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.

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ishpiro | College Teacher | (Level 1) Educator

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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

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will be 

`int_0 ^2pi dphi int_0 ^2 rdr int_0 ^ e^(-r^2) dz = 2pi *int_0 ^2 re^(-r^2)dr`

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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)).`

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