Volume of Solid within Sphere and Outside Cone | Integration Homework Solution

In summary, the problem asks to find the volume of a solid within a sphere, above the xy plane, and outside a cone. The solution can be found using spherical coordinates, with theta varying from 0 to 2pi and phi being the intersection of the sphere and cone. The conversion to spherical polar can help determine the limits of integration for phi.
  • #1
iamwilson
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0

Homework Statement



FInd the volume of the solid that lies within the sphere x^2 + y^2 + z^2 = 1, above the xy plane, and outside the cone z = 3 \sqrt{x^2+y^2}

Homework Equations





The Attempt at a Solution



i tried to use rdzdrd(theta) to integrate this question but i can't get it right, i used 0 to 2pi for theta, z is from 3r to sqrt(1-r^2), and r is from 0 to sqrt(1/10), help
 
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  • #2
I think this might be easier to do in spherical coordinates where the volume element is:

[tex] dV = r^{2}sin(\phi)dr d\theta d\phi [/tex]
 
  • #3
i also tried that way but how do i get phi though and theta
 
  • #4
You know that theta should vary from 0 to 2pi because you aren't restricted to any octant here. To find phi, you should find the intersection of the sphere and the cone. I think the easiest will be to do this in cartesian and then convert them to spherical polar. Upon the conversion, you will see at what phi this happens. Then you know that phi will vary form the point of intersection down to pi/2, which will satisfy the requirement that you are above the xy plane.
 
  • #5
so i did as u said, i changed them to spherical polar, z=3psin(phi), but x^2+y^2+z^2=1, i don't know how to change that to shperical polar, could you please help me. the homework is due tomorrow and I've a test on that day too! thanks!
 

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