- #1

TheBaker

- 19

- 0

I'll use the following equation of a cone to demonstrate:

[tex]x^2 + y^2 = \frac{9}{4}z^2[/tex] (Valid for [tex]0 \leq z \leq 1[/tex])

**Volume - Summing Disks**

[tex]V = \int_0^1 \pi r^2 dz = \int_0^1 \frac{9\pi}{4}z^2 dz = \frac{3\pi}{4}[/tex]

This is the correct answer. However, if instead of summing disks from bottom to top, I sum triangles around the cone...

**Volume - Summing Triangles**

[tex]A = \frac{1}{2} \mbox{base} \times \mbox{height} = \frac{1}{2} \frac{3}{2}[/tex]

Then, integrating this around a circle to form a cone I get:

[tex]\int_0^{2\pi} \frac{3}{4} d\theta = \frac{3 \pi}{2}[/tex]

This is twice as large as the (correct) volume found the other way, but conceptually I can't see where I've made a mistake. A similar thing happens for the surface area - when you sum the hypotenuse of the triangle or the circumference of the disks.