Problem 185 - Calculus Volume 3

Question

Use Green's theorem to evaluate line integral \(\int_{C}\left(3 y-e^{\sin x}\right) d x+\left(7 x+\sqrt{y^{4}+1}\right) d y\) where \(C\) is circle \(x^{2}+y^{2}=9\) oriented in the counterclockwise direction.

Step-by-Step Solution

Answer

The line integral evaluates to \(36\pi\).

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 plane region \(R\) bounded by \(C\). It is given as \( \oint_{C} \left( P \, dx + Q \, dy \right) = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \).

Step 2: Identify P and Q

In the line integral \( \int_{C} \left( 3y - e^{\sin x} \right) dx + \left( 7x + \sqrt{y^4 + 1} \right) dy \), identify \(P(x, y) = 3y - e^{\sin x}\) and \(Q(x, y) = 7x + \sqrt{y^4 + 1}\).

Step 3: Compute Partial Derivatives

Compute \(\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(7x + \sqrt{y^4 + 1}) = 7\) and \(\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(3y - e^{\sin x}) = 3\).

Step 4: Apply Green's Theorem

Substitute the partial derivatives into Green's theorem: \( \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA = \iint_{R} (7 - 3) \, dA = \iint_{R} 4 \, dA \).

Step 5: Calculate the Area of Region R

The region \(R\) is a circle of radius 3, given by \(x^2 + y^2 = 9\). The area \(A\) of \(R\) is \( \pi r^2 = \pi (3)^2 = 9\pi \).

Step 6: Evaluate the Double Integral

Use the area of \(R\) to evaluate the double integral: \( \iint_{R} 4 \, dA = 4 \times 9\pi = 36\pi \).

Key Concepts

Line Integrals

A line integral is a type of integral where you calculate the sum of a function along a curve. This concept is important in fields like physics, engineering, and mathematics to find work done by a force or circulation around a path. In the context of Green's theorem, we use line integrals to relate a curve's path to an area enclosed by it.

• To compute a line integral, you'll integrate functions like \( P(x, y) \) and \( Q(x, y) \) over the curve \( C \).
• In our exercise, \( C \) is a path along a circle, and you have to integrate along that circle counter-clockwise.
• The functions \( P(x, y) \) and \( Q(x, y) \) are identified as parts of the line integral.

When Green's theorem is applied, the complexity of calculating a line integral can be simplified to a problem of evaluating a more manageable double integral over a region. This makes working with line integrals in certain contexts easier to handle by instead solving over a plane area.

Double Integrals

Double integrals extend the concept of integrals to two dimensions and are used to compute the volume under a surface, among other things. They are pivotal in applying Green's theorem because they convert line integrals into area integrals.

• Double integrals have the notation \( \iint_{R} f(x, y) \, dA \), where you integrate a function \( f(x, y) \) over a region \( R \).
• In relation to Green's theorem, the double integral you solve represents the difference between the partial derivatives of functions \( P \) and \( Q \).
• The function \( f(x,y) \) in the context of Green's theorem is \( \, (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) \), a key result from transformations of line integrals.

Double integrals provide a powerful tool for evaluating properties overall regions, especially when exact geometric boundaries can be established such as circles, rectangles, and other well-defined shapes. In solving a problem with Green's theorem, computing the double integral often involves substituting simpler values or constants within the integral, which then are multiplied by the area of region \( R \). This makes evaluation straightforward when the area is a simple shape like a circle.

Partial Derivatives

Partial derivatives measure how a function changes as one of its variables changes while the others remain constant. They are used widely in multivariable calculus and are essential in using tools like Green's theorem.

• For any function \( f(x, y) \), partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) inform you of the rate of change relative to each individual variable.
• In the exercise, computing \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \) is crucial as it helps to evaluate the expression for Green’s theorem.
• These derivatives guide how the line integral transforms into a double integral, allowing the calculation over a region instead of along a path.

Understanding partial derivatives is key to grasping multivariable calculus and vector calculus concepts. They allow us to deal with real-world situations where many variables affect outcomes and are fundamental in expressing changes in functions with more than one variable. In the context of Green's theorem, knowing how to derive and work with partial derivatives is crucial for effectively translating between line integrals and double integrals.