Question

Question #71203

Answer 1

`V = pi (e-1)`

Explanation:

`e^(x^2+y^2)` is a surface of revolution regarding the `z` axis

then the sought volume can be computed as

`V = 2pi int_{x=0}^{x=1}x e^{x^2}dx = pi(e-1)`

Mind that `d/(dx)(e^{x^2}) = 2xe^{x^2}`

Now using polar coordinates.

Using the pass equations

# { (x=r cos(theta)), (y=r sin(theta)) :}#

we have `x^2+y^2=r^2` and
`dx xx dy = r xx d theta xx dr`

Substituting

`int int_D e^{x^2+y^2}dx dy = int_0^{2pi}(int_0^1re^{r^2}dr)d theta = 2pi((e-1)/2)`