Question #71203

1 Answer
Jun 23, 2016

V = pi (e-1)V=π(e1)

Explanation:

e^(x^2+y^2)ex2+y2 is a surface of revolution regarding the zz axis

then the sought volume can be computed as

V = 2pi int_{x=0}^{x=1}x e^{x^2}dx = pi(e-1)V=2πx=1x=0xex2dx=π(e1)

Mind that d/(dx)(e^{x^2}) = 2xe^{x^2} ddx(ex2)=2xex2

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)