# Integrating Over Truncated Sphere

Consider the volume bounded by the sphere of radius R (with center $r_0$ = R/2 z) and the plane z = 0.

Evaluate $\int dS$ over the surface of the truncated sphere.

I'm not sure how to integrate over the partial sphere.

I need to find with to plug in for dS, as well as the limits of integration.

I know in spherical coordinates you would be integrating theta from 0 to π, and phi from 0 to 2π, and dS = $r^2$sinθdθdϕ. I'm just not sure how to go about removing the part of the sphere below the z=0 plane. Since the sphere is centered at R/2 z, the area below the z=0 plane is no longer half of the sphere, so I can't just change the limits of theta to 0 to π/2.

Could put dS in cartesian coordinates (dxdydz)?

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Can you tell me what the bounds of integration should be? – Euler....IS_ALIVE Dec 17 '13 at 8:38
I know in spherical coordinates you would be integrating theta from 0 to π, and phi from 0 to 2π, and dS = r^2sinθdθdϕ. I'm just not sure how to go about removing the part of the sphere below the z=0 plane. – anonymous Dec 17 '13 at 8:45
Well if what you indicated is the full sphere (defined by WHICH ANGLE??) what is that part of the sphere asked in the question? – Euler....IS_ALIVE Dec 17 '13 at 8:47
If the sphere was centered at the origin, then the area below the z=0 plane would just be the lower hemisphere, and I would integrate theta from 0 to π/2. However the sphere is centered at R/2 in the z direction, so the area below the z=0 plane is no longer a hemisphere. – anonymous Dec 17 '13 at 8:56
Just use a polar coordinated system centered at the center of the sphere. The intersection of the plane $z=0$ with the sphere now corresponds to $\cos\theta = -(\frac{R}{2}/R) = -\frac12$ in the new polar coordinate system. – achille hui Dec 17 '13 at 19:08