<s>[INST] 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.
 [/INST] 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
$$\int \limits_0 ^{9} \frac{z}{3} \pi dz + \int \limits_9 ^{18} \frac{18 - z}{3} \pi dz = \pi \left(\frac13 \int \limits_0 ^9 z dz + \int \limits_9 ^{18} 6 dz - \frac13 \int \limits_9 ^{18} zdz \right) = \pi \left(\frac13  \frac{z^2}{2} \big|_0 ^9 + 6 \big|_9 ^{18} - \frac13 \frac{z^2}{2} \big|_9 ^{18} \right) = 27\pi$$
</s>