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?

I get the correct answer when instead I use

\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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1 Answer

1

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Yes, you do interpret it wrongly, most likely as a result of conflicting notations (\psi\psi and QQ that you define are not the same as in the formula). Let’s look at the formula (1). Here xx is the old variable, yy is the new variable and x=\psi(y)x=\psi(y) is the variable substitution that transforms new variable to the old one. The original set in terms of xx is \psi(Q)\psi(Q), and QQ is the integration set in terms of yy, i.e. the new variable.

When you apply it to your case, you define \psi\psi that transforms old variables to the new one, that is it should, in fact, be \psi^{-1}\psi^{-1}. Thus, the determinant in (2) should be \frac12\frac12. Moreover, you get it wrong in (2) (see my comment above). It should be x+y+z=\frac12(u+v+w)x+y+z=\frac12(u+v+w).

Your last edit looks much better.