# volume of a sphere

Discussion in 'Homework Help' started by JasonL, Oct 20, 2013.

1. ### JasonL Thread Starter Active Member

Jul 1, 2011
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Calculate the volume of a sphere of the sphere, x$^{2}$+y$^{2}$+z$^{2}$=16 cut by the planes y=0, z=2, x=1, x=-1.

The problem I have is determining the limits of the integral. I can integrate, its just the setup that is troubling me. I know in Cartesian i would have to integrate dx dy dz. I would think the integral would be,
$\int^{2}_{?}$$\int^{?}_{0}$$\int^{1}_{-1}dxdydz$.

I tried converting everything to spherical coordinates
x$^{2}$+y$^{2}$+z$^{2}$=16 $\rightarrow$ r = 4
y=0 $\rightarrow$ rsinθsin$\phi$ = 0
z=2 $\rightarrow$ rcos$\phi$=2
x=1$\rightarrow$ rsinθcos$\phi$=1
x=2$\rightarrow$ rsinθcos$\phi$=2

$\int^{?}_{?}$$\int^{?}_{?}$$\int^{4}_{0}r^{2}sindr(d theta)(d phi)$

But now, I'm not sure what are the limits for θ and $\phi$

2. ### studiot AAC Fanatic!

Nov 9, 2007
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You seem to have missed the function from the cartesian integral?

As to the cartesian limits, what is the relation between the radius and the limits?

3. ### JasonL Thread Starter Active Member

Jul 1, 2011
47
0
I know the radius of the sphere is 4.
Would the function for the radius be something like this?
x$^{2}$+y$^{2}$ = 16-z$^{2}$
x$^{2}$+z$^{2}$ = 16-y$^{2}$
z$^{2}$+y$^{2}$ = 16-x$^{2}$

Nov 9, 2007
5,005
523
5. ### WBahn Moderator

Mar 31, 2012
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On question is which portion of the sphere are you trying to find the volume of? There are four volumes that use those four surfaces to define them. By symmetry, there are two that are different.

You need to sketch out the sphere and the planes that are cutting it. Doing so in 3D is tricky unless you are either an artist or have appropriate software, but what you can do it draw 2-D sections at different values of the third parameter. Try doing this for each of the three coordinates in turn and you will probably quickly spot which is likely to be the easiest to use as the final (outside) integral and it's limits of integration should become pretty obvious. Then you can look at the 2-D sections and determine the limits of the other two and where they change behavior.

A really good exercise would be to do the integration with all six possible orders of integration and showing that you get the same answer.

6. ### studiot AAC Fanatic!

Nov 9, 2007
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Sorry I missed where you said the radius is 4, that's correct.

OK so what is another name for the plane y=0?

and what does this plane do to the sphere when it cuts it?

Last edited: Oct 20, 2013
7. ### JasonL Thread Starter Active Member

Jul 1, 2011
47
0
y=0 is the XZ plane, it cuts the sphere in half.

8. ### studiot AAC Fanatic!

Nov 9, 2007
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Yeah.

So you have half a sphere sitting on the xz plane.

That's easier to sketch, yes?

Just sketch the whole bowl for the minute and don't worry about the other planes.

Now what is the maximum and minumum values of y for this upturned bowl ?

9. ### JasonL Thread Starter Active Member

Jul 1, 2011
47
0
The maximum would be y=0 and minimum would be y=-4.

10. ### studiot AAC Fanatic!

Nov 9, 2007
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OK you have to choose one half, either minus 4 or plus 4.

Now before you add in the other planes what are the max and min values that x and z can take?

11. ### JasonL Thread Starter Active Member

Jul 1, 2011
47
0
Before adding in the planes, x can range from 0 to 4, and z can range from -4 to 4.

Last edited: Oct 20, 2013
12. ### studiot AAC Fanatic!

Nov 9, 2007
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Yes x can range from -4 to +4 for the (hemi)sphere.

Not sure what the first part of your post meant, did you get the right variable?

13. ### JasonL Thread Starter Active Member

Jul 1, 2011
47
0
sorry, typo. I meant x can be from 0 to 4, and z can be from -4 to 4. Maybe i got the limits switched?

14. ### studiot AAC Fanatic!

Nov 9, 2007
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Yes z can be from -4 to +4 as well.

Why do you think x can't range from -4 to +4?

you have a sphere or hemisphere with centre at the origin.

15. ### JasonL Thread Starter Active Member

Jul 1, 2011
47
0
Actually, nevermind I drew the XY plane wrong earlier. Z is from 0 to 4 and X is -4 to 4. I drew a crude picture on paint.

Thanks a lot for the help so far.

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16. ### JasonL Thread Starter Active Member

Jul 1, 2011
47
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Wow actually, sorry scratch that. I see it now, they both range from -4 to 4.

17. ### studiot AAC Fanatic!

Nov 9, 2007
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The section of a hemisphere sat on the x z plane is a circle of radius 4.

It has 4 cardinal points (0,4), (0,-4), (4,0) and (-4,0)

That is the only way to get a full circle of radius 4.

So -4< x <4 and -4< z <4

These limits will change when we add in the other three planes and chop our sphere down even further.

OK I see you have done that whilst I was posting.

So add in the planes x=1 and x=-1.

these are parallel to which plane (through the origin)?

18. ### WBahn Moderator

Mar 31, 2012
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But the range of x is a function of y and z while the range of z is a function of x and z. Ignoring that can lead one to use -4 and 4 as the limits of integration, which won't work.

19. ### JasonL Thread Starter Active Member

Jul 1, 2011
47
0
x=1 and x=-1 are parallel to the YZ plane

20. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
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Because they are parallel to this plane they are independent of y and z so you can substitute the values into the inequalities defining the limits for x.

Can you follow this on your sketch and see where these planes cut the hemisphere?

You should then be able to do the same for the one plane that is specified parallel to the xy plane. This means that there is no cutting limit on the part of the figure that lies on the negative z axis.

So can you see what the end result on the inequalties defining the limits for z is?