Volume of sphere using integration?

[Prasun-rick]

Is it possible to find the volume of a sphere(i know the formula) using definite integration ? And if possible how to proceed ??
Thanks in advance

 

[BvU]

Is it possible to find the volume of a sphere(i know the formula) using definite integration ? And if possible how to proceed ??
Thanks in advance

Hello Prasun-rick,

Wat is the formula you know ? And what do you know of integration ? Does ##\iiint dV ## mean anything to you ?

 

[Prasun-rick]

Well I only know the geometrical formula of volume of the sphere (i.e frac{4/3}pi*r^3)..and I only happen to know integration in one dimension ..though your integral didn't make any sense ! What I will have to do to understand that??

 

[BvU]

In that case (working the three dimensional integral into a one-dimensional integral): do you know the volume of a disk that is obtained by revolving a rectangle around one of its sides ? (disk thickness a and radius b when revolving around the x-axis in the figure below)

 

[Prasun-rick]

No sir I am afraid I don't know that ! Maybe the derivation is above my scope ! Btw thanks for your valuable comments ! I will get back when I have done the volume Integral of the disk to you for further discussion .

 

[BvU]

https://www.physicsforums.com/attachments/104413
Maybe I confused you. The result of this revolution is a disk like this and that volume is relatively easy to express in a and b ...

 

[Prasun-rick]

https://www.physicsforums.com/attachments/104413
Maybe I confused you. The result of this revolution is a disk like this and that volume is relatively easy to express in a and b ...

Can you just pose the integral equation of the volume of the figure you posted !

 

[BvU]

Can you just pose the integral equation of the volume of the figure you posted !

It's not an integral. It's a disk. I lost the picture, here it is again, with the question: what is the volume, expressed in a and b ?:

 

[Prasun-rick]

Will it be pi*b^2*a??

 

[BvU]

It certainly is ! Now we are going to slice a sphere with radius r into disks of thickness ##dx##. What will be the volume of the disk at ##x## ?

 

[BvU]

Maybe this is better :

 

[Prasun-rick]

∝

Maybe this is better :
View attachment 104424

Maybe pi*(r^2-x^2)*dx ??

 

[BvU]

Perfect. Next step: we are going to add up all these disks from -r to +r and if we take the limit for ##dx\downarrow 0## we get an integral. Can you write down that integral ? (it's an easy question, because you have almost all of it already...)

 

[Prasun-rick]

Perfect. Next step: we are going to add up all these disks from -r to +r and if we take the limit for ##dx\downarrow 0## we get an integral. Can you write down that integral ? (it's an easy question, because you have almost all of it already...)

yeah maybe ∫pi*(r^2-x^2)dx with integral limits from -r to r ??

 

[BvU]

Bingo

 

[Prasun-rick]

Bingo

Thanks

 

[Prasun-rick]

Thanks

But what is that

/iiint dV

and how to solve it ??

 

[BvU]

It means you want to sum up infinitesimal volume elements ##dV##. A volume like a sphere has three dimensions, so three integrations are necessary. In cartesian coordinates you get the volume of a cube at the origin with size ##a## from $$
\int\limits_0^a \int\limits_0^a\int\limits_0^a dxdydz = a^3$$For a sphere the limits are unwieldy in cartesian, but comfortable in spherical coordinates. A volume element is ##r^2drd\phi d\theta## (##\theta## is azimuthal) so you get $$
\int\limits_0^{2\pi} \int\limits_0^\pi \int\limits_0^R \; r^2 \sin\theta \;dr \;d\theta \; d \phi = \\
2\pi \int\limits_0^\pi \int\limits_0^R \; r^2 \sin\theta \;dr \; d\theta =
4\pi \int\limits_0^R \; r^2 \; dr = \ ...
$$

(There are alternative notations, like ##\ \int\limits_{\rm Volume} d^3 V\ ##)

[edit] corrected order of ##d## in expressions but I think I still have it wrong. Need to check if we work outside in or inside out

 

[Prasun-rick]

It means you want to sum up infinitesimal volume elements ##dV##. A volume like a sphere has three dimensions, so three integrations are necessary. In cartesian coordinates you get the volume of a cube at the origin with size ##a## from $$
\int\limits_0^a \int\limits_0^a\int\limits_0^a dxdydz = a^3$$For a sphere the limits are unwieldy in cartesian, but comfortable in spherical coordinates. A volume element is ##r^2drd\phi d\theta## (##\theta## is azimuthal) so you get $$
\int\limits_0^{2\pi} \int\limits_0^\pi \int\limits_0^R \; r^2 \sin\theta \;d\phi d\theta dr = \\
2\pi \int\limits_0^\pi \int\limits_0^R \; r^2 \sin\theta \;d\theta dr =
4\pi \int\limits_0^R \; r^2 \; dr = \ ...
$$

(There are alternative notations, like ##\ \int\limits_{\rm Volume} d^3 V\ ##)

Is it okay for me to learn triple integral now?? And if yes then where to start??

 

[Prasun-rick]

Is it okay for me to learn triple integral now?? And if yes then where to start??

And please teach me how to give such prominent integral sign like the ones you are typing
!?

 

[BvU]

You get to see the ##\LaTeX## source if you right-click on a formula and choose 'Show math as ##TeX## commands' ...

Is it okay for me to learn triple integral now?? And if yes then where to start??

Depends on what book or curriculum you are following; hard for me to answer. But it's always good to satisfy one's curiosity (is my opinion).