Chapter 8: Chapter 8

Problem 209

a) Let $F^{\rightarrow}$ be a vector field on an open set $V$ and $C$ a curve in V defined on the interval $[a, b]$. Prove $\int_{(C)-} F^{\rightarrow}=-\int_{C} F^{-}$, where $C^{-}$ is the reverse path of the curve $C$. b) Then evaluate $\int_{\mathrm{C}} \mathrm{F}^{-} \cdot \mathrm{d} \mathrm{C}^{-}$ where $\mathrm{F}^{-}(\mathrm{x}, \mathrm{y})=\left(\mathrm{x}^{2}, \mathrm{xy}\right)$ along the line segment from the point $(1,1)$ to $(0,0)$ using the reverse path.

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The short version of the answer based on the provided step-by-step solution would be: We first proved the relationship between the line integrals of curve $C$ and its reverse path $C^-$ by parametrizing both paths and comparing the results. Then, we found the parametric representation of the reverse path of the line segment from point (1,1) to point (0,0) as $\vec{r}^-(t) = (1-t,1-t)$ and evaluated the line integral of the given vector field $\vec{F}(x, y) = (x^2, xy)$ along this path. The result of the integral is $\int_{\mathrm{C}} \mathrm{F}^{-}\cdot \mathrm{d} \mathrm{C}^{-} = -\frac{2}{3}$.

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TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: In order to prove the relation between integrals of the curve C and its reverse path, we need to understand what a reverse path is. Given a curve C defined by a parametric representation $\vec{r}(t)$ on the interval $[a,b]$, the reverse path, denoted by $C^-$, is the same curve traced in the opposite direction. Mathematically, this can be represented as $\vec{r}^-(t) = \vec{r}(a+b-t)$, for $t \in [a, b]$. #Step 2: Prove the relation between integrals#

To prove the relationship between the integrals of curve C and its reverse path $C^-$, we will first write down the line integral for both paths. For $C$, the line integral is: \[ \int_{C} \vec{F} \cdot d\vec{r} = \int_{a}^{b} \vec{F}(\vec{r}(t)) \cdot \frac{d\vec{r}}{dt}\,dt. \] For $C^-$, the line integral is: \[ \int_{(C^-)} \vec{F} \cdot d\vec{r}^- = \int_{a}^{b} \vec{F}(\vec{r}^-(t)) \cdot \frac{d\vec{r}^-}{dt}\,dt. \] Since $\vec{r}^-(t) = \vec{r}(a+b-t)$, we can substitute this into the second integral, and calculate the derivative of $\vec{r}^-$ with respect to t. Now let's substitute $\vec{r}^-$ with $\vec{r}(a+b-t)$: \[ \int_{(C^-)} \vec{F} \cdot d\vec{r}^- = \int_{a}^{b} \vec{F}(\vec{r}(a+b-t)) \cdot \frac{d\vec{r}(a+b-t)}{dt}\,dt. \] Then, using the chain rule, we get $\frac{d\vec{r}(a+b-t)}{dt} = -\frac{d\vec{r}(t)}{dt}$: \[ \int_{(C^-)} \vec{F} \cdot d\vec{r}^- = -\int_{a}^{b} \vec{F}(\vec{r}(t)) \cdot \frac{d\vec{r}}{dt}\,dt. \] This shows that $\int_{(C)-} \vec{F} = -\int_{C} \vec{F}$, which is what we wanted to prove. #Step 3: Parametrizing the reverse path for part b#

Step 2: For part b, we have to compute the line integral of the vector field $\vec{F}(x, y) = (x^2, xy)$ along the reverse path of the line segment from point (1,1) to point (0,0). We can parametrize the original path from (0,0) to (1,1) as $\vec{r}(t) = (t,t)$ for $0 \le t \le 1$. To find the reverse path, we simply adjust the parameter so that the endpoints are reversed: $\vec{r}^-(t) = (1-t,1-t)$, where $0 \le t \le 1$. #Step 4: Computing the line integral for part b#

Using the vector field $\vec{F}(x, y) = (x^2, xy)$ and the reverse path $\vec{r}^-(t) = (1-t,1-t)$, we can calculate the line integral: First, find the derivative of the reverse path with respect to t: $\frac{d\vec{r}^-}{dt} = (-1,-1)$. Then, evaluate the vector field along the reverse path: $\vec{F}(\vec{r}^-(t)) = ((1-t)^2, (1-t)^2)$. Now we can write down the integral formula: \[ \int_{C^-} \vec{F} \cdot d\vec{r}^- = \int_{0}^{1} ((1-t)^2, (1-t)^2) \cdot (-1, -1)\,dt. \] Compute the dot product: \[ \int_{0}^{1} (-(1-t)^2 - (1-t)^2)\,dt. \] Combine the terms: \[ \int_{0}^{1} (-2(1-t)^2)\,dt. \] Finally, integrate: \[ -2 \int_{0}^{1} (1-2t+t^2)\,dt = -2\left[t - t^2 + \frac{t^3}{3}\right]_0^1 = -2\left(1 - 1 + \frac{1}{3}\right) = -\frac{2}{3}. \] Therefore, the result $\int_{\mathrm{C}} \mathrm{F}^{-}\cdot \mathrm{d} \mathrm{C}^{-} = -\frac{2}{3}$.

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Chapter 8: Chapter 8

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