Using Green's Theorem, find the outward flux of F across the closed curve C.
-$$\mathbf { F } = - \sqrt { \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } } \mathrm { i } + \sqrt { \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } } \mathrm { j } ; \mathrm { C } \text { is the region defined by the polar coordinate inequalities } 1 \leq \mathrm { r } \leq 4 \text { and } 0 \leq \theta \leq \pi$$
A) 0
B) 15
C) 30
D) 17

Question

Quizplus · Accepted Answer

Using Green's Theorem, we have:
$$\oint_C \mathbf{F} \cdot \mathrm{d}\mathbf{r} = \iint_R \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}ight)\mathrm{d}A$$
where $\mathbf{F}(x,y) = P(x,y)\mathbf{i} + Q(x,y)\mathbf{j}$ and $R$ is the region enclosed by $C$.
Here, $P(x,y) = -\sqrt{x^2+y^2}$ and $Q(x,y) = \sqrt{x^2+y^2}$.
Taking partial derivatives, we get:
$$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} = \frac{x}{\sqrt{x^2+y^2}} + \frac{y}{\sqrt{x^2+y^2}} = \frac{x+y}{\sqrt{x^2+y^2}}$$
Converting to polar coordinates, we have $x = r\cos	heta$, $y = r\sin	heta$, and $\sqrt{x^2+y^2} = r$. Also, $0\leq 	heta \leq \pi$ and $1\leq r\leq 4$. Thus, we have:
$$\frac{x+y}{\sqrt{x^2+y^2}} = \frac{r\cos	heta + r\sin	heta}{r} = \cos	heta + \sin	heta$$
The integral becomes:
$$\iint_R (\cos	heta + \sin	heta) r\mathrm{d}r\mathrm{d}	heta$$
Integrating with respect to $	heta$ from $0$ to $\pi$ and with respect to $r$ from $1$ to $4$, we get:
$$\int_0^\pi \int_1^4 (\cos	heta + \sin	heta) r\mathrm{d}r\mathrm{d}	heta = \int_0^\pi \left(\frac{15}{2}\cos	heta+\frac{15}{2}\sin	hetaight)\mathrm{d}	heta = 15$$
Therefore, the outward flux of $\mathbf{F}$ across the closed curve $C$ is $\boxed{	ext{(B) } 15}$.

Using Green's Theorem, Find the Outward Flux of F Across