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Let the Tetrahedron be bounded by the planes

$x+2y+z=2$, $x=2y$, $y=0$, $z=0$

so the limits for z are easy and would be $0 \leq z \leq 2-x-2y$ I have calculated the intersection point in the $(x,y)$ plane and I get $(1, \frac{1}{2})$. So $y=0$ means that the lower boundary of $y$ should start with zero. I drew a sketch, but I am unsure about the x limits and y limits.

what would the functions $h_1$ and $h_2$ be?

$\int_{0}^{1} \int_{h_1(y)}^{h_2(y)} \int_{0}^{2-2y-x} dzdxdy$

on the other hand, if we had $x=0$ it would be easy to write because I could write it as $\int_{0}^{1} \int_{x/2}^{1-x/2} \int_{0}^{2-2y-x} dzdydx$

but since we have $y=0$ I need to determine the $h_1(x)$, $h_2(x)$, where I am confused how to do it

enter image description here]1

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  • $\begingroup$ As you can see from your figure, $2y \leq x \leq 2-2y$ and $0 \leq y \leq \frac{1}{2}$ $\endgroup$
    – Math Lover
    Nov 29, 2020 at 20:24
  • $\begingroup$ Set up as $dz \, dx \, dy$ $\endgroup$
    – Math Lover
    Nov 29, 2020 at 20:25
  • $\begingroup$ @MathLover: Thank you for your help! I do not know why but for me it is always difficult to see the x limits, how can I see that 2y is the lower bound? $\endgroup$
    – Parinn
    Nov 29, 2020 at 20:35
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    $\begingroup$ You can set up both ways. I will add some details on how to go about it. $\endgroup$
    – Math Lover
    Nov 29, 2020 at 20:39
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    $\begingroup$ I added some details. Please see if it helps and let me know if any questions. $\endgroup$
    – Math Lover
    Nov 29, 2020 at 21:00

2 Answers 2

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enter image description here

y goes from 0 to 0.5

And for a fixed y we know that x goes from $2y$ to $2 - 2y$ because the x values are to the right of the line $x = 2y$ and to the left of the line $x = 2 - 2y$

So we end up with:

$\int_0^{0.5}\int_{2y}^{2-2y}\int_{0}^{2 - x -2y}dzdxdy$

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    $\begingroup$ thank you so much for the graphic $\endgroup$
    – Parinn
    Nov 29, 2020 at 21:21
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It is easier to set up your integral as $dz \, dx \, dy$ given you have one of the equations in just $x, y$.

Here is how you can see it - if you take infinitely thin vertical strips from down to up between $y = 0$ to the plane, you realize you hit upon $x = 2y$ till $x = 1$ and then on it is the other plane $x + 2y + z = 2$. So you will have to break your integral into two.

But if you take horizontal strips, it is between two planes throughout your region finally both planes meeting at $y = \frac{1}{2}$.

$x = 2y, x + 2y + z = 2, y = 0, z = 0$

You rightly set the bounds for $z$.

Now set $z = 0$ to get the upper bound of $x, y$.

So $x = 2y, x + 2y = 2 - 2y$.

That gives you bounds of $2y \leq x \leq 2-2y$

You already know $x$ varies between $0$ and $\frac{1}{2}$ based on the intersection and your graph.

So your integral becomes -

$\displaystyle \int_{0}^{1/2} \int_{2y}^{2-2y} \int_{0}^{2-2y-x} \, dz \, dx \, dy$

You could have also set up as

$\displaystyle \int_{0}^{1} \int_{0}^{x/2} \int_{0}^{2-2y-x} \, dz \, dy \, dx + \int_{1}^{2} \int_{0}^{(2-x)/2} \int_{0}^{2-2y-x} \, dz \, dy \, dx$

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    $\begingroup$ thank you so much! it is very clear now. $\endgroup$
    – Parinn
    Nov 29, 2020 at 21:20
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    $\begingroup$ I am glad to hear that! $\endgroup$
    – Math Lover
    Nov 29, 2020 at 21:21

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