Volume Application of Triple Integrals I have to answer this question:
Before a gasoline-powered engine is started, water must be drained out of the bottom of the tank. Suppose the tank is a right circular cylinder on its side with length 12 feet and a radius of 4 feet. If the water level is 18 inches above the lowest part of the tank, determine how much water must be drained from the tank. Round your final answer to the nearest tenth.
I know this is a volume application, which will use triple integrals. I also get understand that as it's laying on its side, it's centered on the y-axis. 
I just don't know how to start the problem. 
 A: If we take a cross section at any $y$ value, we must 'fill up' the circle of radius $4$ to the line $z = -4+1.5 = 2.5$ (if we are working in feet). To use a triple integral, we need to find the bounds of the $x$ chord at a certain height $z$. These bounds, if one draws a picture and uses the Pythagorean Theorem, are from $-\sqrt{4^2-z^2}$ to $\sqrt{4^2-z^2}$. Now we can already set up our integral. The bounds for $y$ are $0$ to $12$, as given. The bounds for $z$ are $-4$ to $-2.5$, also as given. Now we can set up our triple integral.
$$\int_{0}^{12} \int_{-4}^{2.5}\int_{-\sqrt{4^2-z^2}}^{\sqrt{4^2-z^2}} \ dx \ dz \ dy = \int_{0}^{12} \int_{-4}^{-2.5} 2\sqrt{4^2-z^2} \ dz \  dy$$
To evaluate this, the substitution $z = 4\sin{u} \to dz = 4\cos (u) du$ is quite useful, and it changes the bounds to from $u=\arcsin(-1)$ to $u=\arcsin(-.625)$
$$\int_{0}^{12} \int_{\arcsin(-1)}^{\arcsin(-.625)} \frac{2\sqrt{4^2-4^2\sin(u)}}{4\cos(u)} du \ dy \\ 
=\int_{0}^{12} \int_{\arcsin(-1)}^{\arcsin(-.625)} \frac{2\cos(u)}{\cos(u)}du \ dy \\
= \int_{0}^{12} 2(\arcsin(-.625)-\arcsin(-1) dy \\
= 24(\arcsin(-.625)-\arcsin(-1)) \approx 21.496$$
Which would be the volume. I challenge you do try this problem without calculus, however. It is much cleaner and easier.
