Triple integrals calculus

Question

I have the following integral:

$$\int_0^1\int^0_{1-z}\int^0_{1-y-z}\mathsf e^{x+y+z}\,\mathsf d x\,\mathsf d y\,\mathsf d z$$

I’ve set up the integral, changing it to $e^xe^ye^z$, and integrated with respect to $x$, getting $e^xe^ye^z$, the problem is I’m not entirely sure how to substitute the limits for x into this!

Answer 1

$\int_0^1\int_0^{1-z}\int_0^{1-y-z} (e^x)(e^y)(e^z)\ dx \ dy\ dz$

As we integrate with respect to $x,$ we are going to treat $y,z$ (and $e^y, e^z$) as constants.

$\int_0^1\int_0^{1-z} (e^{x})(e^y)(e^z)|_0^{1-y-z} \ dy\ dz\\ \int_0^1\int_0^{1-z} (e^{1-y-z}- e^0)(e^y)(e^z) \ dy\ dz$

and simplify

$\int_0^1\int_0^{1-z} e - (e^y)(e^z) \ dy\ dz\\ \int_0^1 ey - e^y(e^z)|_0^{1-z} \ dz\\ \int_0^1 e(1-z) - (e^{1-z} - 1)(e^z)\ dz\\ \int_0^1 e(-z) +e^z\ dz\\ (-\frac {z^2}{2})e +e^z|_0^1\\ \frac 12 e - 1 $

Answer 2

Note that

$$\int_0^1 dz \int_{1-z}^0 dy \int_{1-y-z}^0 e^xe^ye^z dx =\int_0^1 dz \int_{1-z}^0 e^ye^z[e^x]_{1-y-z}^0 dy=\\\int_0^1 dz \int_{1-z}^0 e^ye^z(1-e^{1-y-z}) dy =\int_0^1 dz \int_{1-z}^0 (e^ye^z-e)\, dy =\\\int_0^1 [e^ye^z-ey]_{1-z}^0\, dz =\int_0^1(e^z-e+e(1-z))dz=\\=[e^z-e\frac{z^2}2]_0^1=e-\frac{e}2-1=\frac{e}2-1$$

Answer 3

So you have$$\int_0^1\int_{1-z}^0\int_{1-y-z}^0e^{x+y+z}\,\mathrm dx\,\mathrm dy\,\mathrm dz=\int_0^1\int_{1-z}^0\int_{1-y-z}^0e^xe^ye^z\,\mathrm dx\,\mathrm dy\,\mathrm dz.$$But this is equal to$$\int_0^1e^z\int_{1-z}^0e^y\int_{1-y-z}^0e^x\,\mathrm dx\,\mathrm dy\,\mathrm dz.$$Can you take it from here?

Answer 4

$$\int_0^1\int_{1-z}^0\int_{1-y-z}^0 \mathsf e^{x+y+z}\,\mathsf d x\,\mathsf d y\,\mathsf d z ~ {= \int_0^1\int_{1-z}^0 (\mathsf e^0-\mathsf e^{1-y-z})\mathsf e^y\mathsf e^z\,\mathsf d y\,\mathsf d z\\= \int_0^1\int_{1-z}^0 (\mathsf e^z\mathsf e^y-\mathsf e^{1})\,\mathsf d y\,\mathsf d z\\~~\vdots}$$

Answer 5

$$\begin{aligned}\int_0^1\int_{1-z}^0\int_{1-y-z}^0 \mathsf e^{x+y+z}\,\mathsf d x\,\mathsf d y\,\mathsf d z&=\int_0^1\int_{1-z}^0(e^{y+z}-e^{1-y-z+y+z})dy\,dz \\&= \int_0^1\int_{1-z}^0(e^{y+z}-e)dy\,dz\\ &=\int_0^1(e^{z}-e-e(z-1))\,dz\\ &=\int_0^1(e^z-ez)dz\\ &=e-1-\dfrac{e}{2}\\ &=\dfrac{e}{2}-1\end{aligned}$$