Volume bounded by elliptic paraboloids Find the volume bounded by the elliptic paraboloids given by $z=x^2 + 9 y^2$ and $z= 18- x^2 - 9 y^2$.
First I found the intersection region, then I got $x^2+ 9 y^2 =1$. I think this will be area of integration now what will be the integrand. Please help me.

You probably got something looking like this
where the required volume is the dark volume inside the two parabolas. Of course there are many ways to find that volume; I'll stick to the slice method: essentially you fix the third variable $z$ and calculate the area of desired region intersecting the plane $z = z_0$, then you sum up all this areas. Note that if you intersect the region with a plane parallel to the $XY$ plane you get an ellipse, with major semi-axis $\sqrt{\frac{18-z}{9}}$ and minor semi-axis $\sqrt{18-z}$ if $z \ge 9$ (that is when the parabolas intersect each other), and major semi-axis $\sqrt{z}$ and minor semi-axis $\sqrt{\frac{z}{9}}$ if $z \ge 9$ when $z \ge 9$. So the area of the intersection between the region and the plane at height $z$ is
$$
\begin{cases}
\sqrt{\frac{18-z}{9}} \cdot \sqrt{18-z} \cdot \pi = \frac{18 - z}{3} \pi \iff z \ge 9\\
\sqrt{\frac{z}{9}} \cdot \sqrt{z}  \cdot \pi= \frac{z}{3} pi \iff z \le 9
\end{cases}
$$
So the volume desired is
$$\int \limits_0 ^{9} \frac{z}{3} \pi dz + \int \limits_9 ^{18} \frac{18 - z}{3} \pi dz = \pi \left(\frac13 \int \limits_0 ^9 z dz + \int \limits_9 ^{18} 6 dz - \frac13 \int \limits_9 ^{18} zdz \right) = \pi \left(\frac13  \frac{z^2}{2} \big|_0 ^9 + 6 \big|_9 ^{18} - \frac13 \frac{z^2}{2} \big|_9 ^{18} \right) = 27\pi$$
