So, I was wondering if anyone can share some tricks or a systematic way on re-writing a triple integral.
Here's the exercise I'm doing.
$\int _{-1} ^1 \int _0 ^{\sqrt{1-x^2}} \int _0 ^ {y/2} f(x, y ,z)dz\ dy\ dx$
I'm trying to rewrite this to $dy\ dx\ dz$.
I have drawn the "picture" out and realize its sort of a half-cylinder being sliced by the plane $z = (y/2)$ . (Correct me if I'm wrong with the description)
So is there a systematic way to change the order of integration ? I am able to do (quite efficiently) for a double integrals, but I am extremely confused when it comes to triple integral.
Any tips and insights will be deeply appreciated
Best Answer
You can write all the limits as inequalities and then deduce the new bounds from them in order from outside to inside.
In your case, you have
$$ -1\lt x\lt1\;,\\ 0\lt y\lt\sqrt{1-x^2}\;,\\ 0\lt z\lt \frac y2\;. $$
For the new integration order, $z$ is on the outside, so you need the overall bounds for $z$. From the given inequalities, the range of $x$ is $[-1,1]$, thus the range of $y$ is $[0,1]$ and thus the range of $z$ is $[0,\frac12]$.
$x$ is next, so we need to solve the second inequality for $x$, yielding $|x|\lt\sqrt{1-y^2}$. Together with $y\gt 2z$ from the third inequality, this yields $|x|\lt\sqrt{1-4z^2}$, so the range of $x$ is $[-\sqrt{1-4z^2},\sqrt{1-4z^2}]$. Finally, $y$ is bounded both by $y\gt2z$ and by $y\lt\sqrt{1-x^2}$, so its range is $[2z,\sqrt{1-x^2}]$. Thus the integral is
$$ \int_0^{\frac12}\int_{-\sqrt{1-4z^2}}^{\sqrt{1-4z^2}}\int_{2z}^{\sqrt{1-x^2}}f(x,y,z)\,\mathrm dy\,\mathrm dx\,\mathrm dz\;. $$