# Evaluating a given contour integral using Green's Theorem

Let $D$ be the region in $\mathbb{R}^2$ that contains the points $(x,y) : x^2 + y^2 \leq 1$ and $y \geq 0$. Let $C$ be the curve enclosing $D$ oriented against the clock. Evaluate $$\int\limits_C \left(xy + \ln(x^2 + 1) \right) dx + \left(4x + e^{y^2} + 3\arctan(y)\right) dy .$$

This is what I've got:

$$D = \left \{(x,y):-1 \leq x \leq 1, 0 \leq y \leq \sqrt{1 - x^2} \right \}$$

Now if i'm using the fact that: $$\int \limits_C P\:dx + Q\:dy = \iint \limits_D \left(Q_x - P_y \right)\:dxdy = \iint \limits_D \left(4 - x\right)\:dxdy$$ Where $Q_x,\:P_y$ are the partial derivatives.

Now trying to solve this the way i'vs set this up with $D$ i get some expressions that certainly doesn't look like the answer my book gives me $(2\pi)$. Have i set this one up right and calculated wrong? Or is there a much easier way to set this one up? I can use any method as long as it touches on Green's Theorem, i.e., path integral or double, changing of bounds, etc.

Any help would be appreciated.

Your expression leads to an answer of $2\pi$. An easy way is to convert to the polar form. Then, $x=r\cos\theta$ and $y=r\sin\theta$. The determinant of the Jacobian matrix is $r$, thus, $dxdy=rdrd\theta$. $$\iint \limits_D \left(4 - x\right)\:dxdy = \int_{\theta=0}^{\pi}\int_{r=0}^{1}(4r-r^2\cos\theta) drd\theta=\int_{\theta=0}^{\pi}\left[2r^2-\frac13r^3\cos\theta\right]_0^1d\theta$$$$=\int_{\theta=0}^{\pi}2-\frac13\cos\theta d\theta=\left[2\theta - \frac13\sin\theta\right]_0^{\pi}=2\pi$$
• Great idea to switch to polar, your calculations looks good, however in your second step, could you clarify how you got the $4$ multiplied with $\cos(\theta)$ and the boundaries going all the way to $2\pi$ since we have a hemisphere? – Thomas Feb 29 '16 at 16:56
Everything so far looks good to me, and the value of the integral appears to be $w \pi$ as claimed, so the issue must be in the evaluation of the double integral.
Notice that we can evaluate it quickly using the observations that (1) the region $\require{cancel}D$ enclosed by $C$ is symmetric about the $y$-axis and (2) the function $x \mapsto x$ is odd, which together imply the cancellation on the r.h.s. in $$\iint_D (4 - x) \,dA = 4 \iint_D \,dA - \cancelto{0}{\iint_D x \,dA}.$$ By definition, $\iint_D \,dA$ is just the area of $D$. (If for whatever reason you want to evaluate the integral manually, the fact that $D$ is a half-disk centered at the origin suggests changing to polar coordinates.)
• I hope you mean symmetric about $y$-axis?:) Anyways, great you showed how easy this was when pointing out symmetries etc. If i did want to change it to p.c. it would be $D=\left \{(r,\theta):0\leq r \leq 1, 0 \leq \theta \leq \pi \right \}$ right? I'm getting abit confused on these integrals weather i should go a full revolution to capture the curve $C$ or not, but as you set it up it is -as one would probably expect- a half revolution. – Thomas Feb 29 '16 at 17:12
• Oops, yes, I corrected the typo. And yes, that's the right expression for $D$ in polar coordinates, as you can think of $D$ as sector w central angle $\pi$. – Travis Willse Feb 29 '16 at 21:17