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 are wrong about the intersection region. Equating the two expressions for $z$ gives
$$x^2+9y^2=18-(x^2+9y^2)$$
$$2(x^2+9y^2)=18$$
$$x^2+9y^2=9$$
and thus here $z=9$.
The volume of your intersection can be divided into to parts: $0\le z\le 9$ where the restrictions on $x$ and $y$ are $x^2+9y^2\le z$, and $9\le z\le 18$ where the restrictions on $x$ and $y$ are $18-(x^2+9y^2)\ge z$. You can see that those two parts have equal shapes and sizes and thus equal volumes, but that is not necessary to use.
So for each region, for each $z_0$ find the area of the cross-section of your region with the plane $z=z_0$, which is an ellipse so the area is easy to find and is an expression in $z_0$. Then integrate that area over $z$ between the limits I gave. Or if you like, use a triple integral for each region. It is also possible to do a double integral over the area $x^2+9y^2=9$.

Since you ask, I'll give more details. I prefer the single-integral approach, so I'll show that here.
For the lower region $x^2+9y^2\le z$ for $0\le z\le 9$, we can use our knowledge of conic sections to see that for a given $z$ that is an ellipse with major axis $a=\sqrt z$ over the $x$-coordinate and minor axis $b=\frac{\sqrt z}3$ over the $y$-coordinate. We can use the formula for the area of an ellipse
$$A=\pi ab=\pi(\sqrt z)\left(\frac{\sqrt z}3\right)=\frac{\pi z}3$$
We now find the volume of that region with
$$V_1=\int_0^9 \frac{\pi z}3\,dz$$
For the upper region $x^2+9y^2\le 18-z$ for $9\le z\le 18$, we can use our knowledge of conic sections to see that for a given $z$ that is an ellipse with major axis $a=\sqrt{18-z}$ over the $x$-coordinate and minor axis $b=\frac{\sqrt{18-z}}3$ over the $y$-coordinate. We can use the formula for the area of an ellipse
$$A=\pi ab=\pi(\sqrt{18-z})\left(\frac{\sqrt{18-z}}3\right)=\frac {\pi(18-z)}3$$
We now find the volume of that region with
$$V_2=\int_9^{18} \frac{\pi(18-z)}3\,dz$$
Your total volume is then $V_1+V_2$.
I like this approach since it is just a pair of single integrals, each of which is very easy. Your question seems to assume the double-integral approach. Let me know if those are the bounds you really want.

Here is the double-integral, if you really want it.
We saw that the largest possible area for a given $z$ is $x^2+9y^2\le 9$. We get from that
$$-1\le y\le 1, \qquad -3\sqrt{1-y^2}\le x\le 3\sqrt{1-y^2}$$
and the bounds on $z$ from your two original conditions are
$$x^2+9y^2\le z\le 18-x^2-9y^2$$
So the appropriate double integral is
$$\int_{-1}^1 \int_{-3\sqrt{1-y^2}}^{3\sqrt{1-y^2}} [(18-x^2-9y^2)-(x^2+9y^2)]\,dx\,dy$$
Good luck with that!
