Problem 104 - Calculus Volume 3

Question

\([\mathbf{T}] \quad\) Find \(\left.\quad \int_{C} \mathbf{F} \cdot d \mathbf{r}, .\right]\) where \(\mathbf{F}(x, y)=\left(y e^{x y}+\cos x\right) \mathbf{i}+\left(x e^{x y}+\frac{1}{y^{2}+1}\right) \mathbf{j}\) and \(C\) is a portion of curve \(y=\sin x\) from \(x=0\) to \(x=\frac{\pi}{2}\).

Step-by-Step Solution

Answer

The result of the line integral is obtained by evaluating the integral after substituting, dotting, and integrating over the curve's parameter interval.

Step 1: Parameterize the Curve C

The curve is given as \(y = \sin x\) and \(x\) varies from \(0\) to \(\frac{\pi}{2}\). The parameterization of the curve \(C\) is \(\mathbf{r}(t) = t\mathbf{i} + \sin t \mathbf{j}\), where \(t\) ranges from \(0\) to \(\frac{\pi}{2}\).

Step 2: Compute Derivative of r(t)

The derivative of \(\mathbf{r}(t) = t\mathbf{i} + \sin t \mathbf{j}\) with respect to \(t\) is \(\frac{d\mathbf{r}}{dt} = \mathbf{i} + \cos t\mathbf{j}\). This represents the differential element of the curve.

Step 3: Substitute into Line Integral

The line integral \(\int_{C} \mathbf{F} \cdot d\mathbf{r}\) becomes \(\int_{0}^{\frac{\pi}{2}} \mathbf{F}(t, \sin t) \cdot (\mathbf{i} + \cos t \mathbf{j})\, dt\). We substitute \(\mathbf{F}(t, \sin t)= \left( \sin t e^{t \sin t}+\cos t \right) \mathbf{i}+ \left( t e^{t \sin t} + \frac{1}{\sin^2 t + 1} \right) \mathbf{j}\).

Step 4: Evaluate the Dot Product

Calculate \(\mathbf{F} \cdot (\mathbf{i} + \cos t \mathbf{j})\) which simplifies to \(\left( \sin t e^{t \sin t}+\cos t \right) \times 1 + \left( t e^{t \sin t} + \frac{1}{\sin^2 t + 1} \right) \times \cos t\).

Step 5: Integrate with Respect to t

Calculate the integral: \(\int_{0}^{\frac{\pi}{2}} \left( \left( \sin t e^{t \sin t}+\cos t \right) + \cos t \left( t e^{t \sin t} + \frac{1}{\sin^2 t + 1} \right) \right) \, dt\). Solve this definite integral to find the value of the line integral over the curve \(C\).

Step 6: Simplify and Compute the Result

After integrating, evaluate the expression at the limits \(0\) and \(\frac{\pi}{2}\) and simplify to find the result. This requires performing the integration for each term separately and summing the results.

Key Concepts

Vector Calculus

Vector calculus is a branch of mathematics concerned with vector fields and their behavior. It extends calculus concepts like differentiation and integration to vector fields. This is useful for understanding how vector fields, such as magnetic or gravitational fields, act on other objects in the field. A **vector field** is essentially a function that assigns a vector to every point in space. These vectors can be thought of as arrows that have both a magnitude and a direction.
In the context of the given problem, vector calculus allows us to analyze the vector field \( \mathbf{F}(x, y) \). By handling vector fields, vector calculus gives us tools to perform operations on them, such as finding line integrals, which measure the field's effect over a path or curve.
Understanding vector calculus is essential for physics and engineering where movement and forces within fields are analyzed over space and time.

Parametric Curves

A parametric curve is a curve described using parameters instead of traditional Cartesian coordinates \((x, y)\). With parametric equations, different variables are expressed as functions of one or more parameters.
This approach allows us to describe curves in a flexible and powerful way. For example, in the presented problem, the curve \( y = \sin x \) is redefined using a parameter \( t \), resulting in \( \mathbf{r}(t) = t\mathbf{i} + \sin t \mathbf{j} \).

• The parameter \( t \) represents the x-coordinate, ranging from 0 to \( \frac{\pi}{2} \).
• The y-coordinate becomes \( \sin t\), reflecting the curve's geometry.

This parameterization simplifies the calculation of the line integral by transforming the curve into a straightforward form that facilitates differentiation and substitution.

Dot Product

The dot product is an operation that takes two vectors and returns a single scalar (a simple number). This result is obtained by multiplying corresponding components of the vectors and summing these products.
In mathematical terms, for vectors \( \mathbf{A} = A_1\mathbf{i} + A_2\mathbf{j} + A_3\mathbf{k} \) and \( \mathbf{B} = B_1\mathbf{i} + B_2\mathbf{j} + B_3\mathbf{k} \), the dot product \( \mathbf{A} \cdot \mathbf{B} \) is calculated as:\[\mathbf{A} \cdot \mathbf{B} = A_1B_1 + A_2B_2 + A_3B_3.\]For the given problem, we use the dot product within the line integral to combine the force vector field \( \mathbf{F} \) with the differential element \( d\mathbf{r} = \mathbf{i} + \cos t\mathbf{j} \). This summarizes the interaction between the vector field and the curve component-wise, crucial for integrating over a path.
The final expression of the dot product helps translate a vector operation into a form suitable for integration, making it a pivotal step in solving the exercise.

Definite Integral

A definite integral is fundamentally linked to the concept of accumulation. It represents the total area under a curve, from one point to another. In the context of vector calculus and line integrals, it measures the cumulative effect of the vector field along a specific path.
To solve the given line integral \( \int_{C} \mathbf{F} \cdot d \mathbf{r} \), we calculate the integral over a defined range from 0 to \( \frac{\pi}{2} \). It aggregates small contributions from each segment of the curve into a total value.

• The integral involves the computation of the dot product result over the specified interval.
• Performs evaluation at end points to derive the net effect across the path.

The definite integral is vital for compiling the overall influence from localized interactions along the curve, providing a comprehensive result for the integrative process attempted in this problem.