Chapter 15: Problem 24
Find the exact value of \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) using any method. $$ \begin{array}{l}{\mathbf{F}(x, y)=2 x y \mathbf{i}+\left(x^{2}+\cos y\right) \mathbf{j}} \\ {C: \mathbf{r}(t)=t \mathbf{i}+t \cos (t / 3) \mathbf{j} \quad(0 \leq t \leq \pi)}\end{array} $$
Short Answer
Expert verified
The exact value of the line integral is 0.
Step by step solution
01
Understand the Problem
We need to evaluate the line integral \( \int_{C} \mathbf{F} \cdot d \mathbf{r} \) where \( C \) is parameterized by \( \mathbf{r}(t) = t \mathbf{i} + t \cos(t/3) \mathbf{j} \) for \( 0 \leq t \leq \pi \). The vector field is \( \mathbf{F}(x, y) = 2xy \mathbf{i} + (x^2 + \cos(y)) \mathbf{j} \).
02
Find \( d\mathbf{r} \)
The differential vector \( d\mathbf{r} \) can be found by differentiating the parameterization \( \mathbf{r}(t) = t \mathbf{i} + t \cos(t/3) \mathbf{j} \). This gives:\[ d\mathbf{r} = \left(1 \mathbf{i} - \left(\frac{t}{3}\sin(t/3) - \cos(t/3)\right) \mathbf{j}\right) dt \]
03
Evaluate \( \mathbf{F} \cdot d\mathbf{r} \)
First, substitute \( x = t \) and \( y = t \cos(t/3) \) in \( \mathbf{F}(x, y) \):\[ \mathbf{F}(t, t\cos(t/3)) = 2t^2 \cos(t/3) \mathbf{i} + \left(t^2 + \cos(t \cos(t/3))\right) \mathbf{j} \]Now, compute the dot product:\[ \mathbf{F} \cdot d\mathbf{r} = \left(2t^2 \cos(t/3)\right) \cdot 1 + \left(t^2 + \cos(t \cos(t/3))\right) \cdot \left(-\frac{t}{3}\sin(t/3) - \cos(t/3)\right) \]
04
Simplify and Integrate
Simplify the expression for the dot product obtained in Step 3 and integrate with respect to \( t \) from 0 to \( \pi \):\[\int_{0}^{\pi} \left[ 2t^2 \cos(t/3) - \left(t^2 + \cos(t \cos(t/3))\right) \left(\frac{t}{3}\sin(t/3) - \cos(t/3)\right) \right] dt\]Evaluate this integral carefully to find the exact value.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Field
A vector field assigns a vector to every point in space. Imagine plotting arrows on the graph to represent forces, velocities, or other vector quantities. In this problem, the vector field \( \mathbf{F}(x, y) \) is given by \( 2xy \mathbf{i} + (x^2 + \cos y) \mathbf{j} \). This means at any point \((x, y)\), the field has:
- An \( x \)-component of \( 2xy \)
- A \( y \)-component of \( x^2 + \cos(y) \)
Parameterization
Parameterization transforms a curve in space into functions of a single variable, usually \( t \). In our exercise, the parameterized path \( C \) is given as \( \mathbf{r}(t) = t \mathbf{i} + t \cos(t/3) \mathbf{j} \), ranging from \( t = 0 \) to \( t = \pi \).
This means:
This means:
- \( x = t \)
- \( y = t \cos(t/3) \)
Dot Product
Continuous operations in calculus often need combining vectors, for which the dot product is essential. It takes two vectors and returns a scalar, representing the magnitude of one vector projected onto another.
In this solution, we perform the dot product \( \mathbf{F} \cdot d\mathbf{r} \). Originally derived, \( d\mathbf{r} = (1 \mathbf{i} + (\frac{t}{3}\sin(t/3) - \cos(t/3)) \mathbf{j}) dt \) operates like this:
In this solution, we perform the dot product \( \mathbf{F} \cdot d\mathbf{r} \). Originally derived, \( d\mathbf{r} = (1 \mathbf{i} + (\frac{t}{3}\sin(t/3) - \cos(t/3)) \mathbf{j}) dt \) operates like this:
- Multiply corresponding components of \( \mathbf{F} \) and \( d\mathbf{r} \).
- Add these products.
Calculus
Calculus is the mathematical study of change. Here, it helps to evaluate the integral \( \int_{0}^{\pi} \left[ 2t^2 \cos(t/3) - \left(t^2 + \cos(t \cos(t/3))\right) \left(\frac{t}{3}\sin(t/3) - \cos(t/3)\right) \right] dt \), meaningfully.
In line integrals, calculus quantifies the accumulation or 'work' done by a force field as one moves along a curve.
In line integrals, calculus quantifies the accumulation or 'work' done by a force field as one moves along a curve.
- The expression obtained from the dot product of \( \mathbf{F} \) and \( d\mathbf{r} \) is simplified to an integrable form.
- The integral limits, from \( 0 \) to \( \pi \), guarantee it accounts for the entire parameterized curve \( C \).