Find the volume of the solid under the graph of f(x;y) = e^-x2^-y2 over the disk D=f(x;y):x^2 +y^2 less than or equal to 2.
In cylindrical coordinates,
`x^2 + y^2 = r^2`
and the volume element is `dV = rdrdphidz.`
So the volume of the solid under the graph of
z = e^(-x^2 - y^2) = e^(-r^2) over the disk
`int_0 ^2pi dphi int_0 ^2 rdr int_0 ^ e^(-r^2) dz = 2pi *int_0 ^2 re^(-r^2)dr`
This integral can be taken by substitution `t = -r^2`
Then if r = 0 then t = 0 and if r = 2 then t = -4.
`dt = -2rdr`
The integral becomes `int_0 ^(-4) e^t dt/(-2) = -1/2 e^t |_0 ^ -4 = -1/2 (e^(-4) - 1)`
Multiplying this result by `2pi,`
we obtain that the volume of the given solid is `pi(1 - e^(-4)).`