let t>0, compute the triple integral?

Question

let t>0, compute the triple integral

x,y,z>=0, x+y+z<=t int(dx dy xz/(1+(x+y+z)^2)

Answer 1

x,y,z>=0, x+y+z<=t ∫(dx dy xz/(1+(x+y+z)^2)

I'm assuming there's a typo and you meant
x,y,z>=0, x+y+z<=t ∫(dx dy dz/(1+(x+y+z)^2)
(dz instead of xz in the integrand)

In this case, the limits will be x ∈ [0, t], y ∈ [0, t-x] and z ∈[0, t-y]. So we have
∫(0 to t) ∫(0 to t-x) ∫(0 to t-x-y) dz dy dx / (1+(x+y+z)^2)
= ∫(0 to t) ∫(0 to t-x) [arctan (x+y+z)][0 to t-x-y] dy dx
= ∫(0 to t) ∫(0 to t-x) (arctan t - arctan (x+y)) dy dx

Now the integral of arctan x is x arctan x - (1/2) ln (1+x^2) (this is easy to see: use integration by parts on 1.arctan x), so we get
= ∫(0 to t) ((t-x) arctan t - [(x+y) arctan (x+y) - (1/2) ln (1+(x+y)^2)][0 to t-x]) dx
= ∫(0 to t) ((t-x) arctan t - t arctan t + (1/2) ln (1+t^2) + x arctan x - (1/2) ln (1+x^2)) dx
= ∫(0 to t) (-x arctan t + (1/2) ln (1+t^2) + x arctan x - (1/2) ln (1+x^2)) dx
= (-arctan t) (t^2/2 - 0) + (1/2) (t-0) ln (1+t^2) + ∫(0 to t) (x arctan x - (1/2) ln (1+x^2)) dx

We can use integration by parts to find that ∫(x arctan x) = ((1+x^2) arctan x - x) / 2, and ∫(ln (1+x^2)) = x ln (1+x^2) + 2 arctan x - 2x. (Integrate 1 and differentiate ln (1+x^2).) So we get

= -t^2 arctan t / 2 + t ln (1+t^2) / 2 + [((1+x^2) arctan x - x) / 2][0 to t] - (1/2) [x ln (1+x^2) + 2 arctan x - 2x][0 to t]
= (1/2) {-t^2 arctan t + t ln (1+t^2) + arctan t + t^2 arctan t - t - 0 - (t ln (1+t^2) + 2 arctan t - 2t - 0)}
= (1/2) (t - arctan t).

Talk about some heavy-duty integration to get to a really simple answer! I'm going to assume you've just done integration by parts, because this question really taxes that area.

Answer 2

you need first of all to determine the limits of integration.

x + y + z = t is a plane, which intersects the xy plane at the line
x + y = t

So 0 < x < t
0 < y < t - x
0 < z < t - x - y

f(x,y,z) = xz / [ 1 + (x + y + z)^2 ] dz dy dx

You're going to end up with some really complicated integration. But you can use the site below to help with that. Good luck.