Let \(R\) be the region in the first quadrant bounded by the curves \(x^2 + y^2 = 2x\) and \(y = 0\). Let \(C\) be the boundary of the region \(R\), oriented counterclockwise. Evaluate the line integral \[ \int_C x e^x \, dx + (y e^y + x^2)\, dy. \]
Solution
We first note that we can rewrite the equation of the region to obtain something more familiar: \(x^2 + y^2 = 2x \implies (x-1)^2 + y^2 = 1\), which is a translated circle. Integrating over this region will be easy compared to the line integral, so we apply Green’s theorem: \[ \int_C xe^x ~dx + ye^y +x^2 ~dy = \iint_D 2x ~dA. \]
We can parameterize this region as \[ D = \theset{x^2+y^2-2x = 0 \suchthat (x,y) \in \RR^2, y \geq 0} = \theset{(r(1+ \cos\theta), r\sin\theta) \suchthat \theta \in [0, \pi), r\in [0, 1]}. \]
Noting that \(dA = r~dr~d\theta\), we can then integrate \[ \iint_D 2x ~dA = \int_0^{\pi} \int_0^1 2(r(1 + \cos\theta)) r ~dr ~d\theta \\ = 2\int_0^{\pi} \int_0^1 r^2(1+\cos \theta) ~dr ~d\theta \\ = 2\int_0^{\pi} \frac 1 3 r^3(1+\cos \theta) \bigg\rvert_0^1 ~d\theta\\ = \frac 2 3 \int_0^{\pi} (1+\cos \theta) ~d\theta\\ = \frac 2 3 ( \theta + \sin \theta) \bigg\rvert_0^\pi \\ = \frac 2 3 [(\pi + 0) - (0 + 0)] = \frac 2 3 \pi. \qed \]