Consider a cone with height $H$ and radius $R$. Let $dh$ and $dr$ be the respective changes in $H$ and $R$. Therefore $$\text{Volume}=\int H \pi r^2 dh$$ Now $r$ is a function of $h$, so $$h=H−H/R*r$$ Differentiating both sides with respect to $r$ $$dh=−H/ R* dr$$ Substituting the same in the above equation we get the integral as

$$\int_0^R \pi r^2(−H)/R*dr$$ $$\text{Volume}=\frac{1}{3}\pi R^2H$$ If you do the same process for the surface area you will end up getting

$$\text{Surface Area}=\pi RH$$ Which is not true.

The Surface Area of a Cone is $\pi RS$ where $S$ is Slant Height of the Cone.

If we try to argue on this explanation of Surface Area then our explanation for Volume is in contradiction.

  • $\begingroup$ Welcome to MSE! We use MathJax to format mathematics here, so any equations or expressions are much nicer-looking and easier to read. It's pretty easy to learn, so here's a tutorial: math.meta.stackexchange.com/questions/5020/… $\endgroup$ Aug 3, 2018 at 18:27
  • $\begingroup$ Could you please rewrite both integrals properly, in that way it is not clear what you have done. $\endgroup$
    – user
    Aug 3, 2018 at 18:30
  • $\begingroup$ Ok I'll edit it $\endgroup$ Aug 3, 2018 at 18:32
  • $\begingroup$ @NalinYadav Please check if my edits preserve the meaning of the question. $\endgroup$
    – GoodDeeds
    Aug 3, 2018 at 18:40
  • $\begingroup$ Your fractions are ambiguous: does $a/b*c$ mean $\frac{a}{bc}$ or $\ frac{a}{b}c$. It's not clear to me where you got volume = integral (Hπr2dh), and it's not clear what you mean by "do the same process for the surface area". Your confusion probably stems from the fact that the volume of an infinitesimal slice can be taken to be area times height, and thinking that that can be extended to the surface area of an infinitesimal being equal to circumference times height. $\endgroup$ Aug 3, 2018 at 18:55

2 Answers 2


You can get into similar difficulties in just two dimensions.

Consider an isosceles right triangle with legs of length $s.$ If you place one leg horizontal and one vertical, you can slice the triangle by closely-spaced horizontal lines and put as large a rectangle as you can between each pair of lines so that the rectangles nearly fill the triangle. As you reduce the spacing between the lines toward zero, the total area of the rectangles approaches $\frac12s^2,$ which is the correct area of the triangle.

But if you try to measure the length of the hypotenuse by adding up the short edges of the rectangles that are along the hypotenuse, you will get a total length $s,$ whereas the correct answer is $s\sqrt2.$

The reason the rectangles work for area but not for length is that the exact shape of each horizontal slice is a trapezoid, and the rectangle gives a good approximation of the area of the trapezoid but not a good approximation of the length of the slanted side.

When we fill the triangle with thin rectangles, the "missing" parts (the parts of the trapezoids that are outside the rectangles) form a sequence of small triangles along the hypotenuse of the large triangle. As we slice the triangle into greater numbers of slices, the total area of all these triangles goes to zero, but the hypotenuse of every triangle is still $\sqrt2$ times its height, and adding up the heights will always give you $1/\sqrt2$ times the actual length of the hypotenuse.

In your cone, when you fill it with cylindrical slices, the total volume in the "missing" parts of the frustums gets smaller and smaller as the slices get thinner, but the lateral surface area of each frustum is nearly $S/H$ times that of the cylinder inside it (for most of the frustums) and never less than the lateral area of the cylinder, so the total lateral area of the cylinders never approaches the area of the cone.


Your confusion comes from the fact that the element of surface area of a cone is not

$dS=2\pi r dh$

(where $dh$ is the increment in the height of the cone) but

$dS=2\pi r dk$ where

$dk=\frac{dh}{\sin \alpha}$ is the height of the area element along the surface of the cone.

  • $\begingroup$ Yes that's what I wanted to ask why we take a frustum as the element of surface area but while calculating volume taking a cylindrical element works out nicely. $\endgroup$ Aug 4, 2018 at 3:21

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