spherical cordinates

[aronclark]

Describe the surface above the cone and below the sphere
x^2+y^2=z
x^2+y^2+z^2=z
If a is the angle off the z axis and p is the spherical radius I have
0<=a<=Pi/4 and cosa<=p
My question is what does this describe it appears that it is only the cap of the sphere and is there a better way to write an inequallity to describe p

 

[Deleted member 4993]

Describe the surface above the cone and below the sphere
x^2+y^2=z
x^2+y^2+z^2=z^2
If a is the angle off the z axis and p is the spherical radius I have
0<=a<=Pi/4 and cosa<=p
My question is what does this describe it appears that it is only the cap of the sphere and is there a better way to write an inequallity to describe p

Fix that!!

 

[aronclark]

Fixed
So my question .is that just the cap of the sphere it appears that it would be to me since the description of p came only fro. the equation of the sphere also is there a better way to write that description in terms of p?

 

[aronclark]

Anyone?

 

[aronclark]

What throws me off that a cone made simply from 0<=t<=Pi/4
Would bit be the same as the cone x^2+y^2=z ir parabolas the inequallity of p will not describe the cone since it was durived from the sphere so for p on the sphere this must not describe any points in the cone yet this us our solution for above the cone under the sphere

 

[HallsofIvy]

The only "complication is that the sphere is centered at \(\displaystyle (0, 0, \frac{1}{2})\) rather than (0, 0, 0). In spherical coordinates, the sphere is given by \(\displaystyle x^2= y^2+ z^2= \rho^2= z= \rho cos(\phi)\) or \(\displaystyle \rho= cos(\phi)\). The cone is given by \(\displaystyle x^2+ y^2= \rho^2 sin^2(\phi)= z= \rho cos(\phi)\) or \(\displaystyle \rho= \frac{cos(\phi)}{sin^2(\phi)}\).

The volume of the region above the cone but below the sphere is given by
\(\displaystyle \int_{\theta= 0}^{2\pi}\int_{\phi= 0}^{\pi/4}\int_{\rho= \frac{cos(\phi)}{sin^2(\phi)}}^{cos(\phi)} \rho^2 sin(\phi)d\rho d\phi d\theta\)