Taking the determinant like
>>3 does is the right way to do it. However 99% of the time when dealing with volumes/areas in integrals you can remember the answer with a simple geometric argument. Also doing these types of geometric arguments is the only way the proof of the determinant method is going make any sense.
If you take two circles which both have centers at the origin then the area between them is called an _annulus_. If one circle has radius r, the other has raidus r+dr (where dr is a positive real number) then the area of the annulus is approximately 2*pi*r*dr when dr is small. (You can see this by doing the explicity doing the calculation, and you should try to see why a rectangle of length 2 pi r and width dr would have almost the same area as the annulus).
If you want the (approximate) area of a fraction of the annulus then you mutliply that fraction by 2 pi r dr. So a quarter annulus (i.e. 90 degrees worth of the annulus) has area pi * r * dr/2. A fraction of an annulus is also called a _sector_. What if you wanted the area of the sector which corresponds to a d-theta amount of angle (where d-theta is a positive real number)? Then the fraction of the whole annulus is d-theta / 2*pi, and so the area of the sector is r*dr*d-theta.
In your mind's eye you should be able to imagine how for a given dr > 0 and a given d-theta > 0 one can fill a circle with lots of little sectors. The process of making the sectors smaller and smaller is analagous to how you take finer and finer reimman sums when doing 1D integrals which you are probably familar with.