# Integrating 2*pi*R and pi*R^2

• Anzas
In summary, the conversation discusses the concept of integrating 2πr and its relation to finding the volume of a half ball. It is explained that the volume integral takes into account the azimuthal and polar angles, while the area integral only considers the radius. Integrating by the diameter gives the correct formula for the volume of a half ball, but integrating by the radius does not take into account the curvature of the ball. The conversation also touches on the difference between a hemisphere and a half ball, and the use of spherical coordinates in mathematical calculations.

#### Anzas

i noticed that if i integrate
$$2 \pi r$$
i get
$$\int 2 \pi r dr=\pi r^2$$
i figured its because the area of a circle can be seen as the sum of circumference's of circles with radius 0 to radius $$r$$
i was thinking if the half volume of a ball also be seen as made from the sum of areas of circules with radius 0 to radius $$r$$?
integrating $$\pi r^2$$ gives $$\pi \frac{r^3}{3}$$
so multiplying by two should give the ball volume formula but it does not why is that?

The the area integral is a double integral over r and the azimuthal angle. The volume integral is over r, the azimuthal angle and the polar angle. You're missing some parts of the "big picture" about spherical coordinates.

You kind of skipped the intergrating over the azimuthal angle part by including $$2 \pi$$ straight into your integrand. Check http://mathworld.wolfram.com/Sphere.html and http://mathworld.wolfram.com/SphericalCoordinates.html for details.

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In other words, the derivative of the area of a circle, $$\pi r^2$$ ,with respect to r, is the circumference of the circle $$2\pi r$$!

Your formula for the volume of a sphere is incorrect. The volume of a sphere is $$\frac{4}{3}\pi r^3$$ and its derivative is indeed the surface area of the sphere: $$4\pi r^2$$

alright then i have another question
would it be correct to define a half of a ball with diameter D as the set of all circles whose diameters range from 0 to D?

No.That would be a hemisphere.A half of a ball would be a continuous array of disks whose centers lie on a line segment of length R (radius of the ball).

Daniel.

aren't a hemi sphere and a half ball the same things?
maybe i didn't form the question correctly by circles i mean circles with their area not just the circumference i suppose one could call it a disk.
ill reform the question:
can a half ball with diameter D be defined as a set of all discs with diameter's ranging from 0 to D and parallel to each other.

Anzas said:
aren't a hemi sphere and a half ball the same things?

No.
Anzas said:
maybe i didn't form the question correctly by circles i mean circles with their area not just the circumference i suppose one could call it a disk.

One calls that a disk...

Anzas said:
ill reform the question:
can a half ball with diameter D be defined as a set of all discs with diameter's ranging from 0 to D and parallel to each other.

Yes,but it's highly reccomendable to define it as a half of a ball. :tongue2:

Daniel.

im not just asking these questions for nothing I am trying to get to a point
alright then if a half of a ball can be defiend this way and $$A(D)= \pi (\frac{D}{2})^2$$ is a function which gives the area of a disk of diameter D then shouldn't its integral $$\int A(D)$$ which is actually the sum of all the areas of all the discs with diameters ranging from 0 to D which make the half ball give the volume formula of a half ball of diameter D?

It does,doesn't it...?

$$\int_{0}^{D} \pi\frac{D^{2}}{4} \ dD=\pi\frac{D^{3}}{12}=\frac{2\pi R^{3}}{3}=V_{\mbox{half ball}}$$

Daniel.

o right i didnt notice that.
how come if you integrate $$\pi R^2$$ it doesn't come out right what's the diffrence between integrating by the diameter or by the radius?

You've been explained that u should normally integrate the area of the surface (sphere) enclosing the volume (ball) to get it right.It works for sphere-ball and their 1D-2D counterparts:circle-disk...It doesn't for cube surface-cube and neither for square contour-square...

Daniel.

i understand that integrating the surface will work but I am trying to understand why integrating the area function of a disk by diameter gives the correct formula for the volume of a half of a ball but integrating by the radius doesn't even though the idea is the same (adding up the areas of the discs that make the half ball).

Because when you set up the equation for the integral according to the radii, you did not take into account the curvature of the ball... The radius of the disks do not have a linear relationship with the distance from the center (or from the top). You derived a formula of a cone, which is also a bunch of discs centered on the same axis. The difference between the cone and the half-ball is that the half ball bulges out like a circle, while the cone stays linear.

Anzas said:
i understand that integrating the surface will work but I am trying to understand why integrating the area function of a disk by diameter gives the correct formula for the volume of a half of a ball but integrating by the radius doesn't even though the idea is the same (adding up the areas of the discs that make the half ball).

It's because you didn't the change of variable correctly : D=2r->dD=2dr
boundary : D_0=2R

$$\int_0^{D_0}\pi D^2/4dD=\int_0^{2R}2\pi r^2dR=2/3\pi r^3$$

It is just some 'curiosity' that this actually works:
"The sum of all the areas of all the discs with diameters ranging from 0 to D" gives the volume of half a ball

$$\int_{0}^{D} \pi\frac{D^{2}}{4} \ dD=\frac{1}{12}\pi D^3=\frac{2}{3}\pi R^3=V_{\mbox{half ball}}$$

or similarly,

$$\int_{0}^{2R} \pi R^2 \ dR=\frac{2}{3}\pi R^3=V_{\mbox{half ball}}$$

but this is not based on proper reasoning and it does not take the curvature of the ball into account (As Moo of Doom seems to suggest) it is the volume of a cone with base radius R and height 2R (which is the same as the volume of half a ball with radius R)

"The sum of all the areas of all the discs with radi ranging from 0 to R" only gives the volume of a quart of a ball (i.e. of a cone with base radius R and height R)

$$\int_{0}^{R} \pi R^2 \ dR=\frac{1}{3}\pi R^{3}=V_{\mbox{quart ball}}$$

I suggest to go through (Centroid)Theorems of Pappus.

## 1. What is the formula for integrating 2*pi*R and pi*R^2?

The formula for integrating 2*pi*R and pi*R^2 is ∫ 2*pi*R dR = pi*R^2 + C, where C is the constant of integration.

## 2. What does integrating 2*pi*R and pi*R^2 represent?

Integrating 2*pi*R and pi*R^2 represents finding the area under the curve of a circle with radius R. This can be visualized as the process of finding the total amount of space inside the circle.

## 3. How is the integration of 2*pi*R and pi*R^2 related to the circumference and area of a circle?

The integration of 2*pi*R and pi*R^2 is directly related to the circumference and area of a circle. The integral of 2*pi*R represents the circumference of a circle, while the integral of pi*R^2 represents the area of a circle.

## 4. Can the integration of 2*pi*R and pi*R^2 be used to find the volume of a sphere?

No, the integration of 2*pi*R and pi*R^2 only applies to circles. To find the volume of a sphere, the formula V = (4/3)*pi*R^3 can be used.

## 5. Are there any real-world applications of integrating 2*pi*R and pi*R^2?

Yes, integrating 2*pi*R and pi*R^2 has many real-world applications. For example, it can be used in physics to calculate the work done by a rotating object, or in engineering to determine the surface area of a curved structure.