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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

7 Answers 7

See the discussion to a previous question here which might help

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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:

alt text

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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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 '13 at 17:53

When ever you use calculus you need to start with the correct mathematical relationships. You further need to ensure that you integrate over the correct limits and over the correct variables.

You can see integration as the summation of a lot of intesimal small items to give the final answer.

All correctly setup integrals will result in the same answer.

Before one start or rush into solving a problem, think what would be the easiest way to address the problem.

I thought approaching it as a triangle would be quite easy to solve the problem. So this solution will use a triangle as the basic mathematical relationship.

We all know the area of a triangle is:

$$Area = \frac{1}{2}Base * Height$$

I will thus setup the problem as follows. (Solution 1)

Define a triangle with a height of S (defined as the length from point to side of base) and a base of an infinite small length of the base and call it dbase.

As we know that radians is defined in such a way that the circumference of a circle will yield exactly $$2.\pi$$ for a circle of unity (r=1). This relationship helps us to define dbase then as a plain ratio of an infinitely small angle multiplied by the radius. In this case the radius of the cone. Thus we can say:

$$dbase = R. d angle$$

The area of my triangle would then become:

$$darea = \frac{1}{2} S. R. dangle$$

We will integrate between the limits 0 and 2pi to include the total circumference. In degrees this would have been 0 to 360, however the relationship we use is defined for radians. (Note that I couldn't get the upper limit as $$2\pi$$ and thus integrate over half the circumference and multiply the result by 2. I am not familiar with the syntax. The answer however is the same. In some instances we use this as a trick to reduce the effort in calculating the result of integrals etc. )

$$Area = 2\int_0^ \pi \frac{1}{2} S. R. dangle$$

$$Area=2.(\frac{1}{2} S. R . \pi)-2.(\frac{1}{2} S. R . (0))$$

$$Area =\pi S R$$

And the result is exactly as the rest of the world believe the answer should be.

Important is to be always mathematically correct.

In answers above you made basic mathematical errors and end up with wrong answer.

If you follow maths you can't go wrong. If you went wrong you made a mistake, search your assumptions and relationships and try gain.

Try the volume yourself. I did this on my phone and it is somewhat difficult.

Solution 2 - integrate from 0 to H, where H is height of the cone and R the radius.

h =height, dheight is very small increase in height and r is the radius at h.

Angle is the angle between the vertical and side length S.

The mathematical relationship for the area is the circumference multiple by the small increase in the length of S. Note that the length is not dh but do because of the angle. Be very careful here.

$$darea = 2\pi . r. ds$$

To setup the integral correct we need to express r and ds in terms of h to ensure consistency. The relationship is as follows:

$$tan(angle) = r /h $$ $$r = h. tan(angle) $$


$$cos(angle) = dh/ds $$ $$ds = dh. Cos(angle) $$

darea becomes then

$$darea =2\pi.tan(angle). cos(angle) .h.dh$$

We just need to integrate this as:

$$Area = 2\pi. tan(angle). cos(angle) \int_0^ H h. dh$$

$$Area = 2\pi. tan(angle).cos(angle) (( \frac{1}{2} H^2)-(\frac{1}{2} (0)) $$

$$Area = \pi. tan(angle).cos(angle) H^2$$

We only need to get rid of the tan and cos and then we should have the same answer. From the above we can substitute back to get:

$$Area = \pi. S. R$$

It worked again, no magic.

Solution 3 - integrate from 0 to S. Same variables as in solution 2.

Area definition same. This time we need to express r in terms of s.

$$darea = 2\pi . r. ds$$

$$sin(angle) = r /s $$ $$r = s. sin(angle) $$

Replace r in darea, then integrate

$$Area = 2\pi. sin(angle) \int_0^ S s. ds$$

$$Area = 2\pi. sin(angle) (( \frac{1}{2} S^2)-(\frac{1}{2} (0)) $$

$$Area = \pi.sin(angle) S^2$$

Get rid of the sin term and can you believe, once again the same result as the rest of the world.

$$Area = \pi. S. R$$

I love maths.

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Democritus of Abdera, the ancient philosopher of Greece, reasoned over two thousand years ago, in verity there are only 'atoms and void'. He argued, "If a cut were made through a cone parallel to its base, how should we conceive of the two opposing surfaces which the cut has produced - as equal or as unequal? If they are unequal, that would imply (the lateral side of) a cone is composed of many breaks and protrusions like steps. On the other hand if they are equal, that would imply that two adjacent intersecting planes are equal, which would mean a right cone, being made up of equal rather than unequal circles, must have the same appearance as a cylinder; which is utterly absurd."

Consider the following diagram,

enter image description here

By similar triangles, $${H \over S} = {{dy} \over {ds}}$$ Solve for ds, $$ds = {{S(dy)} \over H}$$ The equation of line S, $$y = mx + b$$ where, $$m = - {H \over R}$$ and $$b = H$$ Therefore, $$y = - {H \over R}x + H$$ Next, solve for $x$ $$x = {{ - R(y - H)} \over H}$$ Next, integrate the circles of radii $x$ and thickness differential $ds$ from $y$ = zero to H, so the surface area of a right cone is given by, $$A = 2\pi \int\limits_0^H x ds$$ Substitute $x$ into the integral to get, $$A = 2\pi \int\limits_0^H {{{ - R(y - H)} \over H}} ds$$ Now substitute into the previous equation $$ds = {S \over H}dy$$ To obtain, $$A = 2\pi \int\limits_0^H {{{ - R(y - H)} \over H}} \left( {{S \over H}dy} \right)$$ Cleaning up this equation yields $$A = {{2\pi ( - R)S} \over {{H^2}}}\int\limits_0^H {(y - H)dy} $$ $$A = {{2\pi ( - R)S} \over {{H^2}}}\left[ {{{{y^2}} \over 2} - Hy} \right]_0^H$$ $$A = {{2\pi ( - R)S} \over {{H^2}}}\left[ {{{{H^2}} \over 2} - {H^2}} \right]$$ $$A = {{2\pi ( - R)S} \over {{H^2}}}\left[ {{{{H^2}} \over 2} - {{2{H^2}} \over 2}} \right]$$ $$A = {{2\pi ( - R)S} \over {{H^2}}}\left[ {{{ - {H^2}} \over 2}} \right]$$ $$A = {{2\pi RS} \over {{H^2}}}\left[ {{{{H^2}} \over 2}} \right] = \pi RS$$

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