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Let $R = \{ (x,y,z) : x,y,z \geq 0, \; \; x+y+z \leq 1 \} $. WAnt to find $Vol(R) $. Well, I know that the volume is given by

$$ Vol(R) = \int_0^1 \int_0^{1-x} \int_0^{1-x-y} dzdydx = \frac{1}{6} $$

My teacher said that using the change of variables $u =x $, $v = x+y$ and $w = x+y+z$, it would be easier the computation. I calculated the jacobian of this change of variables which is $1$. However, I have difficulties trying to find the limits of integration. I need some help . thanks

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So we have $$\left\{ \begin{array}{l} u = x \\ v = x + y \\ w = x + y + z \end{array} \right .$$ From the first equation, you get $v = u + y$ in the second, so $y = v - u$ and then in the third one you get $w = u + v - u + z$, so $z = w - v$. We've got $$\left\{ \begin{array}{l} x = u \\ y = v-u \\ z = w-v \end{array} \right .$$

Now, $0 \leq x \leq 1$, thus $0 \leq u \leq 1$.

For the second variable, $0 \leq y \leq 1-x \Rightarrow 0 \leq v - u \leq 1-x \Rightarrow 0 \leq v-u \leq 1-u$ and finally $u \leq v \leq 1$.

For the last one, $z = w-v$ and $0 \leq z \leq 1-x-y$, so $$0 \leq w-v \leq 1-x-y \Rightarrow 0 \leq w-v \leq 1-u-v+u$$ hence $$0\leq w -v\leq 1-v$$ so finally $$v \leq w \leq 1.$$ That being said, the integral is $$\mathrm{Vol}(R)=\int_0^1 \int_u^1 \int_v^1 dwdvdu.$$

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