Problem 29 - Calculus Early Transcendentals

Question

Use Green’s Theorem to find the work done by the force field F on a particle that moves along the stated path. \(\mathbf{F}(x, y)=x y \mathbf{i}+\left(\frac{1}{2} x^{2}+x y\right) \mathbf{j} ;\) the particle starts at \((5,0)\) traverses the upper semicircle \(x^{2}+y^{2}=25,\) and returns to its starting point along the \(x\) -axis.

Step-by-Step Solution

Answer

The work done by the force field is \( \frac{250}{3} \).

Step 1: Identify the force field components

The force field \( \mathbf{F}(x, y) \) is given by \( x y \mathbf{i} + (\frac{1}{2} x^2 + x y) \mathbf{j} \). Identify the components of the vector field as follows: \( P(x, y) = x y \) and \( Q(x, y) = \frac{1}{2} x^2 + x y \).

Step 2: Apply Green's Theorem

Green's Theorem relates the line integral around a simple closed curve \( C \) to a double integral over the plane region \( R \) bounded by \( C \). The theorem is stated as: \[ \oint_C (P \, dx + Q \, dy) = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \]. First, determine the partial derivatives:\ \( \frac{\partial Q}{\partial x} = x + y \) and \( \frac{\partial P}{\partial y} = x \).

Step 3: Simplify the integral expression

Substitute the partial derivatives into Green's theorem:\ \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = (x + y) - x = y \). Therefore, the double integral becomes: \[ \iint_R y \, dA \].

Step 4: Establish the region of integration

Since the path is the upper semicircle of \( x^2 + y^2 = 25 \) and the x-axis, the region \( R \) is the upper half of the circle with radius 5. It can be defined in polar coordinates as \( 0 \leq \theta \leq \pi \) and \( 0 \leq r \leq 5 \).

Step 5: Convert to polar coordinates

The expression for \( y \) in polar coordinates is \( r \sin \theta \). Hence, the double integral is \[ \iint_R y \, dA = \int_0^{\pi} \int_0^5 r \sin \theta \, r \, dr \, d\theta \].

Step 6: Evaluate the integral

First, integrate with respect to \( r \):\ \[ \int_0^5 r^2 \, dr = \left[ \frac{r^3}{3} \right]_0^5 = \frac{125}{3} \].\ Now, integrate with respect to \( \theta \):\ \[ \int_0^{\pi} \frac{125}{3} \sin \theta \, d\theta = \frac{125}{3} \left[ -\cos \theta \right]_0^{\pi} = \frac{125}{3} [1 - (-1)] = \frac{125}{3} \times 2 = \frac{250}{3} \].

Step 7: Conclude the solution

The work done by the force field \( \mathbf{F} \) on the particle as it completes the described path is \( \frac{250}{3} \).

Key Concepts

Line Integrals

Line integrals play a crucial role in vector calculus, especially when dealing with force fields and the concept of work. Imagine a curve in space that a particle follows. The line integral helps us understand the total effect of a vector field along that path. In simpler terms, it's like summing up little bits of work done on the particle as it travels along the path.

To calculate a line integral, we consider both the vector field and the path taken by the particle. The path is often a curve described by a parameter, usually denoted as \( C \). When dealing with force fields represented by \( \mathbf{F} = P \mathbf{i} + Q \mathbf{j} \), the line integral around \( C \) is expressed as \( \oint_C P \, dx + Q \, dy \). This formula allows us to calculate the work done by the force field as the particle moves along the curve.

Understanding line integrals can help you solve problems related to physics, engineering, and other fields that deal with forces and paths.

Vector Fields

A vector field is like a magical map where each point in space has a vector attached to it. These vectors can represent a variety of phenomena, such as wind speed and direction, electric or magnetic fields, or forces acting on particles. In mathematical terms, a vector field in two dimensions can be expressed as \( \mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j} \), where \( P \) and \( Q \) are functions of \( x \) and \( y \).

In our case, the vector field \( \mathbf{F} = xy \mathbf{i} + \left(\frac{1}{2} x^2 + xy\right) \mathbf{j} \) illustrates how the field varies across a plane. The components \( P(x, y) = xy \) and \( Q(x, y) = \frac{1}{2} x^2 + xy \) describe how much force is applied in the \( i \)-direction and \( j \)-direction, respectively.

Grasping vector fields is essential for solving problems in electromagnetism, fluid dynamics, and more, where understanding the distribution and effects of vector quantities is required.

Polar Coordinates

Polar coordinates offer a refreshing alternative to the typical Cartesian coordinates. Instead of using \( x \) and \( y \) to specify a point in a plane, polar coordinates use a radius \( r \) and an angle \( \theta \). This approach is often more intuitive when dealing with circular or radial regions because it directly incorporates the shape of these regions.

For the integration problem in Green's Theorem, converting the area into polar coordinates simplifies calculations. The equation \( x^2 + y^2 = 25 \) represents a circle, which is perfect for polar coordinates where \( 0 \leq \theta \leq \pi \), and \( 0 \leq r \leq 5 \). This rephrases our region of interest in terms of circular properties—radius and angle—making integration more straightforward.

Using polar coordinates can simplify problems involving circular or spherical shapes, such as computing areas and volumes or evaluating integrals for such shapes.

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, especially useful when dealing with functions of two or more variables. They measure how a function changes as one specific variable changes, while the others are held constant. This is crucial in vector calculus and multivariable functions.

In applying Green's Theorem, we compute partial derivatives of the components of a vector field. For example, if we have \( Q(x, y) = \frac{1}{2} x^2 + xy \), the partial derivative of \( Q \) with respect to \( x \) is \( \frac{\partial Q}{\partial x} = x + y \). Similarly, for \( P(x, y) = xy \), the partial with respect to \( y \) is \( \frac{\partial P}{\partial y} = x \).

Using these partial derivatives, Green's Theorem relates a line integral around a closed curve to a double integral over the region it encloses. Thus, partial derivatives become essential tools to interconnect different types of integrals and facilitate problem-solving in multiple dimensions.