I've been given the following problem, and I'm completely unsure how to go about solving it.
$$ \text{Find the volume of the solid enclosed by the}\\ \text{paraboloids } z = 16 \left( x^{2} + y^{2} \right)\text{ and }z = 18 - 16 \left( x^{2} + y^{2} \right). $$
I know that these two paraboloids are two "bowls" essentially. One facing up, the other facing down. Because of this, I think the best thing to do is to probably convert this to polar coordinates.
So the region will need to be defined as:
$$ \begin{align} z &= 16\:(r^2)\\ z &= 18 - 16\:(r^2) \end{align} $$
In my head, I think that integrating this is probably best done by taking a cross-section and then rotating it by integrating $d\theta$ from $0$ to $2\pi$.
So we're looking at something like: $$ \int_0^{2\pi} \int_a^b r*r\:dr\:d\theta $$
I can't for the life of me decide what I need my $a$ and $b$ to be. I can't leave $r$ in my region because I won't have an actual value to integrate around $\theta$. How am I supposed to proceed?
UPDATE: I thought that maybe I could get away with solving for $r$ in $$ \begin{align} 16(r^2) &= 18 - 16(r^2)\\ 32(r^2) &= 18\\ r &= \sqrt{\frac{18}{32}} \end{align} $$ Setting up my equation with $b = \sqrt{\frac{18}{32}}$ and $a = 0$ yielded back $\frac{2\pi \!\left(\sqrt{\frac{18}{32}}\right)^{3}}{3}$, which wasn't the correct answer (and that makes sense, because that number is way too small, and the upper limit is also too small for this to make sense)
Your update of radius is correct.
The problem is your integrand should be $$(18-16r^2)-16r^2$$
This is the upper bound minus the lower bound of the volume. Then you can project it onto the $xy$-plane and use polar coordinate. So the integral should set up like $$\int_0^{2\pi}\int_0^{3/4}((18-16r^2)-16r^2)rdr d\theta$$