Surface of a cone by integration

In summary, we discussed the different methods for calculating the volume and surface area of a cone. While infinitesimal disks can be used for volume calculations, they are not accurate for surface area calculations due to the slanting of the cone. Instead, infinitesimal conical frustums must be considered and integrated using the formula for lateral surface area. We also discussed the importance of ensuring our estimates are valid to the first order in infinitesimals.
  • #1
themagiciant95
57
5
If i want to calculate the volume of a cone i can integrate infinitesimal disks on the height h of the cone.

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I was told that if i want to calculate the surface of the cone, this approximation is not correct and i have to take the slanting into account, this means that instead of infinitesimal disk i have to consider infinitesimal conical frustum and integrate the lateral surface of each infinitesimal conical frustum. Why for the surface i can't use infinitesimal disks as for the volume calculation ?

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Furthermore, the formula for the lateral surface of a conical frustum is :

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Now, let's consider an infinitesimal conical frustum.
For [itex]h\rightarrow 0[/itex] , [itex]R_{1}\approx R_{2}[/itex] so [itex](R_{1} - R_{2})^{2} \rightarrow 0[/itex] and the surface of an infinitesimal conical frustum would be[itex]2\pi R_{2}\,dh[/itex], however i was told that it's not correct. The correct one is [itex]2\pi R_{2}\sqrt{ d(R_{2}-R_{1})^{2}+ dh^2}[/itex], why?
 
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  • #2
As for surfaces of full cone
The bottom circle pf radius r:
[tex]A_b=\pi r^2[/tex]
The side partial circle for cone height h, so with central angle ##\phi##
[tex]A_s=\pi (r^2+h^2) \frac{\phi}{2\pi} [/tex]
the bottom circle and the side partial circle share the periphery
[tex]2\pi r = \phi \sqrt{r^2+h^2} [/tex]
[tex]\phi=\frac{2\pi r}{\sqrt{r^2+h^2}}[/tex]
So
[tex]A_s=\pi r \sqrt{r^2+h^2}[/tex]
From this fulll cone case you can deduce the answer for your case of cut parts.

themagiciant95 said:
For h→0h\rightarrow 0 , R1≈R2R_{1}\approx R_{2} so (R1−R2)2→0(R_{1} - R_{2})^{2} \rightarrow 0 and the surface of an infinitesimal conical frustum would be2πR2dh2\pi R_{2}\,dh, however i was told that it's not correct. The correct one is 2πR2√d(R2−R1)2+dh22\pi R_{2}\sqrt{ d(R_{2}-R_{1})^{2}+ dh^2}, why?

The square of infinitesimal appears in square root. So it should be treated not second but first degree infinitesimal in the formula.
 
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  • #3
themagiciant95 said:
If i want to calculate the volume of a cone i can integrate infinitesimal disks on the height h of the cone.

View attachment 259410

View attachment 259413
View attachment 259414

I was told that if i want to calculate the surface of the cone, this approximation is not correct and i have to take the slanting into account, this means that instead of infinitesimal disk i have to consider infinitesimal conical frustum and integrate the lateral surface of each infinitesimal conical frustum. Why for the surface i can't use infinitesimal disks as for the volume calculation ?

View attachment 259411

Furthermore, the formula for the lateral surface of a conical frustum is :

View attachment 259412

Now, let's consider an infinitesimal conical frustum.
For [itex]h\rightarrow 0[/itex] , [itex]R_{1}\approx R_{2}[/itex] so [itex](R_{1} - R_{2})^{2} \rightarrow 0[/itex] and the surface of an infinitesimal conical frustum would be[itex]2\pi R_{2}\,dh[/itex], however i was told that it's not correct. The correct one is [itex]2\pi R_{2}\sqrt{ d(R_{2}-R_{1})^{2}+ dh^2}[/itex], why?

It's an interesting question of how you know when these estimates are correct. Let's look at the two cases of the volume and the area. If we look at the volume, we can see that:
$$\pi R_1^2 h < V < \pi R_2^2 h$$
If we now imagine an infinitesimal frustrum of height ##dh##, we see that:
$$\pi r^2 dh < dV < \pi (r + dr)^2 dh = \pi (r^2 +2rdr + dr^2)dh$$
And, as we can neglect higher order terms, this gives us a valid equation for the infinitesimal volume element:
$$dV = \pi r^2 dh$$
When we look at the area, however, we see that:
$$2\pi R_2 s < A < 2\pi R_1 s$$
And, for an infinitesimal frustrum:
$$2\pi r s < A < 2\pi (r+dr)s$$
And, we can see that:
$$s = \frac l h dh$$
Where ##l## is the slant height of the full cone.
This gives us the equation for the infinitesimal surface area element:
$$dA = 2\pi r \frac l h dh = \frac{2\pi l}{R}rdr$$
The moral is that you really have to ensure that your estimates are valid to the first order in the infinitesimals.
 
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  • #4
themagiciant95 said:
Now, let's consider an infinitesimal conical frustum.
For h→0h\rightarrow 0 , R1≈R2R_{1}\approx R_{2} so (R1−R2)2→0(R_{1} - R_{2})^{2} \rightarrow 0 and the surface of an infinitesimal conical frustum would be2πR2dh2\pi R_{2}\,dh, however i was told that it's not correct. The correct one is 2πR2√d(R2−R1)2+dh22\pi R_{2}\sqrt{ d(R_{2}-R_{1})^{2}+ dh^2}, why?
[tex] d(R_2-R_1)^2=[d(R_2-R_1)]^2 =(dR_2-dR_1)^2 \neq d[(R_2-R_1)^2]=2(R_2-R_1)(dR_2-dR_1)[/tex]
might be of your help. RHS is first order and LHS is second order infinitesimals.
You do not have to refer the estimation
[tex](R_2-R_1)^2 \rightarrow 0[/tex]
anywhere.
 
  • #5
mitochan said:
[tex] d(R_2-R_1)^2=[d(R_2-R_1)]^2 =(dR_2-dR_1)^2 \neq d[(R_2-R_1)^2]=2(R_2-R_1)(dR_2-dR_1)[/tex]
might be of your help. RHS is first order and LHS is second order infinitesimals.
You do not have to refer the estimation
[tex](R_2-R_1)^2 \rightarrow 0[/tex]
anywhere.

For the frustrum, you can do:
$$dA = 2\pi r dl$$
And, with ##L## as the slant length of the fulcrum, we have:
$$dl = \frac{L}{R_2 - R_1}dr$$
This gives:
$$A = \int_{R_1}^{R_2} \frac{2\pi L}{R_2 - R_1} r dr = \frac{\pi L}{R_2 - R_1}(R_2^2 - R_1^2) = \pi L (R_1 + R_2)$$
And, in terms of ##h##:
$$A = \pi (R_1 + R_2)L = \pi(R_1 + R_2)\sqrt{(R_2-R_1)^2 + h^2}$$
Finally, this makes sense as:
$$A = 2 \pi R_{avg} L$$
 
  • #6
PeroK said:
It's an interesting question of how you know when these estimates are correct. Let's look at the two cases of the volume and the area. If we look at the volume, we can see that:
$$\pi R_1^2 h < V < \pi R_2^2 h$$
If we now imagine an infinitesimal frustrum of height ##dh##, we see that:
$$\pi r^2 dh < dV < \pi (r + dr)^2 dh = \pi (r^2 +2rdr + dr^2)dh$$
And, as we can neglect higher order terms, this gives us a valid equation for the infinitesimal volume element:
$$dV = \pi r^2 dh$$
When we look at the area, however, we see that:
$$2\pi R_2 s < A < 2\pi R_1 s$$
And, for an infinitesimal frustrum:
$$2\pi r s < A < 2\pi (r+dr)s$$
And, we can see that:
$$s = \frac l h dh$$
Where ##l## is the slant height of the full cone.
This gives us the equation for the infinitesimal surface area element:
$$dA = 2\pi r \frac l h dh = \frac{2\pi l}{R}rdr$$
The moral is that you really have to ensure that your estimates are valid to the first order in the infinitesimals.

Why can we neglect those higher-order terms ? Furthemore, if we do this than [itex]\pi r^2 dh < dV < \pi (r + dr)^2 dh = \pi (r^2 +2rdr + dr^2)dh[/itex] becomes pointless
ps: i think i have not understood why you have underlined this chain of disequalities
 
  • #7
themagiciant95 said:
Why can we neglect those higher-order terms ? Furthemore, if we do this than [itex]\pi r^2 dh < dV < \pi (r + dr)^2 dh = \pi (r^2 +2rdr + dr^2)dh[/itex] becomes pointless

Neglecting higher order terms is precisely what I've done. In your OP you also neglected terms that were of first order - in your infinitesimal equation for the area element.
 
  • #8
Yes but i don't understand the rules when dealing with infinitesimal quantities. It's the only thing i can't deeply understand in undergraduate physics and i haven't found a book about it.

ps: i think i have not understood why you have took care of this chain of disequalities: [itex]\pi r^2 dh < dV < \pi (r + dr)^2 dh = \pi (r^2 +2rdr + dr^2)dh[/itex] . For what purpose ?
 
  • #9
themagiciant95 said:
Yes but i don't understand the rules when dealing with infinitesimal quantities. It's the only thing i can't deeply understand in ungraduate physics and i haven't found a book about it.

ps: i think i have not understood why you have took care of this chain of disequalities: [itex]\pi r^2 dh < dV < \pi (r + dr)^2 dh = \pi (r^2 +2rdr + dr^2)dh[/itex] . For what purpose ?

That was to check that the equation ##dV = \pi r^2 dh## is valid to first order.

Whereas, you have:

themagiciant95 said:
the surface of an infinitesimal conical frustum would be [itex] 2\pi R_{2}\,dh[/itex],

But, the width of the area element is ##ds##, which is not equal to ##dh## to first order. Instead, we have:
$$ds = \frac L R dr$$
Where ##L## and ##R## are the slant height and maximum radius of the full cone.
 

Related to Surface of a cone by integration

1. What is the formula for finding the surface area of a cone by integration?

The formula for finding the surface area of a cone by integration is SA = πr√(r^2 + h^2), where r is the radius of the base and h is the height of the cone.

2. How is integration used to find the surface area of a cone?

Integration is used to find the surface area of a cone by breaking the curved surface into infinitesimally small sections and adding them together using the definite integral. This allows for a more accurate calculation of the surface area.

3. Is it necessary to use calculus and integration to find the surface area of a cone?

No, it is not necessary to use calculus and integration to find the surface area of a cone. The formula SA = πr√(r^2 + h^2) can also be used to calculate the surface area without using integration.

4. Can the surface area of a cone be found using only the height and slant height?

Yes, the surface area of a cone can be found using only the height and slant height. The formula for this is SA = πr(r + l), where r is the radius of the base and l is the slant height.

5. How can the surface area of a cone be visualized using integration?

The surface area of a cone can be visualized using integration by imagining the curved surface of the cone being divided into infinitely thin strips of equal width. These strips can then be rearranged into a flat shape, allowing for the use of integration to find the total surface area.

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