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Find the limits for integrals $\int\int f(x,y) \,dy \, dx$ and $\int\int f(x,y)\,dx\,dy$ and compute the integral over the region, based on the function $f(x,y) = 3x^2y$.

Region = triangle inside the lines $x=0$, $y=1$, $y=2x$.


To find what the limits of my inner integral should be, I tried to sketch it. My problem was, how do I sketch a triangle when the only information I have is $x=0, y=1, y=2x$. I have no boundary for $x$?

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  • $\begingroup$ My problem is, I dont know what the inner integral should be - to define what the limits of my integration should be. $\endgroup$ – user1090614 Jun 3 '13 at 23:12
  • $\begingroup$ You might find it helpful to sketch and shade the triangle out to get a better idea of what you're integrating. Do they still teach the "arrows" method to see what your bounds should be? $\endgroup$ – Zen Jun 3 '13 at 23:24
  • $\begingroup$ I tried to sketch it, my problem was, how do I sketch a triangle when the only information I have is $x=0, y=1, y=2x$ I have no boundary for x? $\endgroup$ – user1090614 Jun 3 '13 at 23:37
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I take it that you need to compute

$$\iint_T dx dy \, f(x,y)$$

where $T$ is the triangle you described above. To see what to do, draw a picture. You'll notice that $T$ actually consists of the triangle above the line $y=2 x$, and bounded by the $y$ axis on the left, and the line $y=1$ from above. In this case, I find it easier to integrate over $x$ first, which means that the limit in $x$ is $[0,y/2]$. The integral looks like

$$\int_0^1 dy \, \int_0^{y/2} dx \, f(x,y)$$

When $f(x,y) = 3 x^2 y$ then you have

$$3 \int_0^1 dy \, y \, \int_0^{y/2} dx \, x^2$$

Evaluate from right to left. I take it you can handle the rest.

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  • $\begingroup$ I tried to sketch it, my problem was, how do I sketch a triangle when the only information I have is $x=0, y=1, y=2x$ I have no boundary for x? $\endgroup$ – user1090614 Jun 3 '13 at 23:34
  • $\begingroup$ @user1090614 you don't need another boundary for $x$: the $x$ boundary is at the line $y=2 x$, or alternatively, $x=y/2$. The boundary $y=1$ closes the triangle. $\endgroup$ – Ron Gordon Jun 3 '13 at 23:43
  • $\begingroup$ Ahhhhhh!!! gosh, thanks man! $\endgroup$ – user1090614 Jun 3 '13 at 23:45
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Either

$\displaystyle \int_0^1 dy \int_0^{y/2} dx f(x,y)$

or

$\displaystyle \int_0^{1/2} dx \int_{2x}^1 dy f(x,y) $

In the first, you effectively pick a value of $y$, then integrate in $x$ at that $y$ value from $x=0$ to $x=y/2$. Then you "add up" (i.e. integrate) all the integrals at all the various values of $y$. In the second, you pick a value of $x$, then integrate in $y$ from $y=2x$ to $y=1$, etc.

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I think you may be misreading the question, as the line "Region = triangle inside the lines x=0, y=1, y=2x." provides all the boundary information you need. If you draw the three lines:

 x=0 (for any y) i.e. the vertical line along the y axis,
 y=1 (for any x) i.e. the horizontal line 1 unit above the x axis,
 y=2x i.e. the line of slope 2 that passes through the origin

then you will see they create a triangular region bounded by the points (0,0), (0,1), and (0,1/2). That is the region within which you need to do the integration.

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  • $\begingroup$ Beat me to it while I was typing... $\endgroup$ – MikeFee Jun 3 '13 at 23:48

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