Integral of $xyz$ over $B=\{0<x<1, 0<y, 0<z, 0<x+y<1\}$

145 Views Asked by At

It is looked for $\int_B xyz ~$ where $~ B=\{0<x<1, 0<y, 0<z, 0<y+z<1\}$.

Is it possible to write the limits this way?

$$\int_0^1\int_0^1\int_0^{1-y}xyz~dz~dy~dx$$

After some calculation I mistakely got -11/48 as the answer (1/48 is correct).

So is it correct to say that $z$ goes from $0$ to $1-y$ in the first place?

1

There are 1 best solutions below

1
On BEST ANSWER

Your integration domain differs in your topic and your question text. $$\{0<x<1, 0<y, 0<z, 0<x+y<1\} \not= \{0<x<1, 0<y, 0<z, 0<y+z<1\}$$ For your second domain your ansatz is right and it holds $$\int_0^1\int_0^1\int_0^{1-y}xyz~dz~dy~dx = \frac{1}{48}$$ So you just miscalculated the integrals.

Show your calculations to get your mistake.