In 3 dimensions, RxRxR has infinite volume, but (0,1)x(0,1)x(0,1) has volume 1. Again, these guys are homeomorphic.
So area and volume are not topological invariants.
If you need a more rigid proof (i.e. if you haven't yet established that (0,1)x(0,1) and RxR are homeomorphic), then just compare A=(0,1)x(0,1) and B=(0,2)x(0,2), and set f:A->B as (x,y)->(2x,2y). f is clearly 1-1, onto, and continuous, and the inverse of f is (x,y)->(1/2 x, 1/2 y) which is continuous, so A and B are homeomorphic -- but they have different areas.