Chapter 5: Problem 67
Use Green's theorem to evaluate line integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\), where \(\mathbf{F}(x, y)=\left(y^{2}-x^{2}\right) \mathbf{i}+\left(x^{2}+y^{2}\right) \mathbf{j}\), and \(C\) is a triangle bounded by \(y=0, x=3\), and \(y=x\), oriented counterclockwise.
Short Answer
Expert verified
The value of the line integral \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \) is 9.
Step by step solution
01
Define the vector field and region
First, identify the vector field given, which is \( \mathbf{F}(x, y)=(y^2-x^2) \mathbf{i} + (x^2+y^2) \mathbf{j} \). The path \( C \) is a triangle with vertices at (0,0), (3,0), and (3,3). This triangle is bounded by the lines \( y=0 \), \( x=3 \), and \( y=x \).
02
Set up Green's Theorem
Green's Theorem relates a line integral around a closed curve \( C \) to a double integral over the region \( R \) it encloses. The theorem states: \( \oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \) where \( P(x, y) = y^2 - x^2 \) and \( Q(x, y) = x^2 + y^2 \).
03
Compute partial derivatives
Compute the partial derivatives \( \frac{\partial Q}{\partial x} = 2x \) and \( \frac{\partial P}{\partial y} = 2y \).
04
Set up the double integral
Substitute the partial derivatives into the double integral from Green's Theorem: \( \iint_{R} (2x - 2y) \, dA \).
05
Define limits of integration
The region \( R \) is the triangle with boundaries: from \( y = 0 \) to \( y = x \), and \( x \) from 0 to 3. Thus, the integral is evaluated with limits \( \int_{x=0}^{3} \int_{y=0}^{x} (2x - 2y) \, dy \, dx \).
06
Evaluate the inner integral
The inner integral is \( \int_{y=0}^{x} (2x - 2y) \, dy = [2xy - y^2]_{y=0}^{x} = 2x^2 - x^2 = x^2 \).
07
Evaluate the outer integral
Integrate \( \int_{x=0}^{3} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{3} = \frac{27}{3} - 0 = 9 \).
08
Final result
The evaluated double integral gives the value of the original line integral \( \int_{C} \mathbf{F} \cdot d\mathbf{r} \). Thus, the integral evaluates to \( 9 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Line Integrals
Line integrals are a type of calculus operation that evaluate a function along a curve. Instead of integrating over a simple interval, you integrate along a path, which can be straight or curved.
In the context of vector fields, line integrals help compute the work done by a force field on an object along a path. To evaluate a line integral, follow these general steps:
In the context of vector fields, line integrals help compute the work done by a force field on an object along a path. To evaluate a line integral, follow these general steps:
- Define the vector field and the path of integration.
- Represent the path parametrically, if needed.
- Substitute the parametric representation into the vector field.
- Perform the integration with respect to the path parameter.
Examining Double Integrals
Double integrals extend the concept of integration to functions of two variables. They sum up values over a region in a plane, much like how a single integral sums over an interval on a line.
Here, a double integral offers the combined accumulative value over an area.When using Green's Theorem, a line integral over a closed curve can be turned into a double integral over the region it encloses:
\[\oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA\]In this scenario, calculate partial derivatives and set boundaries for integration to solve the double integral. This method offers a more manageable computation when dealing paths and regions.
Here, a double integral offers the combined accumulative value over an area.When using Green's Theorem, a line integral over a closed curve can be turned into a double integral over the region it encloses:
\[\oint_{C} \mathbf{F} \cdot d\mathbf{r} = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA\]In this scenario, calculate partial derivatives and set boundaries for integration to solve the double integral. This method offers a more manageable computation when dealing paths and regions.
Vector Field Properties
A vector field represents a function that assigns a vector to every point in a plane or space. It helps to visualize quantities that have both direction and magnitude, such as velocity or force.
For the given exercise, the vector field is\[\mathbf{F}(x, y) = (y^2-x^2) \mathbf{i} + (x^2+y^2) \mathbf{j}\]Key considerations for vector fields:
For the given exercise, the vector field is\[\mathbf{F}(x, y) = (y^2-x^2) \mathbf{i} + (x^2+y^2) \mathbf{j}\]Key considerations for vector fields:
- Understand how vector components vary with position.
- Determine how vector fields interact with curves and surfaces in the plane.
- Use Green’s Theorem to simplify calculation of work done along paths encircling regions.
Computing Partial Derivatives
Partial derivatives are a fundamental tool to differentiate functions with multiple variables, analyzing how changes in one variable affect the function while keeping others constant.
In Green’s Theorem, they are crucial in converting a line integral into a double integral.To compute partial derivatives:
In Green’s Theorem, they are crucial in converting a line integral into a double integral.To compute partial derivatives:
- Differentiating with respect to 'x', keep 'y' constant, and vice versa.
- For \( P(x, y) = y^2 - x^2 \), \( \frac{\partial P}{\partial y} = 2y \).
- For \( Q(x, y) = x^2 + y^2 \), \( \frac{\partial Q}{\partial x} = 2x \).