Problem 5 - Calculus Early Transcendentals

Question

Use Green’s Theorem to evaluate the integral. In each exercise, assume that the curve C is oriented counterclockwise. \(\oint_{C} x \cos y d x-y \sin x d y,\) where \(C\) is the square with vertices \((0,0),(\pi / 2,0),(\pi / 2, \pi / 2),\) and \((0, \pi / 2)\)

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Understand Green's Theorem

Green's Theorem relates a line integral around a simple closed curve \(C\) to a double integral over the region \(R\) it encloses. The theorem states: \[ \oint_{C} (L \, dx + M \, dy) = \iint_{R} \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) \, dA. \] In this problem, \(L = x \cos y\) and \(M = -y \sin x\).

Step 2: Compute the Partial Derivatives

Calculate \(\frac{\partial M}{\partial x}\) and \(\frac{\partial L}{\partial y}\). For \(M = -y \sin x\), we have \(\frac{\partial M}{\partial x} = -y \cos x\). For \(L = x \cos y\), \(\frac{\partial L}{\partial y} = -x \sin y\).

Step 3: Set Up the Double Integral

Substitute the partial derivatives into the Green's Theorem expression: \(\iint_{R} \left( -y \cos x + x \sin y \right) \, dA\). The region \(R\) is the square \([0, \pi/2] \times [0, \pi/2]\).

Step 4: Evaluate the Double Integral

Evaluate \[ \iint_{R} ( -y \cos x + x \sin y ) \, dy \, dx. \] The integral becomes \[ \int_{0}^{\pi/2} \int_{0}^{\pi/2} (-y \cos x) \, dy \, dx + \int_{0}^{\pi/2} \int_{0}^{\pi/2} (x \sin y) \, dy \, dx. \] Compute each separately.

Step 5: Compute Integral for \(-y \cos x\)

Evaluate the inner integral: \(\int_{0}^{\pi/2} (-y \cos x) \, dy = -\cos x \int_{0}^{\pi/2} y \, dy = -\cos x \left[ \frac{y^2}{2} \right]_{0}^{\pi/2} = -\cos x \times \frac{(\pi/2)^2}{2} = -\frac{\pi^2 \cos x}{8}\). Then, integrate with respect to \(x\): \[ \int_{0}^{\pi/2} -\frac{\pi^2 \cos x}{8} \, dx = -\frac{\pi^2}{8} \int_{0}^{\pi/2} \cos x \, dx = -\frac{\pi^2}{8} \left[ \sin x \right]_{0}^{\pi/2} = -\frac{\pi^2}{8}. \]

Step 6: Compute Integral for \(x \sin y\)

Evaluate the inner integral: \(\int_{0}^{\pi/2} x \sin y \, dy = x \int_{0}^{\pi/2} \sin y \, dy = x \left[-\cos y \right]_{0}^{\pi/2} = x (1 - 0) = x\). Then, integrate with respect to \(x\): \[ \int_{0}^{\pi/2} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{\pi/2} = \frac{(\pi/2)^2}{2} = \frac{\pi^2}{8}. \]

Step 7: Combine Results

The results from the integrals are \(-\frac{\pi^2}{8}\) and \(\frac{\pi^2}{8}\). Adding these two results gives \(0\). This is the value of the line integral over the curve \(C\).

Key Concepts

Line Integral

A line integral is a way to integrate over a path or curve, taking into account both the function and the path's direction. In this specific exercise, the curve is specified as a square with certain vertices. The integral given is \( \oint_{C} x \cos y \, dx - y \sin x \, dy \), which requires us to compute the integral around this closed curve \( C \).

Here, the notation \( \oint \) represents a line integral over a closed path. The line integral evaluates the function along a path or curve. In the context of Green's Theorem, we use line integrals to calculate a circulation or a flux around a closed loop. To compute a line integral, it’s crucial to follow the curve’s path while accounting for the vector field along the path.

Green’s Theorem connects this line integral over the curve to a double integral over the region it encloses. This allows us to convert the problem into an area integral, which is often easier to solve in certain conditions. Particularly, if you have a well-defined region like a square, Green’s theorem becomes highly practical. This line integral is evaluated around the boundary of such a region, making it central to problems in physics and engineering where fields interact with boundaries.

Double Integral

A double integral allows us to integrate functions over two-dimensional regions. In this task, we use Green’s Theorem to transform a line integral into a double integral over the area enclosed by curve \( C \), which is a square in this case.

With the transformation from the line to the double integral, the theorem provides:
\[ \iint_{R} \left( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \right) \, dA. \]
This is now an area integral over \( R \), the interior of the square. We integrate using the given limits of the region: from \( 0 \) to \( \pi/2 \) over \( x \) and from \( 0 \) to \( \pi/2 \) over \( y \).

Double integrals provide a straightforward method for calculating the volume under a surface or, as in this case, finding results pertinent to physics and engineering problems across two-dimensional spaces. By evaluating this particular integral over two parts—first with respect to \( y \) and then with respect to \( x \)—we simplify the expression needed to find the sum, leveraging the area \( R \).

Partial Derivatives

Partial derivatives show how a function changes as each variable changes, holding others constant. In the context of Green's Theorem, we use partial derivatives to find the circulation or flux through the expression \( \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} \).

In the exercise, \( M \) and \( L \) change as \( x \) and \( y \) change. \( M = -y \sin x \), giving \( \frac{\partial M}{\partial x} = -y \cos x \), while for \( L = x \cos y \), we find \( \frac{\partial L}{\partial y} = -x \sin y \).

These partial derivatives become part of the expression inside the double integral for Green's Theorem. By calculating these, we determine the boundary's influence or the force in physics or gradients in flow problems. This helps in transforming complex surface or line-related problems into manageable calculations by breaking them into two single-variable calculus problems through direct integration.