Solid of revolution question: verify that the volume of the cone is παβh/3

[mclame22]

Homework Statement

Consider a vertical cone of height h whose horizontal cross-section is an ellipse and whose base is the ellipse with major and minor semi-axes α and β. Verify that the volume of the cone is παβh/3.
[ Hint: The area of an ellipse with major and minor semi-axes α and β is παβ. ]

Homework Equations

V = ∫A(y) dy (from c to d)
V = ∫π(radius)² dy (from c to d)

The Attempt at a Solution

It says that the cone is upright, so I'm assuming it wants the cone rotated about the y-axis.
V = ∫A(y) dy
V = ∫π(radius)² dy

Using similar triangles:
x/y = r/h
x = ry/h

V = π∫(ry/h)² dy (the integral is now from 0 to h (c = 0, d = h))
V = π∫(r²y²/h²) dy
V = (πr²/h²)∫(y²) dy (since pi, r, and h are all constants)

At this point I'm not sure where to go. Do I take the integral of y²? How do I incorporate α and β into this integral? (As a side note, I'm very new to these forums and if I've done anything wrong I apologize. I'm not used to writing out integrals on the computer and if the notation is not optimal I'm sorry!)

 

[jackmell]

Personally, I think you need to go out of your way to draw this thing, nicely. The volume via slices is:

$$V=\int_0^h A(y)dy$$

with $A(y)=\pi u(y) v(y)$

and u and v are the minor and major axes as functions of y as you go up the cone starting from the base up to h.

Since the axes at the base are $\alpha,\beta$, then as you said, using similar triangles, I get:

$$u(y)=\frac{\alpha}{h}(h-y)$$

$$v(y)=\frac{\beta}{h}(h-y)$$

Alright, just integrate now.