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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?

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$$\int_0^x (x+y) dy = \left.\left(xy+\frac12 y^2\right)\right|_{y=0}^{y=x} = \frac32 x^2 \neq 2x^2$$

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