MHB How do I evaluate this triple integral for 2ze^(-x^2) over the given bounds?

[harpazo]

Evaluate the triple integral.

Let S S S = triple integral

The function given is 2ze^(-x^2)

We are integrating over dydxdz.

Bounds pertaining to dy: 0 to x

Bounds pertaining to dx: 0 to 1

Bounds pertaining to dz: 1 to 4

S S S 2ze^(-x^2) dydxdz

S S 2yze^(-x^2) from y = 0 to y = x dxdz

S S 2xze^(-x^2) dxdz

I am stuck here.

 

[Greg]

It looks like you can continue by making the sub $u=x^2$.

 

[harpazo]

It looks like you can continue by making the sub $u=x^2$.

You mean u = -x^2 not x^2, right?

 

[MarkFL]

You mean u = -x^2 not x^2, right?

You could use either one, however, I would use the substitution suggested by Greg because then you already have the needed differential without adding a negative sign both within the differential and in the integrand or equivalently, in front of the integral (which would allow you to reverse the now "backwards" limits of integration). However, either substitution will get you to the correct answer. :D

 

[harpazo]

You could use either one, however, I would use the substitution suggested by Greg because then you already have the needed differential without adding a negative sign both within the differential and in the integrand or equivalently, in front of the integral (which would allow you to reverse the now "backwards" limits of integration). However, either substitution will get you to the correct answer. :D

Ok. You know best.

 

[HallsofIvy]

Let, as suggested, u= x^2. Then du= 2xdx so that (1/2)du= x dx. When x= 0, u= 0 and when x= 1, u= 1.

The integral becomes \int_1^4\int_0^1 zxe^{-x^2}dxdz= \left(\int_1^4 z dz\right)\left(\frac{1}{2}\int_0^1 e^{-u} du\right)

 

[harpazo]

It is an interesting problem all around.