The question is as follows:
Find triple integral of f(x,y,z) = xy + z by dxdydz
Over the tetrahedron D created by the following coordinates: $(0,0,0),
(1,0,0), (0,1,0)$ and $(0,0,1)$
My answer doesn't agree with the book's answer. I got $\frac{3}{20}$ while the books gives $\frac{1}{20}$. I suspect that my integrands for $dxdy$ are incorrect somehow, but I still don't quite see what I did wrong >_<
My work:
(1) The region $D$ looks like the following picture
(2) So my thoughts are the following:
a. To work with $dx$, I consider the $xy$ plane (or $yx$ plane). I note that $y = 0$ when $x = 1$ and $y =1$ when $x = 0$. So the upper bound is the line $y = – x + 1$, or $x = 1 – y$. The lower bound is $x = 0$. So $x$ goes from $0$ to $1-y$.
b. With $dy$, I consider the $yz$ plane (or $zy$ plane). Since $z = 0$ when $y = 1$ and $z=1$ when $y = 0$, the equation for upper bound of $y$ is $z = -y + 1$, or $y = 1- z$. So $y$ goes from $0$ to $1-z$
c. With $dz$, lower bound is $0$ and upper bound is $1$
So the triple integral becomes
$$ \int_0^1 \int_0^{1-z} \int_0^{1-y} (xy + z) dxdydz$$
And after some lengthy works, I got $\frac{3}{20}$
But the book's answer is $\frac{1}{20}$. So is my work wrong anywhere?
I strongly suspect that to work with $dx$, I should consider the $xz$ plane instead of $xy$ plane (I'm trying the "shooting arrow" method through the region)
Thank you in advance.
Best Answer
The plane containing $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ is $x+y+z=1$. Therefore, your interval for the x-axis is $x=0$ to $x=1-y-z$, for the y-axis it's $y=0$ to $y=1$, and for the z-axis it's $z=0$ to $z=1$.
Therefore your integral should be:
$$\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{1-y-z}(xy+z)\space{dx}\space{dy}\space{dz}$$