# I Volume of sphere using integration??

1. Aug 7, 2016

### 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 ??

2. Aug 7, 2016

### BvU

Hello Prasun-rick,

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

3. Aug 7, 2016

### 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??

4. Aug 8, 2016

### 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)

5. Aug 8, 2016

### 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 .

6. Aug 8, 2016

### BvU

View attachment 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 ...

7. Aug 8, 2016

### Prasun-rick

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

8. Aug 8, 2016

### BvU

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 ?:

9. Aug 8, 2016

### Prasun-rick

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

10. Aug 8, 2016

### 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 $x$ ?

11. Aug 8, 2016

### BvU

Maybe this is better :

12. Aug 8, 2016

### Prasun-rick

13. Aug 8, 2016

### 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...)

14. Aug 8, 2016

### Prasun-rick

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

15. Aug 8, 2016

### BvU

Bingo

16. Aug 8, 2016

### Prasun-rick

Thanks

17. Aug 9, 2016

### Prasun-rick

But what is that
Code (Javascript):
/iiint dV
and how to solve it ??

18. Aug 9, 2016

### 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\$)

 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

Last edited: Aug 9, 2016
19. Aug 9, 2016

### Prasun-rick

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

20. Aug 9, 2016

### Prasun-rick

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