- #1

MarcMTL

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## Homework Statement

Find the volume of the solid which is contained by 1) [tex]z= \frac{\sqrt{2}}{4}\sqrt{x^2+y^2}[/tex] and 2) [tex]x^2+y^2+z^2= \sqrt{27}z[/tex]

## Homework Equations

I've completed the square on the 2nd equation to obtain [tex]x^2+y^2= 8z^2[/tex]

and also the 1st equation to obtain [tex]x^2+y^2+(x-\frac{1}{2}\sqrt{27})^2=\sqrt{\frac{27}{4}}[/tex]

## The Attempt at a Solution

So we're basically looking for a sphere which is cut off at the bottom by the 1st equation. I though it'd be good to solve using spherical coordinates. For the [tex]\rho[/tex], i've obtained [tex]\rho^2 = \sqrt{27}z [/tex] and [tex]\rho = 27cos(\phi) [/tex]

[tex]\theta[/tex] from 0 to 2[tex]\pi[/tex]

But I can't for the life of me figure out how to find [tex]\phi[/tex]? I figure it wont go from 0 to [tex]\pi[/tex] because of the bottom limit.

As well, is using spherical the best solution in this case?

Thanks,

Marc

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