Problem 130
Question
For the following exercises, $$ \mathbf{F}(x, y)=2 x y^{2} \mathbf{i}+\left(2 y x^{2}+2 y\right) \mathbf{j} $$ \(G(x, y)=(y+x) \mathbf{i}+(y-x) \mathbf{j}, \quad\) and let \(C_{1}\) be the curve consisting of the circle of radius 2 , centered at the origin and oriented counterclockwise, and \(C_{2}\) be the curve consisting of a line segment from (0,0) to (1,1) followed by a line segment from (1,1) to (3,1) . Calculate the line integral of \(\mathbf{F}\) over \(C_{1}\).
Step-by-Step Solution
Verified Answer
The line integral of \(\mathbf{F}\) over \(C_1\) is 0.
1Step 1: Parameterize the Curve C1
The curve \(C_1\) is a circle centered at the origin with radius 2, oriented counterclockwise. We can parameterize this circle using the angle \(t\) (from 0 to \(2\pi\)) as \(x = 2\cos(t)\) and \(y = 2\sin(t)\).
2Step 2: Compute dx and dy
Using the parameterization, we calculate the derivatives: \(dx = -2\sin(t) dt\) and \(dy = 2\cos(t) dt\).
3Step 3: Substitute into the Integral
Substitute \(x = 2\cos(t)\), \(y = 2\sin(t)\), \(dx\), and \(dy\) into the line integral formula \(\int_{C_1} \mathbf{F} \cdot d\mathbf{r}\).
4Step 4: Evaluate the Line Integral
The vector field \(\mathbf{F}(x, y) = 2xy^2 \mathbf{i} + (2yx^2 + 2y) \mathbf{j}\) is substituted into the integral: \[\int_{0}^{2\pi} (2\cdot 2\cos(t) \cdot (2\sin(t))^2) (-2\sin(t)) + ((2 \cdot 2\sin(t) \cdot (2\cos(t))^2 + 2 \cdot 2\sin(t)) \cdot 2\cos(t))\, dt\].
5Step 5: Simplify the Expression
Simplify the expression step-by-step. Combine terms and use trigonometric identities if necessary. The integral involves evaluating trigonometric functions from 0 to \(2\pi\).
6Step 6: Integrate and Final Evaluation
Evaluate the integral over the interval \(0\) to \(2\pi\). Since the circle is symmetric, certain integrals of trigonometric functions over one full period will result in zero.
Key Concepts
Parameterization of CurvesVector FieldsTrigonometric Identities
Parameterization of Curves
Parameterizing a curve is about expressing the path of that curve using a set of equations.The essence of this process is to substitute coordinates like \(x\) and \(y\) into another parameter, often denoted as \(t\).This covers how the curve relates to this parameter within a certain range.
For circle curves, like the one in the exercise, parameterization uses trigonometric expressions.The angle \(t\), ranging from \(0\) to \(2\pi\), represents a complete loop around the circle.For instance, the circle centered at the origin with radius 2 can be described by \(x = 2\cos(t)\) and \(y = 2\sin(t)\).
This provides a set of equations that capture the path of the circle in terms of \(t\).Additionally, it's crucial to calculate the changes in \(x\) and \(y\) with respect to \(t\), termed \(dx\) and \(dy\). These derivatives, \(dx = -2\sin(t)dt\) and \(dy = 2\cos(t)dt\), help when substituting into the line integral.Parameterization simplifies complex paths into more manageable mathematical forms, making evaluation easier.
For circle curves, like the one in the exercise, parameterization uses trigonometric expressions.The angle \(t\), ranging from \(0\) to \(2\pi\), represents a complete loop around the circle.For instance, the circle centered at the origin with radius 2 can be described by \(x = 2\cos(t)\) and \(y = 2\sin(t)\).
This provides a set of equations that capture the path of the circle in terms of \(t\).Additionally, it's crucial to calculate the changes in \(x\) and \(y\) with respect to \(t\), termed \(dx\) and \(dy\). These derivatives, \(dx = -2\sin(t)dt\) and \(dy = 2\cos(t)dt\), help when substituting into the line integral.Parameterization simplifies complex paths into more manageable mathematical forms, making evaluation easier.
Vector Fields
A vector field assigns a vector to every point in space. Imagine it as a map that shows directions and magnitudes at each location.In the given exercise, the vector field \( \mathbf{F}(x, y) = 2xy^2 \mathbf{i} + (2yx^2 + 2y) \mathbf{j} \) shows how vectors affect the plane.
The components, \( \mathbf{i} \) and \( \mathbf{j} \), correspond to the x-direction and y-direction, respectively. It interprets how the field varies based on position \((x, y)\).To intuitively understand, imagine the field affecting the movement of particles or flows in a certain direction.
When calculating line integrals, you evaluate how this field interacts with a defined path, like \(C_1\) in our problem.Line integrals help determine the total effect of this vector field along a path.It considers how the components of the field, multiplied by the path's direction, sum up over the curve.
Such fields arise in various applications, from physics with electromagnetic fields, to fluid dynamics exploring how water flows in a bath.
The components, \( \mathbf{i} \) and \( \mathbf{j} \), correspond to the x-direction and y-direction, respectively. It interprets how the field varies based on position \((x, y)\).To intuitively understand, imagine the field affecting the movement of particles or flows in a certain direction.
When calculating line integrals, you evaluate how this field interacts with a defined path, like \(C_1\) in our problem.Line integrals help determine the total effect of this vector field along a path.It considers how the components of the field, multiplied by the path's direction, sum up over the curve.
Such fields arise in various applications, from physics with electromagnetic fields, to fluid dynamics exploring how water flows in a bath.
Trigonometric Identities
Trigonometric identities are equations relating the angles and sides of triangles in terms of sine, cosine, and other trig functions.These identities simplify complex trigonometric expressions and are particularly useful in integrals like the one in our exercise.
The exercise often uses identities such as:
By using these identities, integrals simplify, making them manageable and eliminating unnecessary complexity.In this case, they're employed to break down the sine and cosine components interwoven in the vector field's expression.
Understanding and applying trigonometric identities is crucial for effective problem-solving in calculus as they convert intricate trigonometric integrals into basic forms.These identities help reveal underlying patterns, providing clarity and solutions efficiently.
The exercise often uses identities such as:
- \(\sin^2(t) + \cos^2(t) = 1\): A core identity that relates sine and cosine functions.
- Double angle formulas, like \(\sin(2t) = 2\sin(t)\cos(t)\): Useful for breaking down expressions where angles are under multiplication.
By using these identities, integrals simplify, making them manageable and eliminating unnecessary complexity.In this case, they're employed to break down the sine and cosine components interwoven in the vector field's expression.
Understanding and applying trigonometric identities is crucial for effective problem-solving in calculus as they convert intricate trigonometric integrals into basic forms.These identities help reveal underlying patterns, providing clarity and solutions efficiently.
Other exercises in this chapter
Problem 127
For the following exercises, evaluate the integral using the Fundamental Theorem of Line Integrals. [T] Evaluate \(\int_{C} \nabla f \cdot d \mathbf{r}\), where
View solution Problem 128
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View solution Problem 131
For the following exercises, $$ \mathbf{F}(x, y)=2 x y^{2} \mathbf{i}+\left(2 y x^{2}+2 y\right) \mathbf{j} $$ \(G(x, y)=(y+x) \mathbf{i}+(y-x) \mathbf{j}, \qua
View solution Problem 132
For the following exercises, $$ \mathbf{F}(x, y)=2 x y^{2} \mathbf{i}+\left(2 y x^{2}+2 y\right) \mathbf{j} $$ \(G(x, y)=(y+x) \mathbf{i}+(y-x) \mathbf{j}, \qua
View solution