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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 $\frac{47}{3}$ My answer is wrong.

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We have $$\min(8 - x - y,3) = \begin{cases} 8-x-y & \text{if } x+y\ge5, \\ 3 & \text{if } x+y \le 5. \end{cases}$$ when the triangle $x+y \leq 5 $ contains some points out of the rectangle, to fix this we have to split it into two parts: $1\leq y \leq 3 $ when $1\leq x \leq 2$ and $1 \leq y \leq 5-x$ when $2\leq x \leq 3$ So:

$$\int_1^3 \int_1^3 \displaystyle \int_1^{\min(8-x-y,3)} 2\,dz\,dy\,dx \qquad= \int_1^2 \int_1^3 \int_1^3 2\,dz\,dy\,dx \qquad + \int_2^3 \int_1^{5-x} \int_1^3 2\,dz\,dy\,dx +\int_2^3 \int_{5-x}^3 \int_1^{8-x-y} 2\,dz\,dy\,dx \qquad $$ I think that you can take it from here.

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    $\begingroup$ When $x=1$ the second integral in the first triple integral has bounds $1$ and $5-x=4$. $\endgroup$ – bjorn93 Dec 30 '19 at 1:50
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    $\begingroup$ Could I suggest that instead of $\displaystyle \int_1^{min(8-x-y,3)},$ you write $\displaystyle \int_1^{\min(8-x-y,3)},$ and instead of $2 dz dy dx$ you write $2\,dz\,dy\,dx. \qquad$ $\endgroup$ – Michael Hardy Dec 30 '19 at 1:53
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    $\begingroup$ @MichaelHardy, Done, thank you $\endgroup$ – ahdahmanii Dec 30 '19 at 1:58
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    $\begingroup$ @ahdahmanii now you go outside the bounds in the second triple integral. When $x=1$, $y$ goes from 4 to 3. $\endgroup$ – bjorn93 Dec 30 '19 at 2:08
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    $\begingroup$ @bjorn93 fixed, thank you. $\endgroup$ – ahdahmanii Dec 30 '19 at 2:11
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Draw a picture of the rectangle $1\leq x\leq 3, 1\leq y\leq 3$ and of the boundary line $8-x-y=3 \Leftrightarrow y=5-x$. Notice that below the line (and staying in the rectangle) $\min(8-x-y,3)=3$ and above the line in the rectangle $\min(8-x-y,3)=8-x-y$. Also notice that $\min(8-x-y,3)=3$ in the rectangle $1\leq x\leq 2,1\leq y\leq 3$. So, we can split it up as: $$\int_1^2\int_1^3\int_1^32\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x+\int_2^3\int_1^{5-x}\int_1^32\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x+\int_2^3\int_{5-x}^3\int_1^{8-x-y}2\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x $$

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  • $\begingroup$ How sure are you that your answer is right? $\endgroup$ – Bob Dec 30 '19 at 15:29
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    $\begingroup$ @Bob It seems like my answer evaluates to what WolframAlpha gives, so I guess it's OK. $\endgroup$ – bjorn93 Dec 30 '19 at 15:38
  • $\begingroup$ I have evaluated your answer using WolframAlpha and the answer checks. Therefore, I believe your answer is correct. $\endgroup$ – Bob Dec 30 '19 at 16:02
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$$ \min\{8 - x - y,3\} = \begin{cases} 8-x-y & \text{if } x+y\ge5, \\ 3 & \text{if } x+y \le 5. \end{cases} $$ $$ {} $$ \begin{align} & \overbrace{ \iint\limits_{\begin{smallmatrix} (x,y) \,:\, 1\,\le\,x,y\,\le 3 \\[3pt] \&\ \ x\,+\,y\,\le\, 5 \end{smallmatrix} } \left( \int_1^3 2 \, dz \right) \, d(x,y)}^\text{This one says “$\le~5$''.} {} \,\,+\,\, {} \overbrace{ \iint\limits_{\begin{smallmatrix} (x,y) \,:\, 1\,\le\,x,y\,\le 3 \\[3pt] \&\ \ x\,+\,y\,\ge\, 5 \end{smallmatrix}} \left( \int_1^{8-x-y} 2\,dz \right) \, d(x,y)}^\text{This one says “$\ge~5$''.} \\[10pt] = {} & \int_1^3 \left( \int_1^{\min\{3,5-x\}} \left( \int_1^3 2\,dz \right) \, dy \right) \, dx + \int_1^3 \left( \int_{\min\{3,5-x\}}^3 \left( \int_1^{8-x-y} 2 \, dz \right) \, dy \right) \, dx. \end{align}

And then $$ \int_1^3 \left( \int_1^{\min\{3,5-x\}} \cdots \, dy \right) \,dx = \int_1^2\left( \int_1^3 \cdots \, dy \right) \, dx + \int_2^3 \left( \int_1^{5-x} \cdots \, dy \right) \, dx $$ and $$ \int_1^3 \left( \int_{\min\{3,5-x\}}^3 \cdots \, dy \right) \, dx = \int_2^3 \left( \int_{5-x}^3 \cdots\,dy \right) \, dx. $$

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  • $\begingroup$ Well, the same issue here, right? When $x=1$ the second integral in the first triple integral has bounds $1$ and $5-x=4$. $\endgroup$ – bjorn93 Dec 30 '19 at 1:54
  • $\begingroup$ @bjorn93 The same issue as what? $\endgroup$ – Michael Hardy Dec 30 '19 at 1:57
  • $\begingroup$ As the other answer. The second integral in the first triple integral (on the last line) goes outside the bounds $1\leq y\leq 3$. When $x=1$, they are $1$ and $4$. $\endgroup$ – bjorn93 Dec 30 '19 at 2:01
  • $\begingroup$ @bjorn93 : And how would I have known that that is what you meant by "same issue"? $\endgroup$ – Michael Hardy Dec 30 '19 at 2:04

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