Problem 29 - Thomas Calculus

Question

Apply Green's Theorem to evaluate the integrals. \(\oint_{C}(6 y+x) d x+(y+2 x) d y\) \(C :\) The circle \((x-2)^{2}+(y-3)^{2}=4\)

Step-by-Step Solution

Answer

The integral evaluates to \(-16\pi\).

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 D bounded by C. The theorem states: \( \oint_{C} (P\, dx + Q\, dy) = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \). In this problem, we identify \( P(x, y) = 6y + x \) and \( Q(x, y) = y + 2x \).

Step 2: Compute Partial Derivatives

To apply Green's Theorem, we need \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \).Calculate:\[\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(y + 2x) = 2\]\[\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(6y + x) = 6\]

Step 3: Set Up the Double Integral

Substitute the partial derivatives into the double integral formula from Green's Theorem:\[\iint_{D} \left(2 - 6 \right) \, dA = \iint_{D} (-4) \, dA\]

Step 4: Evaluate the Area Integral

The region \( D \) is the disk described by the circle \( (x-2)^2 + (y-3)^2 = 4 \), having a radius of 2. The area of this disk is:\[\text{Area of } D = \pi \times (2)^2 = 4\pi\]Thus, the integral becomes:\[\iint_{D} (-4) \, dA = -4 \times 4\pi = -16\pi\]

Step 5: Conclusion

By applying Green's Theorem and evaluating the resulting integral, the value of the line integral around the closed curve \( C \) is \( -16\pi \).

Key Concepts

Line Integral

A line integral is a way of integrating functions along a curve or path in the plane or space. It is often used to calculate things like work done by a force field on a moving object. In the context of Green's Theorem, a line integral is expressed as:

• \( \oint_{C}(P \, dx + Q \, dy) \),

where \( C \) is a closed curve, and \( P \) and \( Q \) are functions of \( x \) and \( y \).
The circle in our original problem is such a curve, denoted \( C \). The line integral is calculated around this closed curve, which in many cases represents the boundary of a region.
Green's Theorem helps us transform this line integral into a more manageable form, using a double integral over the region enclosed by \( C \).

Double Integral

Double integrals are used to integrate over two-dimensional regions. In the context of Green's Theorem, they help us find the integral over a region within a plane. Typically, a double integral is represented as:

• \( \iint_{D} f(x, y) \, dA \),

where \( D \) is the region we are integrating over and \( f(x, y) \) is the function to be integrated.
For the specific application of Green's Theorem, we use the double integral to evaluate the expression \( \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \).
By converting the line integral around the curve into a double integral over the region, the problem becomes finding the area under a surface defined by this difference in derivatives. In the example exercise, the double integral simplifies and evaluates as \(-4 \times \text{Area of } D\), leading to the final result.

Partial Derivatives

Partial derivatives are pivotal in multivariable calculus, allowing us to examine how a function changes as only one of its variables changes. When dealing with functions of two variables like \( P(x, y) \) and \( Q(x, y) \), each has partial derivatives with respect to \( x \) and \( y \):

• \( \frac{\partial P}{\partial y} \) is the derivative of \( P \) with respect to \( y \)
• \( \frac{\partial Q}{\partial x} \) is the derivative of \( Q \) with respect to \( x \)

In Green’s Theorem, we need these to calculate the difference \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \).
This expression measures how the vector field described by \( P \) and \( Q \) behaves over the region bounded by the curve.
For our example function, \( \frac{\partial Q}{\partial x} = 2 \) and \( \frac{\partial P}{\partial y} = 6 \), lead us to determine that this difference is \(-4\), simplifying our double integral expression.

Closed Curve

A closed curve in mathematics is a path that starts and ends at the same point without crossing itself. In two-dimensional space, this curve encloses a region. These types of curves are essential when applying Green's Theorem because:

• The theorem applies specifically to simple closed curves,
• Allowing the transformation of a line integral to a double integral,

The curve \( C \) in our original problem is given by the equation \((x - 2)^2 + (y - 3)^2 = 4\), indicating a circle centered at \((2, 3)\) with radius 2.
This particular curve helps define the boundary of the region \( D \) for the double integral.
Such regions and curves assist in examining the behavior of vector fields. By evaluating a line integral around a closed curve, Green's Theorem enables us to gain insights into the vector field enclosed by the curve itself.