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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.

Thanks in advance!

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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, $dxdy=rdrd\theta$. $$\iint \limits_D \left(4 - x\right)\:dxdy = \int_{\theta=0}^{\pi}\int_{r=0}^{1}(4r-r^2\cos\theta) drd\theta=\int_{\theta=0}^{\pi}\left[2r^2-\frac13r^3\cos\theta\right]_0^1d\theta$$$$=\int_{\theta=0}^{\pi}2-\frac13\cos\theta d\theta=\left[2\theta - \frac13\sin\theta\right]_0^{\pi}=2\pi$$

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  • $\begingroup$ 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? $\endgroup$ – Thomas Feb 29 '16 at 16:56
  • $\begingroup$ @Thomas Very serious typo. I am very sorry. Thank you for pointing it out. I have corrected my answer. $\endgroup$ – GoodDeeds Feb 29 '16 at 16:58
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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.)

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  • $\begingroup$ 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. $\endgroup$ – Thomas Feb 29 '16 at 17:12
  • $\begingroup$ 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 $\pi$. $\endgroup$ – Travis Willse Feb 29 '16 at 21:17

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