Question

Use Green's Theorem to evaluate the line integral. $$ \begin{array}{l} \int_{C} 2 \arctan \frac{y}{x} d x+\ln \left(x^{2}+y^{2}\right) d y \\ C: x=4+2 \cos \theta, y=4+\sin \theta \end{array} $$

Short answer

The integral evaluated using Green's Theorem is 4\pi.

Step-by-step solution

- Write down the given line integral form

The given line function represents a line integral in the form \( \int_C Mdx + Ndy \), where \(M = 2\arctan(y/x)\) and \(N = \ln(x^2+y^2)\). The curve C is parameterized by \(x = 4 + 2\cos\theta\) and \(y = 4 + \sin\theta\).

- Set up Green's Theorem

By Green's Theorem, the line integral is equal to the double integral of \((\partial N/\partial x - \partial M/\partial y) dA \) over the region D enclosed by the curve C. The next step is to compute the partial derivatives.

- Calculate the partial derivatives

We first find the partial derivative of N with respect to x, which is \((2x)/(x^2+y^2)\), and the partial derivative of M with respect to y, which is \(-2/(x^2+y^2)\). Then subtract to get \( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = \frac{2x}{x^2+y^2} + \frac{2}{x^2+y^2} = \frac{2(x^2+y^2)}{x^2+y^2} = 2\).

- Setup the double integral

Green's Theorem tells us that the line integral equals the double integral over the region D of 2 dA. Because D is a disc centered at (4,4) with radius 2, it is convenient to use polar coordinates to evaluate this integral. Thus, the integral becomes \(\int_{0}^{2\pi} \int_{0}^{2} 2 r dr d\theta\).

- Solve the double integral

Upon integrating and evaluating, we obtain the result as \(\int_{0}^{2\pi} \int_{0}^{2} 2 r dr d\theta = 2\pi[ \frac{r^2}{2}]_0^2 = 4\pi\).