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the integral is as follows:

find the volume between these regions bounded by : $z = x^2 + 3y^2$ and $z = 9 - x^2$

I discovered that this would be the space bounded by the elliptic paraboloid and the cylinder but can not find how could define the limits of integration. Please Help me

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  • $\begingroup$ Is finding the region a step in doing an integral? I don't see any integration in your question! $\endgroup$ – Rory Daulton May 24 '15 at 22:02
  • $\begingroup$ I'm sorry, what I need is to find the volume between these regions $(z = x^2 + 3y^2$ and $z = 9 - x^2 )$ $\endgroup$ – Richard May 24 '15 at 22:07
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The second function does not represent the graph of a cylinder, rather an open parabola (graph it!).

Now, the first function is a paraboloid---a deformed "bowl". You want the volume under the open parabola and inside the bowl. So:

1). Integrate from $z=x^{2}+3y^{2}$ to $z=9-x^{2}$.

Now you have $\int \int \int_{x^{2}+3y^{2}}^{9-x^{2}}dzdydx$.

2). Next, project the solid region onto the $xy$-plane by setting $z$ equal to $0$ and you get the region

$2x^{2}+3y^{2}=9$.

This rest of this exercise is now an ordinary double integral over an ellipse. Using symmetry, you need only integrate over the upper (or lower) half of the ellipse and multiply the result by two.

Can you take it from here?

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We find where the two surfaces intersect by setting the $z$'s equal to each other:

$$x^2+3y^2=9-x^2$$ $$2x^2+3y^2=9$$ $$\frac{x^2}{\left(3/\sqrt 2\right)^2}+\frac{y^2}{\left(\sqrt 3\right)^2}=1$$

So the projection of the intersection onto the $xy$ plane is an ellipse centered at the origin. You should know several ways to express that bounds of that region in a double integral.

The surface of $z=9-x^2$ will be above the surface of $z=x^2+3y^2$. So your volume is the integral of the difference of those two over the elliptical region given above.

There are several ways to represent that volume. Here is one way. The overall limits for $y$ in the elliptical region is $-\sqrt 3$ to $\sqrt 3$. For a given $y$, the $x$ limits are $-\sqrt{9/2-3y^2/2}$ to $\sqrt{9/2-3y^2/2}$. Therefore one integral you can use is

$$\int_{-\sqrt 3}^{\sqrt 3}\int_{-\sqrt{9/2-3y^2/2}}^{\sqrt{9/2-3y^2/2}}[(9-x^2)-(x^2+3y^2)] \,dx\,dy$$

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