I'm given a line integral $$\int_{C} \left(\frac{\sin(3x)}{x^2+1}-6x^2y\right) dx + \left(6xy^2+\arctan\left(\frac{y}{7}\right)\right) dy$$ where C is the circle $$x^2+y^2=8$$ oriented in the counter clockwise direction. I'm supposed to solve it with Green's Theorem.

What I have so far is the parametrization of C:$$\vec{r}=\left \langle 2\cos(t), 2\sin(t) \right \rangle , 0\leq t\leq 2\pi$$

What I am confused about is how to proceed after this step. When I try to substitute $x$ and $y$ into the integral, I get a very complicated integral. I feel like I am missing something to simplify the integral to something more nice to solve. Any insights would be greatly appreciated.

  • $\begingroup$ Green's Theorem converts the line integral into an area integral over the disc of radius $\sqrt{8}$. You don't need a parametrization of $C$ anymore. $\endgroup$ – Christian Blatter Apr 11 '16 at 7:50

Green's theorem converts your line integral into a double integral over the region bounded by the (closed) curve, in your case the circle.

$$\oint_{C^+} (L\, dx + M\, dy) = \iint_{D} \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)\, dx\, dy$$

Calculating the line integral itself is hard, but notice that the integrand becomes a lot simpler since in your case:

$$\frac{\partial M}{\partial x} = 6y^2 \quad \mbox{and} \quad \frac{\partial L}{\partial y} = -6x^2$$

The integral is then simply:

$$6 \iint_{D} \left( x^2+y^2 \right) \, dx\, dy$$

where $D$ is the disc $x^2+y^2 \le 8$; this is begging for polar coordinates!


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