A right elliptical cone is 4m high and has an elliptical base with half axes lengths of 1m and 2m. Find its volume. A right elliptical cone is $4$m high and has an elliptical base with half axes lengths of $1$m and $2$m. Find its volume. [The area of an ellipse with half axes lengths $a$ and $b$ is $πab$.]
I know that $V = \int A(x) \ dx$. I've usually used this in the context of solids of revolution using circles, but I have never encountered a problem like this. I have also sketched an image of an upside-down cone, with the x-axis going from the apex of the cone to the base, such that the base is located at $x = 4$. Therefore, the x-axis is representing height. 
However, I am unsure of how to proceed from here. I would greatly appreciate it if people could please take the time to help me with this problem.
 A: 
In the figure above, $a = 1$ and $b=2$. I'll call the figure a cone even though its base is not circular.
Sit the cone in question pointing upward with its elliptical base on the plane $xy$-plane and its apex sitting on the $z$-axis four units above the origin. 
A plane parallel to the $xy$-plane at a height $0 \leqslant h \leqslant 4$ units above the origin will intersect the cone at an ellipse $E(h)$ with major half-axis $b(h)$ and minor half-axis $a(h)$ - I haven't actually drawn $E(h)$ but I did draw its half-axes. For instance, $E(0)$ is the base of the cone with $a(0) = 1$ and $b(0) = 2$. 
To find $a(h),b(h)$ in terms of $a,b$, use similar triangles: $$\frac{a(h)}{4-h} = \frac{a}{4} = \frac{1}{4} \Longrightarrow a(h) = \frac{1}{4}(4-h)$$ $$\frac{b(h)}{4-h} = \frac{b}{4} = \frac{2}{4} \Longrightarrow b(h) = \frac{1}{2}(4-h).$$
Thus, $$A(h) := \mathsf{area}(E(h)) = a(h)b(h)\pi = \frac{\pi}{8}(4-h)^2.$$
To find the area of the cone, integrate:
$$ \int_0^4 A(h) dh = \frac{\pi}{8} \int_0^4 (4-h)^2 dh $$ $$= -\frac{\pi}{8} \int_4^0 t^2 dt = \frac{\pi}{8}\int_0^4 t^2 dt = \frac{\pi}{8}\left( \frac{(4)^3}{3} - \frac{(0)^3}{3} \right) = \frac{8\pi}{3}. $$
Note that I used the substitution $t = 4-h$ to solve the integral. Note also that $$\frac{8 \pi}{3} = (1/3)(2\pi)(4) = (1/3)(\mathrm{area \,\, of \,\, base})(\mathrm{height}).$$
