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Triple integral?

  • Thread starter pivoxa15
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Find the volume of the solid inside the cylinder x^2+y^2=2x
and bounded by the sphere x^2+y^2+z^2=4

It appears that cyclindrical is out the question because there is no symmetry about the centre of the cylinder. So only spherical coords are applicable. Any clues?
 
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Have you tried making a rough sketch of the described solid?
 

GCT

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I think that you can solve this with a double integral. The cylinder describes the x and y boundaries within the sphere, that is, the top and bottom boundaries are defined by the cylinder. Define the boundaries, x and y, with respect to the polar coordinates. Then use the equation of the sphere with respect to z for the function, and subsequently the polar coordinate transformation.
 
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I wonder if someone's going to post the n-fold integral question any time.:rolleyes:

Anyway, cylindrical is not out of the question. In fact, it is far more easier to use cylindrical than any other coordinate system for this question. The bounds shouldn't be too difficult to obtain if you draw the xy plane showing what the circle looks like. By the way, when doing these sorts of problems don't always look for symmetry. A lot of the time the trick is to just get simpler expressions to work with.
 
2,234
1
Benny said:
I wonder if someone's going to post the n-fold integral question any time.:rolleyes:

Anyway, cylindrical is not out of the question. In fact, it is far more easier to use cylindrical than any other coordinate system for this question. The bounds shouldn't be too difficult to obtain if you draw the xy plane showing what the circle looks like. By the way, when doing these sorts of problems don't always look for symmetry. A lot of the time the trick is to just get simpler expressions to work with.
If that is the case, I tried putting in the values of the integral. I ended up needing to integrate 2r(sqrt(3-r^2+2rcos(theta))) with respect to r which is proving to be difficult. How would you go about this integral?
 

GCT

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pivoxa15 said:
Find the volume of the solid inside the cylinder x^2+y^2=2x
and bounded by the sphere x^2+y^2+z^2=4

It appears that cyclindrical is out the question because there is no symmetry about the centre of the cylinder. So only spherical coords are applicable. Any clues?
the equation for the cylinder becomes

simplyfing the cylinder equation results in

r=2cos@

thus the boundaries with respect to r is from 0 to 2cos@

the boundaries with respect to @ is from 0 to pi

We will need to multiply the double integral by 4, since the 0 to pi takes advantage of the x axis symmetry and also the xy plane symmetry

the equation for the function which should be included within the integral is

z=sqrt(4-x^2 -y^2), which with the polar coordinate application, becomes

z=sqrt(4-r^2)

that all, so let's see you write this up and solve the integral
 
2,234
1
GCT said:
the equation for the cylinder becomes

simplyfing the cylinder equation results in

r=2cos@

thus the boundaries with respect to r is from 0 to 2cos@

the boundaries with respect to @ is from 0 to pi

We will need to multiply the double integral by 4, since the 0 to pi takes advantage of the x axis symmetry and also the xy plane symmetry

the equation for the function which should be included within the integral is

z=sqrt(4-x^2 -y^2), which with the polar coordinate application, becomes

z=sqrt(4-r^2)

that all, so let's see you write this up and solve the integral

I think I worked it out using cyclindrical coords using essentially your method except I multiplied by 2 instead of 4. I got 16pie/3 - 64/9. If I multiply it by another factor of 2, I get a number greater than half the volume of the sphere and also greater than the full volume of the cyclinder which can't be right.

This example showed me how versatile the coords systems can be. I wonder how it is possible to do it in spherical coords.
 

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