# Volume of a defined region using triple integrals

I was doing some excercises and I came upon this one, but I couldn't define the limits of integration. The problem says the following:

Find the volume of the region defined by:
$$z = x^2 + 3y^2 ~,\quad z = 9 - x^2$$

I already know that these figures are a paraboloid and a parabola respectively. I have tried fixing one of the axis and then defining the upper limit of the integral along the curves but I found this impossible to define.

Any help would be gladly appreciated

After plotting the surfaces and looking at them, it seems like letting the innermost integral be with respect to $$z$$ is a bad choice. This is also backed by the fact that for both surfaces, $$z$$ is a function of $$x$$ and $$y$$.
So the innermost integral is easily seen to be $$L(x, y) = \int_{x^2 + 3y^2}^{9-x^2}dz$$ (I call it $$L$$ because it represents the length of the line segment through the region given a value for $$x$$ and $$y$$.)
Next, we tackle $$y$$. It turns out that $$x$$ and $$y$$ give basically the same result at this stage, so it doesn't matter which one we pick. Given a value of $$x$$, the $$z$$-value of the intersection points of the two surfaces is $$z = 9-x^2$$. And so the $$y$$ values of the two intersection points are given by $$9-x^2 = x^2 + 3y^2\\ y^2 = \frac{9-2x^2}{3}\\ y = \pm\sqrt\frac{9-2x^2}{3}$$ which means that the middle integral should be $$A(x) = \int_{-\sqrt{(9-2x^2)/3}}^{\sqrt{(9-2x^2)/3}} L(x, y)dy$$ (I call it $$A$$ because it represents the area of the section of the region for any given value of $$x$$.)
Finally, we get to $$x$$. And the extreme values of $$x$$ over this solid is given by the extreme values of $$x$$ along the intersection between the surfaces. This happens at a point where $$y = 0$$, so to find it, we have to solve $$x^2 + 3\cdot 0^2= 9-x^2\\ 2x^2 = 9\\ x = \pm \sqrt{4.5}$$ which gives $$V = \int_{-\sqrt{4.5}}^{\sqrt{4.5}} A(x)dx$$ (I call it $$V$$ because it represents the volume of the region.)