Find volume on a shape with base of an ellipse I have an ellipse with area $\pi ab$ $a = 6$, $b = 4$ these are the axis lengths. I am suppose to compute the volume of a cone of height 12.
I tried many solutions but none of them worked and I don't know why. I would type them up but I doubt much could be learned. Basically I have been trying to use the fact that $\pi ab$ is the area that I can find $a$ or $b$ and then that ratio is my radius at any point. This doesn't really work though and I don't know why.
How do I do this?
 A: Assume the cone is vertical, its axis is the $z$-axis and its base is an
ellipse on the $xy$-plane, centered at $(x,y,z)=(0,0,0)$, with semi-major
axis $a$ and semi-minor axis $b$. The equation of the base is thus
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad z=0.$$
Let $h$ be the height of the cone. The
cone cross section at height $z$ is an ellipse with semi-major axis $$x_1=a\left( 1-\frac{z}{h}\right) $$ and semi-minor axis $$x_2=b\left( 1-\frac{z}{h}\right) $$ by similarity of (right) triangles, as shown in the following sketch:

The equation of the boundary of this cross section is 
$$\frac{x^2}{x_1^2}+\frac{y^2}{x_2^2}=1,\qquad z=z.$$

Basically I have been trying to use the fact that $\pi ab$ is the area

The area $A(z)$ of this cross section is thus
$$
\begin{equation*}
A(z)=\pi x_1 x_2=\pi ab\left( 1-\frac{z}{h}\right) ^{2}.
\end{equation*}
$$
The volume is the integral of the area $A(z)$ from $z=0$ to $z=h$
$$
\begin{eqnarray*}
V &=&\int_{0}^{h}A(z)\, dz=\int_{0}^{h}\pi ab\left( 1-\frac{z}{h}\right) ^{2}dz
\\
&=&\pi ab\int_{0}^{h}\left( 1-2\frac{z}{h}+\frac{z^{2}}{h^{2}}\right) dz
\\
&=&\pi ab\left( h-h+\frac{h}{3}\right) =\frac{h}{3}\pi ab,
\end{eqnarray*}
$$
i.e. 
$$V=\frac{1}{3}A_{\text{base}}\times \text{height},$$ 
as expected. For $a=6,b=4,h=12$, we have:
$$
\begin{equation*}
V=\frac{h}{3}\pi ab=96\pi.
\end{equation*}
$$
A: Hint: Assume the cone is tip up. By similarity, the area of horizontal cross-section at distance $w$ down from the apex of the cone is 
$$\left(\frac{w}{12}  \right)^2 (\pi ab).$$
Integrate, $w=0$ to $12$.
Better, imagine that the cone is tip down, with tip at the origin. Then the area of cross-section a distance $z$ up from the $x$=$y$ plane is $(z/12)^2(\pi ab)$. For the volume, integrate from $z=0$ to $z=12$. 
If you prefer (I don't) you can have the cone in conventional position, and measure distance up from the $x$-$y$ plane.  Then the area of cross-section at distance $z$ above the $x$-$y$ plane is $\left(  \frac{12-z}{12}\right)^2 (\pi ab)$.
Added: For the integration, we want 
$$\int_0^{12} (\pi ab)\frac{w^2}{12^2}\,dw.$$
Calculation gives $(\pi ab)\frac{(12)^3}{3\cdot 12^2}$.
