I'm given the following integral, with some advice to switch the limits of integration by switching the "$dx$" and the "$dy$" (easier said than done).
$$\iiint f(x,y,z) \, dV = \int^1_0\int^{\pi z}_0\int^z_\frac y\pi \sin(y/x) \,dx\,dy\,dz$$
It's a very ugly integral, but I had some initial thoughts about the domain of integration. Since the region is defined by linear constraints only, it's safe to say the region of integration should be a tetdrahedron bounded by certain planes. The corners should be (I think) $(0,0,0)$, $(0,0,1)$, $(1,0,1)$, and $(1,\pi, 1)$. So I should integrate first with respect to $y$. From here, I'm stuck. Is there a better way to approach this problem? Any help would be much appreciated.
Best Answer
$$0 \le z \le 1, 0 \le y \le \pi z, \frac{y}{\pi} \le x \le z$$
is equivalent to
$$0 \le z \le 1, 0 \le x \le z, 0 \le y \le \pi x$$
\begin{align} \int_0^1 \int_0^{\pi z}\int_{\frac{y}\pi}^z \sin \left( \frac{y}{x}\right) \, dx\, dy\, dz &= \int_0^1 \int_0^{ z}\int_{0}^{\pi x} \sin \left( \frac{y}{x}\right) \, dy\, dx\, dz\\ &= \int_0^1 \int_0^{ z}\left[ -x\cos \left( \frac{y}{x}\right) \right]_{y=0}^{y=\pi x} dx\, dz\\ &= \int_0^1 \int_0^{ z}2x dx\, dz\\ \end{align}