Find volume on a shape with base of an ellipse

Question

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?

Answer 1

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: $$ \begin{equation*} V=\frac{h}{3}\pi ab=96\pi. \end{equation*} $$

Answer 2

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 $(\pi ab)\frac{(12)^3}{3\cdot 12^2}$.