I am having issues switching an iterated integral from dxdy to dydx:

Switch the integral $\int_0^2 \int_y^{2y}6xy$ dx dy to a dy dx integral in the form of $\int_0^2 \int_{...}^{...}6xydydx$ + $\int_2^4 \int_{...}^{...}6xydydx$

What I'm having issues with here is trying to break it up into pieces-- is the first one just from 0 to y, and the next one from y to 2y?

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    $\begingroup$ How about drawing the region of integration? $\endgroup$ – Lord Shark the Unknown Oct 25 '17 at 0:22
  • $\begingroup$ The hard part for me is that it is in 3 dimension-- do I draw the traces, for instance, holding X constant? $\endgroup$ – Scott B Oct 25 '17 at 0:24
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    $\begingroup$ But it's not in three dimensions. The domain is a subset of the plane. Draw the set $$\left\{(x,y) : 0 \leq y \leq 2,\ y \leq x \leq 2y\right\}$$ $\endgroup$ – Matthew Leingang Oct 25 '17 at 0:40
  • $\begingroup$ Right, so x goes from 0 to 4. So then, for the y, would it be from x to x/2 ? $\endgroup$ – Scott B Oct 25 '17 at 0:58

The domain of integration is the inside of the triangle with vertices at the points $(0,0)$, $(2,2)$ and $(4,2)$, bounded by the lines:$y=x$, $y=x/2$ and $y=2$. The integral of an arbitrary function $f(x,y)$ in this domain can be done first in $x$ and then in $y$, or first in $y$ and then in $x$:

$$\int_0^2dy\int_y^{2y}f(x,y)\,dx = \int_0^2dx\int_{x/2}^xf(x,y)\,dy+\int_2^4dx\int_{x/2}^2f(x,y)\,dy$$


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