I am stuck on a practice exam problem.
A wedge of Manchego can be modeled as the solid region in $R^3$ bounded by the following surfaces:
z = 0,
z = y,
y = 9-$x^2$.
Express the volume of this region as an iterated integral in dy dz dx order.
I am specifically stuck on the dy integral. My guess is that its lower bound is z and its upper bound is 9-$x^2$, but I only deduced this through taking x-y cross sections at different z values.
If I am incorrect, please explain the best way to approach the problem. I do find iterated integrals a bit trickier when one of the variables is dependent on both of the others instead of just one.
There is a subset of $\mathbb{R}^3$ bounded by the given surfaces for which we have $y\leq z$ and a subset of $\mathbb{R}^3$ bounded by the given surfaces for which we have $y\geq z.$ Therefore, the question about the lower bound and the upper bound of $y$ can only be answered by additional assumptions. In this case, our additional assumption is that we do not have an infinite amount of Manchego. We must choose the sides of surfaces such that we get a finite amount of cheese.
The only choice that ensures a finite volume is (as you already guessed) $y\geq z.$ In the other case ($y\leq z$), we can easily find $(x,y,z)\in\mathbb{R}^3$ between the given surfaces for any $y<0.$
So you get $$ 0\leq z\leq y\leq 9-x^2 $$ We observe that $9-x^2$ is the upper bound not only for $y$ but also for $z$. Furthermore, we can easily see that $x\in [-3,3].$ This results in $$ V=\int\limits_{-3}^{3} \;\;\int\limits_{0}^{9-x^2} \int\limits_{z}^{9-x^2} 1\, dy\,dz\,dx $$