1. phi

the first equation is a sphere centered at (0,0,1/2) with radius 1/2. to see this, complete the square on the z variable $x^2 + y^2 + z^2 =z \\ x^2 + y^2 + z^2-z +(1/4) = 1/4 \\ x^2+ y^2 + (z-\frac{1}{2})^2 = \frac{1}{4}$ in cylindrical coords, $r^2 + \left(z-\frac{1}{2}\right)^2 = \frac{1}{4} \\ r= \sqrt{ \frac{1}{4} -\left(z-\frac{1}{2}\right)^2 }$

2. phi

the second equation is a cone with its apex at the origin $z= \frac{1}{\sqrt{3} }\sqrt{x^2+y^2}$ in cylindrical coords $z = \frac{r}{\sqrt{3} } \\ r = \sqrt{3} \ z$

3. phi

the composite shape is similar to an ice-cream cone: |dw:1438214470899:dw|

4. phi

the cone intersects the sphere at a particular z. fix y=0. the second equation gives us $$x = \sqrt{3} z$$ plug this (and y=0) in the first equation to get $x^2 + y^2 + z^2 =z \\ 3z^2 + z^2 = z\\ 4z^2 - z = 0$ we find z=0 (i.e the cone's apex intersects the sphere at the origin) and 4z-1=0 or $$z= \frac{1}{4}$$ which is what we want.

5. phi

we can now find the volume of the bottom cone, from z=0 to 1/4 in cyl coords the area element is r dr d$$\theta$$ and the iterated integral is $\int_0^\frac{1}{4} \int_0^{2\pi} \int_0^{\sqrt{3}\ z} r \ dr\ d\theta\ dz$ visually we are doing |dw:1438215237071:dw|

6. phi

above the cone, we integrate volume of the sphere $\int_{\frac{1}{4}}^1 \int_0^{2\pi} \int_0^{\sqrt{ \frac{1}{4} -\left(z-\frac{1}{2}\right)^2 }} r \ dr\ d\theta\ dz$ visually |dw:1438215437879:dw|

7. phi

the sum of those two integrals will be 5pi/32