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Volume of a solid by triple integration

  • Thread starter kasse
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  • #1
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Homework Statement



Find the volume of the solid inside the sphere x^2 + y^2 + z^2 = 4 and over the paraboloid 3z = x^2 + y^2

The Attempt at a Solution



This should be easy to calculate using polar coordinates. The limits for z is [r^2/2, sqrt(4-r^2)] and for tetha: [0, 2*pi], but how do I find the limits for r? The intersection between the two surfaces is: sqrt(4 - r^2) = r^2/3. By inspection I can see that the answer is r = sqrt(3), but what is the mathematical method to find this limit?
 

Answers and Replies

  • #2
Defennder
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You meant cylindrical coordinates, right? Firstly you have the equations r^2 + z^2 = 4 and 3z = r^2. Express z in terms of r and substitute it into the other equation. Then solve for r. You'll get the answer.
 
  • #3
HallsofIvy
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Homework Statement



Find the volume of the solid inside the sphere x^2 + y^2 + z^2 = 4 and over the paraboloid 3z = x^2 + y^2

The Attempt at a Solution



This should be easy to calculate using polar coordinates. The limits for z is [r^2/2, sqrt(4-r^2)]
There must be a typo here. You said the parabola was given by 3z= r2.

and for tetha: [0, 2*pi], but how do I find the limits for r? The intersection between the two surfaces is: sqrt(4 - r^2) = r^2/3. By inspection I can see that the answer is r = sqrt(3), but what is the mathematical method to find this limit?
Don't look at z, look at z2. z= r2/3 implies z2= r4/9. So you have r4/4/9= 4- r2 or r4+ 9x- 36= 0.

That can be factored as (r2- 3)(r2+ 12)= 0 so r2= 3 or r2= -12. The four roots to that equation are [itex]r= \pm\sqrt{3}[/itex] and [itex]r= \pm 2i\sqrt{3}[/itex]. Of course, r must be positive so [itex]r= \sqrt{3}[/itex].
 

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