Calculate the double integral $I=\iint_Dxy\,{\rm d}x\,{\rm d}y$ The problem is as follows: Calculate the double integral
$$I=\iint_Dxy\,{\rm d}x\,{\rm d}y$$
when the region $D$ is defined by $\{(x,y):0\le x\le1, 0\le y\le1, xy\le\frac{1}{2} \}$. 
The region $D$ looks like this 1. 
Let $D$ be subdivide into three regions called $R_1=[0,\frac{1}{2}]\times [0,1]$, $R_2=[\frac{1}{2},1]\times [0,\frac{1}{2}]$ and $R_3=[\frac{1}{2},1]\times [\frac{1}{2},\frac{1}{2x}]$.
For $R_1$ we get that 
$$I_1=\int_0^\frac{1}{2}(x \int_0^1  \!y  \ dy) \ dx =\frac{1}{16},$$ 
for $R_2$ we get 
$$I_2=\int_\frac{1}{2}^1(x \int_0^\frac{1}{2}  \!y  \ dy) \ dx =\frac{3}{64}$$
and for $R_3$ we get $$I_3=\int_\frac{1}{2}^1(x \int_\frac{1}{2}^\frac{1}{2x}  \!y  \ dy) \ dx =\frac{\ln(2)}{8}-\frac{3}{64}.$$
Now, adding up $I_1+\ I_2+ \ I_3$ we get $$I=\frac{1}{16}+\frac{3}{64}+\frac{\ln(2)}{8}-\frac{3}{64}=\frac{1}{16}+\frac{\ln(2)}{8}.$$
Is this correct?
Now, I realise that it would be easier to calculate the full region $R=[0,1]\times [0,1]$ and then subtract the region $R_\Delta=[\frac{1}{2},1]\times [\frac{1}{2x},1].$
For the region $R$ we get $$I_R=\int_0^1(x \int_0^1  \!y  \ dy) \ dx =\frac{1}{4}.$$
For the region $R_\Delta$ we get $$I_{R_\Delta}=\int_\frac{1}{2}^1(x \int_\frac{1}{2x}^1  \!y  \ dy) \ dx =\frac{3}{16}-\frac{\ln(2)}{8}.$$
Continue to subtract and we get that
$$I_R-I_{R_\Delta}=\frac{1}{4}-(\frac{3}{16}-\frac{\ln(2)}{8})=\frac{1}{16}+\frac{\ln(2)}{8}.$$
So, we get the same answer as in the previous calculation. Is this really correct? If not, please show me how to do it properly.
 A: Alternatively, you can consider the symmetric line $y=x$, which crosses $xy=\frac12$ at $(\frac1{\sqrt{2}},\frac1{\sqrt{2}})$. Refer to the graph:
$\hspace{5cm}$
Hence the integral is:
$$S=2(A_1+A_2)=\\
2\left(\int_0^{1/\sqrt{2}}\int_0^x xy\, dy\, dx+\int_{1/\sqrt{2}}^1\int_0^{1/(2x)} xy\, dy\, dx\right)=\\
2\left(\int_0^{1/\sqrt{2}}x\cdot \frac{x^2}2\, dx+\int_{1/\sqrt{2}}^1 x\cdot \frac1{8x^2}\, dx\right)=\\
2\left(\frac1{32}+\frac1{16}\ln 2\right)=\frac1{16}+\frac{\ln 2}{8}.$$
Note: It works because the integrand function is symmetric with respect to $y=x$ line.
A: Yup, your working is fine.
Actually there is no need to separate $R_2$ from $R_3$.
You could have computed 
$$\int_\frac12^1 \int_0^\frac{1}{2x}xy\, dy\, dx = \int_\frac12^1\frac{x}{2}\left( \frac1{4x^2}\right)\, dx = \frac18\int_{\frac12}^1x^{-1}\, dx=\frac{\ln 2}8$$
Adding it with integration over $R_1$ gives you the result.
A: While both methods are correct, I definitely prefer the second method to get $$I_{R_{Δ}}=\frac{1}{4}-(\frac{3}{16}-\frac{ln(2)}{8})=\frac{1}{16}+\frac{ln(2)}{8}$$
