Question

Use Green's Theorem to evaluate the line integral. $$ \begin{array}{l} \int_{C}\left(x^{2}-y^{2}\right) d x+2 x y d y \\ C: r=1+\cos \theta \end{array} $$

Short answer

The line integral evaluated on the specified curve using Green's Theorem is 0.

Step-by-step solution

Compute the Curl of the Vector Field

The curl of a two-dimensional vector field \(f = (M,N)\) is given by \(\frac{dN}{dx} - \frac{dM}{dy}\). So for our vector field \(f = (x^2 - y^2, 2xy)\), the curl (or z-component of the curl, in 2D), is \(\frac{d}{dx}(2xy) - \frac{d}{dy}(x^2 - y^2) = 2y - (2x + 2y) = -2x\).

Compute the Double Integral Over the Region

Now that we have the curl of \(f\), \(-2x\), we can compute the double integral \(\int\int_D -2x\, dA\). Since our region \(D\) is defined in polar coordinates, we should convert to polar coordinates by substituting \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\), and noting that \(dA = rdrd\theta\). So, our double integral becomes \(\int_0^{2\pi} \int_0^{1+\cos(\theta)} -2r\cos(\theta) rdrd\theta\).

Evaluate the Polar Double Integral

To evaluate the polar double integral, we first integrate with respect to \(r\), then with respect to \(\theta\). First, we find the antiderivative of \(-2r\cos(\theta)\) with respect to \(r\), which is \(-r^2\cos(\theta)\). When evaluating this from \(r=0\) to \(r=1+\cos(\theta)\), we get \(-((1+\cos(\theta))^2)\cos(\theta)\). Now, we integrate this from \(\theta=0\) to \(\theta=2\pi\). Using a standard integral table or a computer algebra system, we get 0.