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Related is Why is the volume of a cone one third of the volume of a cylinder?, but it does not outline finding the volume of a cone using solids of revolution.

I am reviewing Calculus II for the Math GRE Subject Test. From Stewart, section 6.2., #47, I would like to find the volume of a right circular cone with height and base radius $r$.

How I approached this was I thought, suppose the tip of the cone is at the origin. Then a cone is just a triangle rotated with length $r$ (parallel to the positive $y$-axis) and width $h$ (a side of length $r$ on the $x$-axis), so that the area is $A(r)$, defined below. As Stewart states:

$A(x)$ is the area of a moving cross-section obtained by slicing through $x$ perpendicular to the $x$-axis.

So a cross section of a cone is a circle, with area $A(r) = \pi r^{2}$. But I know that computing the integral $$V= \int\limits_{0}^{h}\pi r^{2}\text{ d}r$$ does not give the correct answer, which is $\dfrac{1}{3}\pi r^{2}h$.

What am I missing?

Community
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Clarinetist
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2 Answers2

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I'm not sure about your setup, but in any case you should be integrating in a direction perpendicular to the $r$ in your cross section.

I'll use a setup that's hopefully what you had in mind: the tip of the cone is at the origin, and the cone is lying on its side. Suppose the radius of the base is $r$ (this is the $y$ value at the end of the triangle you're rotating) and the height is $h$ (this is the $x$ value at the end of the triangle).

Then, consider a circular cross section at some point $x$. The radius is the corresponding $y$ value, which by similar triangles is $\frac{rx}{h}$. So, the area is $$ A(x) = \frac{\pi r^2}{h^2} x^2, $$ and the volume is $$ V = \int_0^h \frac{\pi r^2}{h^2} x^2\ dx = \frac{\pi r^2 h^3}{3h^2} = \frac{1}{3} \pi r^2h. $$ Note the cone lies on its side, so the $x$ values we integrate over range from $0$ to the "height" of the cone, $h$.

Platehead
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The volume you calculated is that of a cylinder. Note that your radius $r$ is not changing as your height at $x$. If you want a height of $h$ and a base of $r$, your radius $r(x)$ will change along the line connecting $(0,r)$ and $(h,0)$.

If you integrate this $\int_0^h \pi [r(x)]^2dx$ should give exactly what you want.

K4000
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    I don't think they've calculated the area of a cylinder - the $r$ does vary in the calculation. It's just that the integrand corresponds to rotating the line $y = x$ around the $x$ axis (which has been renamed $r$), meaning that they've assumed $r = h$. Indeed, the answer we'd end up with using that integral will be $\frac{\pi h^3}{3}$, which is what we get if we set $r = h$ in the correct formula. – Platehead Oct 15 '14 at 02:39
  • ah you are right, i misread their notation. if this is how they want to think of it, then the line should connect the origin and $(h,r)$ instead, which would yield the same answer. – K4000 Oct 15 '14 at 02:48