# Vector calculus double integrals

1. Sep 23, 2007

Find the volume V of the solid under the surface z=4-x^2-y^2 and over the rectangle R consisting of all points (x,y) such that 0<=x<=1 and 0<=y<=2.

I have started, but am unsure if my approach is correct or not.

x = 4-x^2-y^2
$$\int^{2}_{0}\int^{1}_{0} 4-x^{2}-y^{2} dx dy$$

is this correct?

2. Sep 23, 2007

### Staff: Mentor

3. Sep 23, 2007

### HallsofIvy

Staff Emeritus
I could have sworn I had answered this before! I particularly remember the peculiar "x= 4- x^2- y^2" term to which I noted that it should be "z=" rather than "x= ".

But that post also asked about changing to polar coordinates! Rectangle, as here, uses "rectangular" coordinates. Disks and circles use polar coordinates.