Integrate $x+y$ over the triangle with the vertices $(0,0),(1,0),(0,1)$

Question

Integrate $f(x,y) = x+y$ along the triangle with the vertices $(0,0),(1,0),(0,1)$

I have set $\gamma_{trinagle}=\gamma_1+\gamma_2+\gamma_3$ where:

$$\gamma_1(t)=(t,0), t\in[0,1]$$ $$\gamma_2(t)=(1-t,t), t\in[0,1]$$ $$\gamma_3(t)=(0,t), t\in[1,0]$$

so $$\int_{\gamma}f(x,y)ds=\int_{\gamma_1}f(x,y)ds+\int_{\gamma_2}f(x,y)ds+\int_{\gamma_3}f(x,y)ds=$$

$$=\int_{0}^{1}tdt+\int_{0}^{1}(1-t+t)\sqrt{2}dt-\int_{0}^{1}tdt=\frac{1}{2}+\sqrt{2}-\frac{1}{2}=\sqrt{2}$$

But the answer in the book is $\sqrt{2}+1$ where did I get it wrong?

Answer 1

$\gamma_3$ goes from $(0,0)$ to $(0,1)$, you need one from $(0,1)$ to $(0,0)$.

Answer 2

You titled this "Integrate x+ y over the triangle with vertices (0, 0), (1, 0), (0, 1) and I at first, like the previous poster, thought you meant over the area of the triangle. But if you mean a path integral, then then you also must state the direction.

Going around the triangle counter-clock-wise, the integral is from (0, 0) to (1,0), then from (1, 0) to (0, 1), and finally from (0, 1) TO (0, 0).

Going around the triangle clockwise, the integral is from (0, 0) to (0, 1), then from (0, 1) to (1, 0), and finally from (1, 0) to (0, 0).

Swapping the direction of a line integral changes to sign.