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.)