How to use Green's theorem for a line integral over a triangular region

Key takeaway: When a curve encloses a triangle, Green's theorem often turns a line integral into a double integral over the interior. The only real work is describing the triangular region correctly, especially the moving boundary.

Let

$P(x,y)=\sqrt{1+x^3},\qquad Q(x,y)=2xy.$

Green's theorem says

$\oint_C P\,dx+Q\,dy=\iint_D\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA.$

1) Compute the derivatives

$\frac{\partial Q}{\partial x}=2y,\qquad \frac{\partial P}{\partial y}=0.$

So the integral becomes

$\iint_D 2y\,dA.$

2) Describe the triangular region

The triangle has vertices $(0,0)$, $(1,0)$, and $(1,3)$. The slanted edge is the line from $(0,0)$ to $(1,3)$, so its equation is

$y=3x.$

That means the region can be written as

$0\le x\le 1,\qquad 0\le y\le 3x.$

So yes, the inner bound for $y$ is not from $0$ to $3$; it depends on $x$.

3) Evaluate the double integral

$\iint_D 2y\,dA=\int_0^1\int_0^{3x}2y\,dy\,dx.$

First integrate in $y$:

$\int_0^{3x}2y\,dy=y^2\Big|_0^{3x}=9x^2.$

Now integrate in $x$:

$\int_0^1 9x^2\,dx=9\cdot\frac{1}{3}=3.$

So

$\oint_C \sqrt{1+x^3}\,dx+2xy\,dy=3.$

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Pitfalls the pros know 👇 A common mistake is to set the region as $0\le y\le 3$ and then forget that the triangle does not extend that way for every $x$. The top edge changes with $x$, so the correct upper bound is the line $y=3x$.

Another easy slip is mixing up the boundary orientation. Green's theorem requires counterclockwise orientation; if the curve were clockwise, the answer would change sign.

What if the problem changes? If you reverse the order of integration, the same triangle can be written as

$0\le y\le 3,\qquad \frac{y}{3}\le x\le 1.$

Then

$\iint_D 2y\,dA=\int_0^3\int_{y/3}^1 2y\,dx\,dy,$

which gives the same value $3$.

So the bounds are not unique; what matters is that they match the actual triangular region.

Tags: Green's theorem, double integral, triangular region