Confusion with change of (multiple) variables theorem

Problem: Calculate I

Q={x,y,z∈R3:x+y>0,y+z>0,x+z>0}Q = \{x,y,z \in \mathbb{R}^3 : x+y > 0, y+z>0, x+z > 0\}

I=∫Qe−x−y−zdVI =\int_{Q}e^{-x-y-z}dV

My attempt:

ψ(x,y,z)=(x+y,y+z,x+z)\psi(x,y,z) = (x+y, y+z, x+z)

The change of variables theorem states that (for suitable psi, Q)

∫ψ(Q)f(x)dx=∫Qf(ψ(y)) det(D(ψ)(y))dy\int_{\psi(Q)}f(x)dx = \int_{Q} f(\psi(y)) \ det(D(\psi)(y)) dy\tag{1}

Equivalently

∫ψψâپ»1Qf(x)dx=∫ψ−1Qf(ψ(y)) det(D(ψ)(y))dy\int_{\psi \psi^{âپ»1}Q}f(x)dx = \int_{\psi^{-1}Q} f(\psi(y)) \ det(D(\psi)(y)) dy

So…
I =\int_{Q}e^{-x-y-z}dxdydz = 2\int_{\mathbb{R_+}^3}e^{-2u-2v-2w}dudvdw = \frac{1}{4}\tag{2}I =\int_{Q}e^{-x-y-z}dxdydz = 2\int_{\mathbb{R_+}^3}e^{-2u-2v-2w}dudvdw = \frac{1}{4}\tag{2}

However the correct answer is 44. What is wrong with my reasoning here?

\int_{Q}f(x)dx = \int_{\psi^{-1}Q} f(\psi^{-1}(y)) \ det(D(\psi^{-1})(y)) dy\int_{Q}f(x)dx = \int_{\psi^{-1}Q} f(\psi^{-1}(y)) \ det(D(\psi^{-1})(y)) dy

But i’m confused how that follows from the original equation. Am i interpreting it wrong somehow?

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u+v+w=2(x+y+z)u+v+w=2(x+y+z)
– A.G.
2 days ago

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