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.
On
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 $$\int \limits_0 ^{9} \frac{z}{3} \pi dz + \int \limits_9 ^{18} \frac{18 - z}{3} \pi dz = \pi \left(\frac13 \int \limits_0 ^9 z dz + \int \limits_9 ^{18} 6 dz - \frac13 \int \limits_9 ^{18} zdz \right) = \pi \left(\frac13 \frac{z^2}{2} \big|_0 ^9 + 6 \big|_9 ^{18} - \frac13 \frac{z^2}{2} \big|_9 ^{18} \right) = 27\pi$$
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!