x^2}{25}+\frac{y^2}{9}\le 9\}$. Thus, we have $$\iiint_V (z+2)\,dx\,dy\,dz=\iint_D\left(\int_1^3 (z+2)\,dz \right)dx\,dy=8\iint_D dx\,dy$$
and this is now easy to compute (I just need to make the substituion $x=5r\cos \theta$, $y=3r\sin \theta$), but I am not really sure whether what I did is correct. So, please let me know if I reduced the triple integral to a double integral correctly.
No, that is not correct. Note that at no point you have use that fact that, when $z=1$, then you will be dealing only with the circle $\left\{(x,y)\in\Bbb R^2\,\middle|\,\frac{x^2}{25}+\frac{y^2}9=1\right\}$. However, your idea of doing $x=5r\cos\theta$ and $y=3r\sin\theta$ is fine. Doing this, your integral becomes$$\int_0^{2\pi}\int_1^3\int_0^{\sqrt z}15r(z+2)\,\mathrm dr\,\mathrm dz\,\mathrm d\theta=250\pi.$$