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## Main Question or Discussion Point

This is probably just me being a bit of an idiot (I'm going to blame exam stress), but why do the following ways of calculating the volume and lateral area of cones produce different results?

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])

[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...

[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.

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.