Q:

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 $x$? 

A:

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.

