I am trying to evaluate $$\iint_{R} x+y \:d A$$, where $R$ is the region formed by the vertices $$(0,0),(5,0),\left(\frac{5}{2}, \frac{5}{2}\right) \text { and }\left(\frac{5}{2},-\frac{5}{2}\right)$$.

My try:

Here is the picture of the region which has two triangular regions.

Let the top traingle is $R1$ and bottom triangle is $R2$

We have

$$\iint _{R}(x+y)dA=\iint_{R1}(x+y)dA+\iint_{R2}(x+y)dA$$

Now we have:

$$\iint_{R1}(x+y)dA=\int_{x=0}^{5}\int_{y=x}^{5-x}(x+y)dydx=\frac{-125}{6}$$

Also $$\iint_{R2}(x+y)dA=\int_{x=0}^{5}\int_{y=-x}^{x-5}(x+y)dydx=\frac{125}{6}$$

Adding both i am getting zero. But that is not the answer. What's wrong in this approach?

## Best Answer

You are first integrating over the region bound by lines $y = x$ and $x + y = 5$ above x-axis. If you are setting up one integral, you will have to integrate wrt $x$ first. So the integral should be,

$ \displaystyle \iint_{R1}(x+y)dA=\int_{y=0}^{5/2}\int_{x=y}^{5-y}(x+y) \ dx \ dy = \frac{125}{6}$

Similarly for the region below x-axis,

$ \displaystyle \iint_{R2}(x+y)dA=\int_{y= - 5/2}^{0}\int_{x=-y}^{5+y}(x+y) \ dx \ dy = \frac{125}{12}$

But please note the symmetry of the region about x-axis $(y = 0)$ and as $y$ is an odd function i.e $f( - y) = - f(y)$, its integral over the region $(R1+R2)$ would be zero. So you could as well avoid integrating $y$.

Also given the symmetry, the integral of $x$ over $R1$ and $R2$ would be the same.

So you could as well write the integral as,

$ \displaystyle 2 \int_{y=0}^{5/2}\int_{x=y}^{5-y} x \ dx \ dy = \frac{125}{4}$

Also while it is straightforward without change of variable, you could use $x + y = u, x-y = v$ and that translates to a simple square region aligned to coordinate axes $0 \leq u \leq 5, 0 \leq v \leq 5$ and with $|J| = \frac{1}{2}$, the integral becomes

$ \displaystyle \int_0^5 \int_0^5 \frac{u}{2} ~ du ~ dv = \frac{125}{4}$