Problem 14 - Advanced Engineering Mathematics

Question

Use Green's theorem to evaluate the given line integral. \(\oint_{C} x y^{2} d x+3 \cos y d y\), where \(C\) is the boundary of the region in the first quadrant determined by the graphs of \(y \quad x^{2}, y \quad x^{3}\)

Step-by-Step Solution

Answer

The value of the line integral is \(\frac{1}{24}\).

Step 1: Identify the Vector Field and Functions

The line integral given is \(\oint_{C} xy^2 \, dx + 3\cos y \, dy\). We can express this in terms of a vector field \( \vec{F} = (M, N) \) where \( M = xy^2 \) and \( N = 3\cos y \).

Step 2: Calculate the Partial Derivatives

To apply Green's theorem, find the partial derivative \( \frac{\partial N}{\partial x} = 0\) and the partial derivative \( \frac{\partial M}{\partial y} = 2xy\).

Step 3: Apply Green's Theorem

Green's theorem states \(\oint_{C} M \, dx + N \, dy = \iint_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \). Substituting the partial derivatives found, we get the double integral \(\iint_{R} -2xy \, dA\).

Step 4: Determine the Region of Integration

Since \(C\) is the boundary of the region in the first quadrant between \(y = x^2\) and \(y = x^3\), the integration bounds are \(x^2 \leq y \leq x^3\) integrating along \(0 \leq x \leq 1\).

Step 5: Set Up the Double Integral

The double integral \(\iint_{R} -2xy \, dA\) becomes \(\int_{0}^{1} \int_{x^2}^{x^3} -2xy \, dy \, dx\).

Step 6: Integrate with Respect to \(y\)

Evaluate the inner integral: \(\int_{x^2}^{x^3} -2xy \, dy = -2x \left[ \frac{y^2}{2} \right]_{x^2}^{x^3} = -x(y^2)|_{x^2}^{x^3}\). Substituting the limits of \(y\), get \(-x((x^3)^2 - (x^2)^2) = -x(x^6 - x^4)= -x^7 + x^5\).

Step 7: Integrate with Respect to \(x\)

Now evaluate \(\int_{0}^{1} (-x^7 + x^5) \, dx\). This integral becomes \(\left[ -\frac{x^8}{8} + \frac{x^6}{6} \right]_{0}^{1} = \left(-\frac{1}{8} + \frac{1}{6}\right)\).

Step 8: Simplify and Compute the Result

Calculate the resulting expression: \(-\frac{1}{8} + \frac{1}{6} = -\frac{3}{24} + \frac{4}{24} = \frac{1}{24}\). Therefore, \(\oint_{C} xy^2 \, dx + 3\cos y \, dy = \frac{1}{24}\).

Key Concepts

Vector Fields

A vector field describes a situation where at every point in space, there is a vector (having both direction and magnitude) associated. It is often denoted by \( \vec{F} = (M, N) \) for planar scenarios. Here, \( M(x, y) \) and \( N(x, y) \) represent the components of the vector in the horizontal and vertical directions, respectively. In the context of Green's theorem, the goal is to relate a region's area bounded by a curve to a circulation or flux via a line integral of these vector field components.

For our exercise, we identified the vector field \( \vec{F} = (xy^2, 3\cos y) \). This decomposition helps transform the line integral into a format suitable for applying Green's theorem.

Calculations involving vector fields are foundational in fields such as fluid dynamics and electromagnetism, as they allow the exploration of how quantities change in space.

Partial Derivatives

Partial derivatives are derivatives of functions of multiple variables with respect to one variable, keeping others constant. In our context, for the vector field components \( M(x, y) \) and \( N(x, y) \) defined earlier, partial derivatives help us prepare for applying Green's theorem.

To apply Green's theorem, we need:

• \( \frac{\partial N}{\partial x} \): Differentiating \( N = 3\cos y \) with respect to \( x \), gives \( 0 \) since there is no \( x \) in \( N \).
• \( \frac{\partial M}{\partial y} \): Differentiating \( M = xy^2 \) with respect to \( y \), gives \( 2xy \).

These partial derivatives will then be used in setting up the double integral over the region R, which is the crux of using Green's theorem.

Double Integration

Double integration is an extension of single integration and is geared towards integrating over a two-dimensional region. It allows the evaluation of functions involving two variables over a specified area.

In the exercise provided, after finding the partial derivatives, we transform the line integral using Green's theorem into a double integral. The goal becomes to evaluate:

• \( \iint_{R} -2xy \, dA \)

This requires setting up limits that reflect the region we are integrating over. For our case, these limits are determined by:

• The curves \( y = x^2 \) and \( y = x^3 \) bounding the region in the first quadrant.
• The \( x \)-bounds are from 0 to 1.

This double integral accounts for integrating first with respect to \( y \), and then \( x \). Double integrals are pivotal because they create a link between multi-variable calculus and geometry.

Line Integrals

Line integrals enable the integration of functions along a curve or path, capturing the accumulation of values. This is especially useful for functions that describe physical quantities like work done by a force field or flux through a curve.

For Green's theorem, line integrals are interpreted over a closed path \( C \). Our exercise's original line integral \( \oint_{C} xy^2 \, dx + 3\cos y \, dy \) represents the work done by the vector field \( \vec{F} \) along \( C \). Using Green's theorem, the challenge is to convert this line integral into a more straightforward double integral over the region \( R \) enclosed by \( C \).

This conversion taps into how line integrals aggregate local properties along a path into a global context, echoing principles present in fields like thermodynamics and dynamics. Line integrals emphasize the importance of path-dependence when calculating integrals of vector fields.