Problem:
Evaluate the following triple integral: $$ \int_1^3 \int_1^3 \int_1^{\min(8 - x - y,3)} 2 \, dz \, dy \, dx $$ Answer:
The problem is the bound $\min(8 - x - y,3)$. I would like to write it as the sum of two triple integrals with simple bounds. I could try something like: $$ \int_1^3 \int_1^3 \int_1^{\min(8 - x - y,3)} 2 \, dz \, dy \, dx = \int_1^2 \int_1^2 \int_1^{3} 2 \, dz \, dy \, dx + \int_2^3 \int_2^3 \int_1^{8 - x - y} 2 \, dz \, dy \, dx $$ but I know that is wrong. What is the right approach to evaluate this integral?
Based upon comments I received, I am updating my post. Using Wolfram, I find: $$\int_1^3 \int_1^3 \int_1^{\min(8 - x - y,3)} 2 \, dz \, dy \, dx = \frac{47}{3} $$
Using Wolfram, I find: $$ \int_2^3 \int_1^{5-x} \int_1^3 2\,dz\,dy\,dx = 2 $$
Using Wolfram, I find: $$ \int_2^3 \int_{5-x}^3 \int_1^{8-x-y} 2\,dz\,dy\,dx = 4 $$
Since $4 + 2 = 6$ not $\frac{47}{3}$ I am questioning the correctness of the first answer I received.
from Hot Weekly Questions - Mathematics Stack Exchange
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