# Why the answer for double integral is coming as zero

definite integralsmultiple integralmultivariable-calculus

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?

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

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