Question

calculus continued

Answer 1

Alright, good. Here goes the explanation, and I apologize in advance for the probably atrocious drawing.

Answer 2

When we evaluate integrals, we break up domains into tiny chunks, as we talked about previously. When it's a line integral, we break it up into little tiny lines. When it's a surface integral, we break it up into little tiny bits of surface area, and for a volume integral, we break it up into little tiny volumes. Therefore, fundamentally, we want to find this:
\[\iiint F\space dV\]
Now, when we use cartesian coordinates, we need to express both the function F and the volume element dV in terms of the rectangular coordinates x,y, and z. We then get
\[\iiint F(x,y,z) \space dxdydz \]

Answer 3

right!

Answer 4

But who says I want to use cartesian coordinates? Spherical coordinates are also lovely. Transforming the function is one thing, which you can do already, but what about transforming the volume element? When we use cartesian coordinates, the volume element is a cube, with side lengths dx,dy, and dz. What about in spherical coordinates?

Answer 5

This is a bit more complicated, so lets look at a drawing: