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Calculate the volume of the body bounded by the following surface:

$$x-y+z=6, \ x+y=2, \ x=y, \ y=0, \ z=0$$

I would do this with a triple integral. For this, however, we need to find the boundaries first. I would do it this way:

$$\int_{0}^2\int_0^{x} \int_0^{6-x+y} 1 \ dzdydx$$

Is this correct? I found the boundary for $x$ by the Geogebra plot below.

enter image description here

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The region in question lies over the triangle with vertices $(0,0)$, $(1,1)$, and $(2,0)$ in the $xy$-plane. If you're going to set up an iterated integral over that triangle, you'll need two separate integrals if you use the order $dy\,dx$. I recommend you switch to $dx\,dy$. Then $0\le y\le 1$ and for each fixed $y$, what is the range on the $x$ values? Your $z$ limits are correct.

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By making a simple hand drawing you can realize the object is a prism with base in the triangle you said and top in the plane $x-y+z=6$ so your volume integral admits the form $V=\int_0^1\left(\int_{y}^{2-y}(6-x+y)dx\right)dy$.

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  • $\begingroup$ Good observation, simple and nice :) $\endgroup$ – Vuk Stojiljkovic Jan 28 at 0:33
  • $\begingroup$ Anytime we see z's involved we can just fix them as a function to the integral , if ther is no function given, and work with other things. $\endgroup$ – Vuk Stojiljkovic Jan 28 at 0:34
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Note that you never incorporated the boundary $x+y=2$ in your setup. It is the upper limit for $y$ over the region $x\in[1,2]$. Thus, the integral is

$$\int_{0}^1\int_0^{x} \int_0^{6-x+y} 1 \ dzdydx+\int_{1}^2\int_0^{2-x} \int_0^{6-x+y} 1 \ dzdydx$$

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