Bounding Surface of Volume Integral: Sphere at Origin

[rafaelpol]

Homework Statement

Say I have a sphere at the origin with radius "a". If I am integrating over a region in which r is greater than "a", how can the bounding surface of this volume be the spherical surface? This comes from an explanation in Greiner Classical Electrodynamics, in which he says the volume integral was done in a region such that r is greater than "a", but at the same time he says the bounding surface of this volume is the sphere...

The Attempt at a Solution

I feel like I am missing something on the defintion of a bounding surface, but the only definition I can find is the one that says that it is the surface that "binds" the whole volume. Hence, the spherical surface at "a" might be a part of the surface that is part of the volume "v", but there should be more, since the integration wasn't done over the sphere.

Any help will be appreciated.

Thanks

 

[cragar]

Maybe you integrate from a<R<r , But I am not sure