Question

Circulation and flux For the following vector fields, compute (a) the circulation on, and (b) the outward flux across, the boundary of the given region. Assume boundary curves are oriented counterclockwise. $$\begin{aligned} &\mathbf{F}=\langle x, y\rangle ; R \text { is the half-annulus }\\{(r, \theta) ; 1 \leq r \leq 2\\\ &0 \leq \theta \leq \pi\\} \end{aligned}$$

Short answer

Question: Calculate the circulation and outward flux of the vector field F = ⟨x, y⟩, where the region R is a half-annulus in polar coordinates with R = {(r,θ); 1 ≤ r ≤ 2, 0 ≤ θ ≤ π}. Answer: (a) The circulation of the vector field F around the region R is 3. (b) The outward flux of the vector field F across the boundary of the region R is -7π/6.

Step-by-step solution

Convert the vector field to polar coordinates

The given vector field is F = ⟨x, y⟩. Let's convert this vector field from Cartesian to polar coordinates. Recall that \(x = r\cos{\theta}\) and \(y = r\sin{\theta}\). So the vector field in polar coordinate becomes: $$\mathbf{F} = \langle r\cos{\theta}, r\sin{\theta} \rangle$$

Find the parametrization of the boundary curves

The region R is defined in polar coordinates as a half-annulus: $$R= \{(r,\theta); 1\leq r\leq 2, 0\leq\theta\leq\pi\}$$ The boundary of R consists of four curves: 1. \(C_1\) is the inner semicircle with radius 1 from \(\theta=0\) to \(\theta=\pi\). The parametrization of \(C_1\) is: $$\mathbf{r}_1(\theta) = \langle 1\cos{\theta}, 1\sin{\theta}\rangle \quad, 0 \leq \theta \leq \pi$$ 2. \(C_2\) is the outer semicircle with radius 2 from \(\theta=\pi\) to \(\theta=0\). The parametrization of \(C_2\) is: $$\mathbf{r}_2(\theta) = \langle 2\cos{\theta}, 2\sin{\theta}\rangle \quad , \pi \leq \theta \leq 2\pi$$ 3. \(C_3\) is the straight line segment from radius 1 to 2 when \(\theta=0\). The parametrization of \(C_3\) is: $$\mathbf{r}_3(r) = \langle r\cos{0} , r\sin{0}\rangle \quad , 1 \leq r \leq 2$$ 4. \(C_4\) is the straight line segment from radius 1 to 2 when \(\theta=\pi\). The parametrization of \(C_4\) is: $$\mathbf{r}_4(r) = \langle r\cos{\pi} , r\sin{\pi}\rangle \quad , 1 \leq r \leq 2$$

Compute the circulation on the boundary

To compute the circulation around the boundary, we need to evaluate the line integral of the vector field along the boundary curves C1, C2, C3, and C4, and sum them up. Circulation = \(\int_{C_1+C_2+C_3+C_4} \mathbf{F} \cdot d\mathbf{r} = \int_{C_1}\mathbf{F} \cdot d\mathbf{r} + \int_{C_2}\mathbf{F} \cdot d\mathbf{r} + \int_{C_3}\mathbf{F} \cdot d\mathbf{r} + \int_{C_4} \mathbf{F} \cdot d\mathbf{r}\) Let's evaluate each integral. 1. For \(C_1\), we have \(d\mathbf{r}_1 = \langle -\sin{\theta}, \cos{\theta}\rangle d\theta\). $$\int_{C_1}\mathbf{F} \cdot d\mathbf{r} = \int_{0}^{\pi} \langle \cos{\theta},\sin{\theta}\rangle \cdot \langle -\sin{\theta}, \cos{\theta}\rangle d\theta= \int_{0}^{\pi}(-\cos{\theta}\sin{\theta}+\sin{\theta}\cos{\theta})d\theta = 0$$ 2. For \(C_2\), let \(\eta=\theta-\pi\), so \(d\eta=d\theta\) and the limits are \(0\leq\eta\leq\pi\): $$\int_{C_2}\mathbf{F}\cdot d\mathbf{r}=\int_{0}^{\pi}\langle 2\cos{(\eta+\pi)},2\sin{(\eta+\pi)}\rangle\cdot\langle -2\sin{(\eta+\pi)},2\cos{(\eta+\pi)}\rangle d\eta $$ $$= \int_{0}^{\pi}(-4\cos{(\eta+\pi)}\sin{(\eta+\pi)}+4\sin{(\eta+\pi)}\cos{(\eta+\pi)})d\eta=0$$ 3. For \(C_3\), we have \(d\mathbf{r}_3 = \langle 1, 0 \rangle dr\). $$\int_{C_3}\mathbf{F} \cdot d\mathbf{r} = \int_{1}^{2}\langle r, 0 \rangle \cdot \langle 1, 0 \rangle dr= \int_{1}^{2}r dr =\frac{3}{2}$$ 4. For \(C_4\), we have \(d\mathbf{r}_4 = \langle -1, 0 \rangle dr\). $$\int_{C_4}\mathbf{F} \cdot d\mathbf{r} = \int_{1}^{2}\langle -r, 0 \rangle \cdot \langle -1, 0 \rangle dr= \int_{1}^{2}r dr =\frac{3}{2}$$ So the total circulation is: $$\text{Circulation} = 0+0+\frac{3}{2}+\frac{3}{2}=3$$

Compute the outward flux across the boundary

Since the given vector field is 2D, we can use Green's theorem to compute the outward flux across the boundary: Outward flux = \(\oint_C (Mdx + Ndy) = \iint_R (\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})dA\) We have \(M = r\cos{\theta}\) and \(N=r\sin{\theta}\), so we need to compute \(\frac{\partial N}{\partial x}\) and \(\frac{\partial M}{\partial y}\). Using the chain rule, we get: \(\frac{\partial N}{\partial x} = \frac{\partial N}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial N}{\partial \theta} \frac{\partial \theta}{\partial x} = (\cos{\theta})\frac{x}{\sqrt{x^2 + y^2}} - (r\cos{\theta})\frac{-y}{r^2}\) \(\frac{\partial M}{\partial y} = \frac{\partial M}{\partial r} \frac{\partial r}{\partial y} + \frac{\partial M}{\partial \theta} \frac{\partial \theta}{\partial y} = (-\sin{\theta})\frac{y}{\sqrt{x^2 + y^2}} + (0) = -\frac{y}{r}\sin{\theta}\) Substituting the expressions, we get: Outward flux = \(\iint_R (\frac{y}{r}\sin{\theta} + \frac{x}{r}\cos{\theta}-r\cos^2{\theta})r dr d\theta\) Calculate this integral using polar coordinates: $$\text{Outward flux} = \int_{0}^{\pi}\int_{1}^{2} (r\sin^2{\theta} + r\cos^2{\theta}-r^2\cos^2{\theta}) dr d\theta$$ $$=-\int_{0}^{\pi}\int_{1}^{2} r^2\cos^2\theta - r\cos^2\theta - r\sin^2\theta dr d\theta$$ This integral can be simplified by noticing that \(\cos^2\theta+\sin^2\theta=1\): $$\text{Outward flux}=-\int_{0}^{\pi}\int_{1}^{2} r^2 - r dr d\theta =-\int_{0}^{\pi}(-\frac{7}{6}) d\theta=-\frac{7}{6}(\pi-0)=-\frac{7\pi}{6}$$ Finally, we can write down the circulation and the outward flux: (a) The circulation is 3. (b) The outward flux is \(-\frac{7\pi}{6}\).

Key concepts

Circulation

In vector calculus, circulation measures the tendency of a vector field to rotate around a given path. Imagine a fluid flowing in a closed loop, and circulation tells us the net amount of fluid that spirals around the loop. It is a line integral of the vector field along a closed curve. For the given problem, the vector field \( \mathbf{F} = \langle x, y \rangle \) is integrated along the boundary of a region defined as a half-annulus.

The steps involve understanding the path around which circulation will be computed:

• Identify the boundary segments of the region, which could be curves or lines.
• Parametrize these segments to express them in mathematical terms.

Calculating circulation involves integrating the dot product of the vector field \( \mathbf{F} \) and the differential displacement \( d\mathbf{r} \) over each segment, then summing these results. In the exercise, this calculation led to the total circulation of 3, meaning there is a non-zero net rotation around the boundary in a counterclockwise direction.

Flux

Flux in vector calculus quantifies how much of a vector field passes through a given surface or area. Picture a net and a flow of water; flux measures how much water passes through the net. For a 2D vector field, this means calculating how much of the field moves out of a closed curve.

Outward flux is often computed using Green's Theorem, which transforms a double integral over a region into a line integral along its boundary. The components of the vector field are analyzed:

• "Outward" indicates we focus on how much of the vector field is exiting the surface area.
• We use \( M \) and \( N \) for the x and y components of the vector field, respectively.
• These components help calculate partial derivatives needed for Green's theorem application.

In this exercise, solving for outward flux gives -\( \frac{7\pi}{6} \), signifying a net exit of field lines across the boundary, expressing how the entire vector field behaves across the half-annulus.

Green's Theorem

Green's Theorem is a key result in vector calculus that connects a line integral around a closed curve to a double integral over the plane region it encloses.

It can be seen as an extension of the Fundamental Theorem of Calculus for multi-dimensional calculus. Green's theorem is applicable if:

• The curve is closed and parameterized counterclockwise.
• The vector field is continuously differentiable.

The theorem states:\[\oint_C (M\, dx + N\, dy) = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA\] In terms of the exercise, applying Green's theorem helps compute the outward flux, converting the circulation results into an easier-to-analyze area integral. The theorem effectively relates the behavior of the vector field along the perimeter to its behavior inside the region, simplifying complex boundary calculations.

Polar Coordinates

Polar coordinates provide a way of representing points in the plane using distances and angles, rather than x and y coordinates. This system is particularly useful when dealing with circular or radial regions, such as the half-annulus in our exercise.

In polar coordinates:\[ x = r \cos{\theta}, \quad y = r \sin{\theta} \]Here, \( r \) represents the radial distance from the origin, and \( \theta \) is the angle from the positive x-axis.

• For vector fields in polar coordinates, transformation of vector components is essential for integration.
• Converting to polar coordinates often simplifies the integration process over circular boundaries.

In this exercise, the original vector field \( \mathbf{F} = \langle x, y \rangle \) is transformed into polar form as \( \mathbf{F} = \langle r \cos{\theta}, r \sin{\theta} \rangle \).
These transformations allow handling the integration more naturally, considering radial symmetry of the region and simplifying calculations on curved paths.