# Triple integral/tetrahedron

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

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• As you can see from your figure, $2y \leq x \leq 2-2y$ and $0 \leq y \leq \frac{1}{2}$ Nov 29, 2020 at 20:24
• Set up as $dz \, dx \, dy$ Nov 29, 2020 at 20:25
• @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? Nov 29, 2020 at 20:35
• You can set up both ways. I will add some details on how to go about it. Nov 29, 2020 at 20:39
• I added some details. Please see if it helps and let me know if any questions. Nov 29, 2020 at 21:00

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:

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

• thank you so much for the graphic Nov 29, 2020 at 21:21

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.

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