$$\int_0^1dx\int_0^{1-x}dy\int_0^{x+y}f(x,y,z)dz$$
From this it is obvious that $x\in[0,1],y\in[0,1-x],z\in[0,x+y]$. For it asks for the order to be in $$\int dz\int dx\int f(x,y,z)dy$$ . My method of doing this is for start from the outer variable ,in this case z, and from the original one can deduct that $z\in [0,1]$
But the professor goes on to make a assisting plane that has the points $(1,0,1)(0,1,0)(0,0,0).$ Can anyone explain how to integrate this? I understand the picture of the shape, just need help in the steps in integrating.
Why was there a need to integrate with the assisting plane. Could it have been done without it ? And in general how to i go about integrating shapes in 3d if i know how it looks?
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The shaded shape is the assisting plane.. The answer is in the picture..
Why bother drawing a picture? I preferred to do it algebraically.
Since $0\le x\le1$, $0\le y\le1-x$, and $0\le z\le x+y$, we know $0\le z\le 1$. For fixed $z$, we have $0\le x\le1$, since no other restriction given by $z$ is put on $x$. Finally, for fixed $z,x$, we find $0\le y\le1-x$ and $y\ge z-x$.
Thus, we have to write the integral in two parts, namely, $$\int_0^1{dz\int_0^z{dx\int_{z-x}^{1-x}{f(x,y,z)dy}}} + \int_0^1{dz\int_z^1{dx\int_{0}^{1-x}{f(x,y,z)dy}}}.$$