# Setting Up an Integral to Find A Cone's Surface Area

I tried proving the formula presented here by integrating the circumferences of cross-sections of a right circular cone: $$\int_{0}^{h}2\pi sdt, \qquad\qquad s = \frac{r}{h}t$$ so $$\int_{0}^{h}2\pi \frac{r}{h}tdt.$$ Integrating it got me $\pi h r$, which can't be right because $h$ isn't the slant height. So adding up the areas of differential-width circular strips doesn't add up to the lateral surface area of a cone?

EDIT: I now realize that the integral works if I set the upper limit to the slant height - this works if I think of "unwrapping" the cone and forming a portion of a circle. The question still remains though: why won't the original integral work? Won't the value of the sum of the cylinders' areas reach the area of the cone as the number of partitions approaches infinity?

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The problem is that you need to scale your surface area element appropriately, see en.wikipedia.org/wiki/Surface_integral, en.wikipedia.org/wiki/Surface_of_revolution and also en.wikipedia.org/wiki/Pappus%27s_centroid_theorem I hope you're aware of the fact that you need absolutely no calculus here: just observe that the cone can be built out of a circular sector of the plane. – t.b. Jan 20 '11 at 3:07
You can't integrate circumferences to get a surface area for the same reason you can't integrate points to get a length. – Qiaochu Yuan Jan 20 '11 at 3:31
Yuan: But the circumferences are multiplied by $dt$ – G.P. Burdell Jan 20 '11 at 3:43

You seem to be ignoring the fact that s and r vary as the segment you consider varies. By using the same variable names it appears that you are confusing them to be constants...

Anyway, for a derivation, look at the following figure:

This is a cross-section of the cone.

The area of the strip of width $\displaystyle dh$ that corresponds to $\displaystyle h$ (from the apex) is $\displaystyle 2\pi r \frac{dh}{\cos x}$

Now $\displaystyle r = h \tan x$

Thus $\displaystyle dA = 2 \pi h \frac{\tan x}{\cos x} dh$

Thus the total area

$$= \int_{0}^{H} 2 \pi h \frac{\tan x}{\cos x} dh = \pi H^2 \frac{\tan x}{\cos x} = \pi (H \tan x) \left(\frac{H}{\cos x}\right) = \pi R S$$

Hope that helps.

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See the discussion to a previous question here which might help

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I appreciate the thought process and did something similar as did by Mr G.P Burdell. To add to the confusion (I apologise!) let me put my point forward as this:

Consider a Cone with height H and radius R. Let dh and dr be the respective changes in H and R. Therefore
$$Volume = \int_0^H \pi{r^2} dh$$

Now “r” is a function of “h” so $$h = H- \frac{H}{R}r$$

$$dh = - \frac{H}{R}dr$$

Substituting the same in the above equation we get the integral as

$$\int_R^0 \pi{r^2} \frac{(-H)}{R}dr$$ $$Volume = \frac{\pi}{3}R^2H$$

If you do the same process for the surface area you will end up getting

$$Surface Area = \pi RH$$

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.

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Ok, so I've been thinking about this for a few days, and I asked the same question on physicsforums.com. And luckily, someone posted an answer that explained the problem. Here is the question that I posted: http://www.physicsforums.com/showthread.php?p=4031752. If you scroll all the way to the second to last post, you will see a person who says,

"The problem I think is that you wouldn't get the surface area if you used cylinders whose sides aren't parallel to the sides of the shape. For example, consider trying to "square" the perimeter of a circle. For illustration: http://qntm.org/trollpi

"The sides of the square are cut into many pieces, and then "steps" are created from it, but you can always combine the steps back into the side of the square. Consider the fourth picture in the link, look at the upper half of the circle. The whole upper side of the square is there, it is just in pieces. So you can jag them all you want, the perimeter is the same and doesn't start approximating a circle.

"In the cross section of the cone, considering it as a 2-d object, you can also try to calculate the perimeter of the slice. If you add up the sides of the rectangles, they will always add up to 2H, no matter how small you make them, however you can check using Pythagoras that it should be more. because its twice (for two sides of the slice) the square root of H^2+R^2"

That answer made a lot of sense to me, and I hope that it makes sense to everyone else here :)

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The diagram makes sense to me. The area around our slice clearly seems to me to be the circumference at that point multiplied by dS. Not dh, as we might be tempted to do just because dh is easily defined in terms of r. handily dS is just a constant*dr: dS=dr/cos(x)

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Welcome to MSE! It helps readability to format questions/answers using MathJax (see FAQ). Regards – Amzoti Apr 30 at 17:53