Vector calculus double integrals

[braindead101]

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
[tex]\int^{2}_{0}\int^{1}_{0} 4-x^{2}-y^{2} dx dy[/tex]
is this correct?

or should i be putting it in polar coordinates, in which case how do i set up the limit of integration
[tex]\int\int_{\Omega} f(x,y) dA[/tex]
[tex]\Omega = \left\{ (x,y) | 0<=x<=1, 0<=y<=2 \right\}[/tex] ? is this right?

 

[Mindscrape]

You're looking for the volume under the surface, within the bounds of the rectangle, right? Your first one is fine, though I think you meant to say f(x,y) = 4-x^2-y^2.

 

[HallsofIvy]

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

I'm not sure why you would write that: z= 4- x²- y², not x.

[tex]\int^{2}_{0}\int^{1}_{0} 4-x^{2}-y^{2} dx dy[/tex]
is this correct?

or should i be putting it in polar coordinates, in which case how do i set up the limit of integration
[tex]\int\int_{\Omega} f(x,y) dA[/tex]
[tex]\Omega = \left\{ (x,y) | 0<=x<=1, 0<=y<=2 \right\}[/tex] ? is this right?

Of course, your first calculation is correct. In fact, the second formula,
[tex]\int\int_{\Omega} f(x,y) dA[/tex]
is also correct- it's just a general statement of the formula for volume.
But there is no reason to change to polar coordinates. Polar coordinates would be appropriate if you were finding the volume over a disk.