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I have this problem, evaluate this integral:

$$\iiint\limits_T\mathrm dx\mathrm dy\mathrm dz$$ $$T=\{x\gt 0,y\gt 0,z\gt 0, x+y+z\lt 2\}$$

So saw the figure and I tried to solve it as follows:

$$\int_0^2\int_0^2\int_0^{2-x-y}\mathrm dx\mathrm dy\mathrm dz=0$$

But I think I failed to set the extreme of the integration, because the result of the integral is wrong. Can you tell where is the mistake? Where am I failing?

2 Answers 2

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Note that when z=0

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$0 \le x \le 2$ and $0 \le y \le 2-x $ and z is correct $0 \le z \le 2-x-y$

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In this case, we have $$V=\iiint\limits_T\mathrm dV$$ Now let's define region $T$ as $$\{(x,y,z):(x,y)\in D, 0\leq z\leq 2-x-y\}$$ Which yields the following bounds for each variable $$0\leq z\leq 2-x-y$$ $$0\leq y\leq 2-x$$ $$0\leq x\leq 2$$ So now we have $$\iint\limits_D\left(\int_0^{2-x-y}\mathrm dz\right)\mathrm dA$$ $$=\int_0^2\int_0^{2-x}\int_0^{2-x-y}\mathrm dz\mathrm dy\mathrm dx$$ $$=\int_0^2\int_0^{2-x}(2-x-y)\ \mathrm dy\mathrm dx$$ $$=\frac12\int_0^2 (2-x)^2\ \mathrm dx$$ $$=\frac12\int_0^2 u^2\ \mathrm du$$ $$=\frac{4}{3}$$

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    yes, but i know the result is $\frac{2}{3}$ – 2017-01-13
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    @Mazzorca, the result is indeed $\frac43$ as shown above. Can you update your question and show how you got $\frac23$? – 2017-01-15
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    This is a problem of a univerity test, i know only the result but not the method. I try with the method showed in the question, but don't work, and your method, seem correct. – 2017-01-15
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    @Mazzorca Perhaps the original problem was$$\int_0^2\int_0^{2-x}\int_0^{2-x-y}x\ \mathrm dz\mathrm dy\mathrm dx$$ Which is in fact, equivalent to $\frac23$. I'll update my answer if this is the case. Please confirm. – 2017-02-08