Let $D$ be the triangle with vertices $(0,0)$, $(1,0)$ and $(1,1)$.
I want to evaluate the following integral
$$I = \iint_D (x+y)\, dy\,dx$$
using two methods: by direct integration, and by application of Green's Theorem.
Method I: Direct integration
This triangle is bound between the three curves $y=0$, $y=x$ and $x=1$, so
$$I = \int_0^1\int_0^x(x+y)\, dy\,dx = \int_0^12x^2 \, dx = \frac 23$$
Method II: Green's Theorem
Observe that
$$x+y = \frac{\partial}{\partial x}\bigg(\frac 12(x^2+2xy-3y^2)\bigg)-\frac{\partial}{\partial y}(0)$$
So by Green's Theorem, we have
\begin{align}
I = & \oint_{\partial D} \bigg(0 \, , \, \frac 12(x^2+2xy-3y^2)\bigg) \cdot d\mathbf r \\
= & \int_0^1 \bigg(0 \, , \, \frac 12(x^2+2xy-3y^2)\bigg)\bigg|_{(x,y) = (t,0)} \cdot (1,0) \, dt \\
& \qquad + \int_0^1 \bigg(0 \, , \, \frac 12(x^2+2xy-3y^2)\bigg)\bigg|_{(x,y) = (1,t)} \cdot (0,1) \, dt \\
& \qquad + \int_1^0 \bigg(0 \, , \, \frac 12(x^2+2xy-3y^2)\bigg)\bigg|_{(x,y) = (t,t)} \cdot (1,1) \, dt \\
= & \int_0^1 \bigg(0 \, , \, \frac 12 t^2 \bigg) \cdot (1,0) \, dt \\
& \qquad + \int_0^1 \bigg(0 \, , \, \frac 12 (1+2t-3t^2) \bigg) \cdot (0,1) \, dt \\
& \qquad + \int_1^0 \bigg(0 \, , \, 0 \bigg) \cdot (1,1) \, dt \\
= & 0 + \int_0^1 \frac 12 (1+2t-3t^2) \, dt + 0 \\
= & \frac 12
\end{align}
The two answers are not equal. Can someone please tell me what I have done wrong?
Best Answer
$$\int_0^x (x+y) dy = \left.\left(xy+\frac12 y^2\right)\right|_{y=0}^{y=x} = \frac32 x^2 \neq 2x^2$$