Calculate triple integral(zdV) over the area $E$, where $E$ is bounded by the planes $y=0 ~ z=0$ and $x+y=2$ and the cylinder $y^2 + z^2 = 1$ in the first octant.
I have seen the correct answer already, however I have difficulty in understanding why the limits in the triple integral are taken in that way. Could anyone give me advice on what strategy I should use to determine the boundries of the integrals in such relatively complex cases?
Here is the answer is the correct for the sake of completeness:
$$\displaystyle \int\limits_0^1 \int\limits_0^{\sqrt{1-y^2}} \int\limits_0^{2-y} \ z~dx~dz~dy$$
Thanks in advance
Best Answer
You want it in the first octant so we have $$x\ge0, y\ge0,z\geq0$$ The lower limit of $y$ is zero but another limit of it is ruled by $y^2+z^2=1$. Since $z=0$ is one of our boundary , so $y^2=1$ and so $y=1$. Clearly, the latter equation gives us the restriction for $z$, that is: $$z\big|_0^{\sqrt{-y^2+1}}$$ Note that here again we have $z\ge 0$. For $x$ is the same, that is $$x\big|_0^{2-y}$$ since we already know that $x+y=2$.