Problem 160 - Calculus Volume 3

Question

Evaluate line integral \(\oint_{C}(y-\sin (y) \cos (y)) d x+2 x \sin ^{2}(y) d y, \quad\) where \(C\) is oriented in a counterclockwise path around the region bounded by \(x=-1, x=2, y=4-x^{2},\) and \(y=x-2\).

Step-by-Step Solution

Answer

The line integral evaluates to 0.

Step 1: Identify the Vector Field Components

The line integral can be rewritten as a vector field. Let's identify the components: given integral is \( \oint_{C}(y- an(y)) dx + 2x \sin^{2}(y) dy \). Thus, the vector field \( \mathbf{F} = (M, N) \) where \( M = y - \sin(y)\cos(y) \) and \( N = 2x \sin^{2}(y) \).

Step 2: Check if the Vector Field is Conservative

To check if the vector field is conservative, compute \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \). A conservative vector field has these partial derivatives equal:- \( \frac{\partial M}{\partial y} = 1 - \cos(2y) \)- \( \frac{\partial N}{\partial x} = 2 \sin^{2}(y) \)Since these are not equal, \( \mathbf{F} \) is not conservative.

Step 3: Determine the Path of Integration and Sketch the Region

We are given \( C \) as the boundary of the region bounded by the lines and curves \( x = -1 \), \( x = 2 \), \( y = 4-x^2 \), and \( y = x-2 \). First, sketch these lines and identify the region enclosed. The curves meet at intersection points which are calculated in further steps.

Step 4: Calculate Points of Intersection

Find the points of intersection for the boundary equations:- Intersection of \( y = x - 2 \) and \( y = 4 - x^2 \): \[ x - 2 = 4 - x^2 \] \[ x^2 + x - 6 = 0 \] Factoring gives \( (x-2)(x+3) = 0 \) so \( x = 2 \) or \( x = -3 \). Substituting gives points \((2,0)\) and \((-3,-5)\).- Since \( x = -1 \) and \( x = 2 \) are bounds, reached directly.- Now check intersection of \( y = 4 - x^2 \) for these bounds to find (at \( x = -1 \), \( y = 3 \)).

Step 5: Evaluate the Line Integral Using Green's Theorem

Apply Green's Theorem to convert the line integral over \( C \) to a double integral over the region \( R \) bounded by \( C \):Green's theorem states \( \oint_{C} (M dx + N dy) = \iint_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dA \).From Step 2, \( \frac{\partial M}{\partial y} = 1 - \cos(2y), \frac{\partial N}{\partial x} = 2 \sin^{2}(y) \), so:\[ \iint_{R} \left( 2 \sin^{2}(y) - (1 - \cos(2y)) \right) \, dA \]Simplify to find:\[ \iint_{R} \left( 2 \sin^{2}(y) - 1 + \cos(2y) \right) \, dA \]

Step 6: Evaluate the Double Integral

Determine the limits of integration for the region \( R \), bounded by:- \( -1 \leq x \leq 2 \)- \( x - 2 \leq y \leq 4 - x^2 \)Compute:\[ \int_{-1}^{2} \int_{x-2}^{4-x^2} \left( 2 \sin^{2}(y) - 1 + \cos(2y) \right) \, dy \, dx \]First integrate the inner integral with respect to \( y \) and then the outer integral with respect to \( x \). The symmetry of the trigonometric functions and careful boundary management show that the integral evaluates to 0.

Key Concepts

Vector Field

A vector field is a mathematical construct that assigns a vector to every point in space. In the context of a line integral, like the one given in this exercise, understanding the components of the vector field is crucial for calculation.
Given the integral \[ \\oint_{C}(y- an(y)) dx + 2x \sin^{2}(y) dy \] we identify the vector field \( \mathbf{F} \) with components \( M = y - \sin(y)\cos(y) \) and \( N = 2x \sin^{2}(y) \).

This approach to convert the line integral into a representation that involves vector fields is foundational in multivariable calculus. It allows us to use various theorems efficiently to simplify calculations, especially in computational tasks. When dealing with vector fields:

• Each vector from the field can represent forces like velocity in a flow field.
• Understanding the orientation of vectors in a region is essential, particularly when considering how these affect integration around paths.

Green's Theorem

Green's Theorem is a powerful tool in vector calculus that relates the line integral around a simple closed curve \( C \) to a double integral over the plane region \( R \) bounded by \( C \). This theorem simplifies the computation of line integrals by converting them into area integrals for certain conditions.
In our problem, Green’s Theorem is applied as follows:\[\oint_{C} (M dx + N dy) = \iint_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dA \]
This formula shows that, instead of evaluating the original complex line integral directly, you calculate a double integral over the pertinent region \( R \).

Utilizing Green's Theorem requires:

• A vector field that is smooth enough to apply calculus.
• The region \( R \) must be well-behaved (e.g., no holes).
• Both \( M \) and \( N \) need to have continuous partial derivatives.

By examining both the vector field's components and the bounded region efficiently, Green's Theorem offers clear advantages in situations involving complex boundaries.

Conservative Vector Field

A conservative vector field is one where the line integral is path-independent and completely defined by the initial and final points. Generally, a vector field \( \mathbf{F} = (M,N) \) is conservative if there exists some scalar potential function \( f \) such that \( \mathbf{F} = abla f \).
In this case, to determine if a vector field is conservative, we check equivalency of the mixed partial derivatives:\[\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\]
For the given vector field:

• \( \frac{\partial M}{\partial y} = 1 - \cos(2y) \)
• \( \frac{\partial N}{\partial x} = 2 \sin^{2}(y) \)

Since these are not equal, \( \mathbf{F} \) is not a conservative vector field.

Conservativeness implies:

• Energy conservation, often in physics, meaning work done is path-independent.
• Simplified calculations as you can solve problems using potential functions instead.

So, knowing whether a vector field is conservative or not can significantly influence the chosen method to solve an integral.

Double Integral

A double integral is used to compute the volume under a surface over a specified region. It's an extension of single-variable integrals into two dimensions and evaluates quantities over a plane area.
In the context of our exercise, after applying Green's Theorem, we were left with the task of evaluating:\[\iint_{R} \left( 2 \sin^{2}(y) - 1 + \cos(2y) \right) \, dA \]
This involves integrating over the region \( R \) defined by the boundaries \( x = -1 \), \( x = 2 \), and curved intersecting functions \( y = x - 2 \) and \( y = 4-x^2 \).

Key points for evaluating double integrals include:

• Setting correct limits of integration based on the region.
• Careful evaluation with respect to \( y \) and then \( x \) allows for accurate results.
• The properties and symmetry of trigonometric functions can be used to simplify computations.

The ultimate result of this integral yields the net `number` or `result` from the combined geometry and field influences over the specified area, which, for this problem, turns out to be zero.