91Ó°ÊÓ

We need to calculate the flow of the vector field \(\mathbf{F} = (x+y)\mathbf{i}-(x^2+y^2)\mathbf{j}\) along given paths. Flow, in this context, means calculating the line integral \(\int_C \mathbf{F} \cdot d\mathbf{r}\), where \(C\) is the path, and \(d\mathbf{r}\) is the differential element along the path. We'll assess the problem separately for each curve.

For the upper half of the circle \(x^2 + y^2 = 1\), we use the parameterization \(x = \cos t\), \(y = \sin t\), which covers \(t\) from 0 to \(\pi\). Thus, \(d\mathbf{r} = (-\sin t \mathbf{i} + \cos t \mathbf{j}) dt\).

Compute \(\int_C \mathbf{F} \cdot d\mathbf{r}\) for path A using the parameterization: \[ \mathbf{F} \cdot d\mathbf{r} = ((\cos t + \sin t)(-\sin t) + (-(\cos^2 t + \sin^2 t))\cos t) \, dt = (-\cos t \sin t - \sin^2 t - \cos t) \, dt. \]Integrate from 0 to \(\pi\): \[ \int_{0}^{\pi} (-\cos t \sin t - \cos t) \, dt = \int_{0}^{\pi} (-\cos t(\sin t + 1)) \, dt. \]This evaluates to 0 after integration.

For the line segment from (1,0) to (-1,0), the parameterization is \(x = 1 - 2t\), \(y = 0\) with \(t\) from 0 to 1. Then, \(d\mathbf{r} = (-2 \mathbf{i}) dt\). Compute \(\int_C \mathbf{F} \cdot d\mathbf{r}\), substituting in \(\mathbf{F}\) and \(d\mathbf{r}\): \[ \mathbf{F} \cdot d\mathbf{r} = ((1-2t)-1)(-2) dt = -2 (1-2t) dt. \]Integrate from 0 to 1: \[ \int_{0}^{1} -2 (1-2t) \, dt = -2 \left[t - t^2 \right]_0^1 = 0. \]

Path C consists of two line segments. First, (1,0) to (0,-1): parameterize as \(x = 1-t\), \(y = -t\), with \(t\) from 0 to 1. Then, \(d\mathbf{r} = (-\mathbf{i} - \mathbf{j}) dt\).Compute \(\int \mathbf{F} \cdot d\mathbf{r}\) for this segment: \[ \mathbf{F} = (1-2t)\mathbf{i} - (1 + t^2), \mathbf{F} \cdot d\mathbf{r} = (1-2t)(-1) - (1+t^2)(-1) dt = (-1+2t + 1 + t^2) dt = (2t + t^2) dt. \]Integrate: \[ \int_0^1 (2t + t^2) dt = \left[t^2 + \frac{t^3}{3}\right]_0^1 = 1 + \frac{1}{3} = \frac{4}{3}. \]For the segment from (0,-1) to (-1,0), parameterize as \(x = -t\), \(y = -1+t\), with \(t\) from 0 to 1. Then, \(d\mathbf{r} = (-\mathbf{i} + \mathbf{j}) dt\).Compute \(\int \mathbf{F} \cdot d\mathbf{r}\) for this segment: \[ \mathbf{F}\cdot d\mathbf{r} = ((-t+(-1+t))(1)) + (-(t^2 + (1-t)^2))(-1)dt. \]Simplify and integrate: \(\int_0^1 (-2t + 1)\) dt which evaluates to 0.Thus, the total flow for Path C is \(\frac{4}{3}.\)

The flow integral is evaluated for each path:- Path A (half-circle): 0.- Path B (line): 0.- Path C (poly-line): \(\frac{4}{3}.\)