The solid is upper bounded by $z=16-y^2-4x^2$ and lower bounds by $z=7$ What I've done up until now is
$$\int_0^{2\pi}\int_2^{4} \int_7^{16-r^2\sin(\theta)-4r^2\cos(theta)} r\sqrt{(16-r^2\sin(\theta)-4r^2\cos^2(\theta))}dzdrd\theta$$
What I've thought is that maybe the exercise is proposed wrong in the first place due to it makes more sense (and gets easier to solve ) by having the parabolloid equation like this $z=16-4y^2-4x^2$
The answer the sheet has is $(\frac{1664+197 \sqrt(7)}{30}) \pi$ And as you can see, I don't know how to get to that answer.
First analyse the bounds of the different variables : $$z\ge7\iff 16-y^2-4x^2\ge7\iff 4x^2+y^2\le 9$$ So the volume is determine by : $$\left\{\begin{matrix}-\frac32\le x\le \frac32 \\ -\sqrt{9-4x^2}\le y\le \sqrt{9-4x^2} \\ 7\le z\le 16-y^2-4x^2\end{matrix}\right.$$ Now use Fubini theroem : $$\iiint_V dxdydz = \int_{-\frac32}^{\frac32} \left(\int_{-\sqrt{9-4x^2}}^{\sqrt{9-4x^2}} \left(\int_7^{16-y^2-4x^2} dz\right)dy\right)dx$$ I leave you the rest of the work :-)