Calculate double integral of function over triangle
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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$\begingroup$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.
$\endgroup$ 3 $\begingroup$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.
$\endgroup$ $\begingroup$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 originthen 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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