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?
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$\begingroup$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, $h$.
$\endgroup$ $\begingroup$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.
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