Question #71203

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Answer 1 · 2016-06-23T13:44:03

#V = pi (e-1)#

#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)#