Problem 14 - The Calculus with Analytic Geometry

Question

\(\int_{C}(\sin y \sinh x+\cos y \cosh x) d x+(\cos y \cosh x-\sin y \sinh x) d y ; A\) is \((1,0)\) and \(B\) is \((2, \pi) ;\) Exercise 4

Step-by-Step Solution

Answer

\( \int_{C} ( \sin y \sinh x + \cos y \cosh x ) dx + ( \cos y \cosh x - \sin y \sinh x ) dy = 0 \)

Step 1 - Understand the Components

Analyze the integrand to identify its structure: Given integral: \int_{C} \( \sin y \sinh x + \cos y \cosh x \) \ dx + \( \cos y \cosh x - \sin y \sinh x \) \ dy. Both terms appear to be structured in a specific form. Identify any connection to known theorems, such as Green’s theorem.

Step 2 - Recall Green's Theorem

Green's Theorem relates a line integral around a simple closed curve C to a double integral over the plane region D bounded by C. The theorem is given as: \oint_{C} \( P dx + Q dy \) = \iint_{D} \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \) dA. Identify P and Q in our integral.

Step 3 - Identify P and Q

From the given integral, identify P and Q: P(x, y) = \( \sin y \sinh x + \cos y \cosh x \) Q(x, y) = \( \cos y \cosh x - \sin y \sinh x \)

Step 4 - Verify P and Q Conditions for Green’s Theorem

Compute \( \frac{\partial P}{\partial y} \) and \( \frac{\partial Q}{\partial x} \). \( \frac{\partial P}{\partial y} = \cos y \sinh x - \sin y \cosh x \) \( \frac{\partial Q}{\partial x} = \cos y \sinh x - \sin y \cosh x \). Notice that both partial derivatives are equal, thus satisfying Green’s Theorem's condition where \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 \). Hence the line integral reduces to zero.

Step 5 - Conclusion

Since \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0 \), the integral simplifies to \( \int_{C} P dx + Q dy = 0 \) by Green’s Theorem.

Key Concepts

Line Integral

A line integral is a type of integral where a function is evaluated along a curve. Unlike regular integrals that sum values over an interval, line integrals sum values over a path in a plane or space. The basic idea is to integrate a function along a curve, which could be a straight line or a more complex path. In our exercise, we have a line integral of the form \(\int_{C}(\sin y \sinh x+\cos y \cosh x) dx+(\cos y \cosh x-\sin y \sinh x) dy\).
Line integrals are crucial in fields such as physics and engineering. They often represent work done by a force field over a path, for instance.

• Identify the path or curve C.
• Recognize the functions involved along this path, also known as the integrand.
• Evaluate the integral along this curve to find the sum of the given function's values along the path.

By transforming a line integral into a double integral using the powerful Green's Theorem, complex computations become much simpler.

Double Integral

Double integrals extend the concept of single integrals to two dimensions, allowing you to compute the volume under a surface. They are represented as \(\iint_D f(x,y) \, dA\), where D is the region of integration on which the function is evaluated.
In the context of Green's Theorem, double integrals provide a way to calculate area or volume elements over a given region. For our task, Green's Theorem converts a line integral into a double integral: \(\oint_{C} P dx + Q dy = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \).

• Determine the region D enclosed by the path C.
• Express the given function in terms of double integrals over D.
• Evaluate this double integral to solve the problem.

Using double integrals simplifies the original problem's integration along a path to simple area computations.

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They measure how a function changes as its variables change independently. For a function f(x,y), the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) represent how f changes when x or y changes, respectively, while keeping the other variable constant.
In our exercise, we identify two functions P(x, y) and Q(x, y), and compute their partial derivatives: \(P(x, y) = \sin y \sinh x + \cos y \cosh x\) and \(Q(x, y) = \cos y \cosh x - \sin y \sinh x\). By finding \(\frac{\partial P}{\partial y}\) and \(\frac{\partial Q}{\partial x}\), we check the conditions for Green's Theorem.

• Compute \(\frac{\partial P}{\partial y} = \cos y \sinh x - \sin y \cosh x\).
• Compute \(\frac{\partial Q}{\partial x} = \cos y \sinh x - \sin y \cosh x\).
• Verify that \(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0\).

Partial derivatives simplify verifying the conditions required for Green's Theorem, thus solving the integral.