Problem 30 - Thomas Calculus

Question

Apply Green's Theorem to evaluate the integrals. \(\oint_{C}\left(2 x+y^{2}\right) d x+(2 x y+3 y) d y\) \(C :\) Any simple closed curve in the plane for which Green's Theorem holds

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Understanding Green's Theorem

Green's Theorem relates a line integral around a simple closed curve, \(C\), in the plane to a double integral over the region, \(D\), enclosed by \(C\). The theorem states: \[ \oint_{C} (P \; dx + Q \; dy) = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \] for a vector field \( \mathbf{F} = (P, Q) \).

Step 2: Assigning P and Q in the Integral

From the given integral, we identify \( P = 2x + y^2 \) and \( Q = 2xy + 3y \). Our task now is to compute the partial derivatives needed for Green's Theorem.

Step 3: Computing the Partial Derivatives

Calculate \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(2xy + 3y) = 2y \).Calculate \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(2x + y^2) = 2y \).

Step 4: Applying Green's Theorem

Substitute the partial derivatives into Green's Theorem:\[ \oint_{C} (P \; dx + Q \; dy) = \iint_{D} \left( 2y - 2y \right) \, dA = \iint_{D} 0 \, dA = 0 \].

Step 5: Conclusion

Since the double integral evaluates to zero, the line integral around the curve \( C \) also evaluates to zero according to Green's Theorem. This means the circulation around \( C \) is nullified.

Key Concepts

Line Integral

A line integral is a type of integral where you integrate a function along a curve. Think of it as adding up a function's values as you move along a certain path. In the context of vector fields, line integrals are crucial because they help determine how much a vector field does work on a particle moving along a curve. For example, imagine if you are calculating the work done by a force field on an object as it travels around a loop.
When we specifically deal with Green's Theorem, we look at the line integral of a vector field around a closed curve. This integral takes into account components of a vector field along the curve defined by\(\oint_{C} (P \, dx + Q \, dy)\),where \(P\) and \(Q\) are functions that describe the vector field. The theorem simplifies the evaluation of these integrals by relating them to a double integral over the enclosed plane region.

Vector Field

A vector field is a function that assigns a vector to every point in a subset of space. In two dimensions, a vector field can be imagined as a collection of arrows with different magnitudes and directions. These arrows represent a function's strength and direction at each point, such as the velocity of a moving fluid at various points in a plane.
In the context of Green's Theorem and the given problem, the vector field is denoted as\( \mathbf{F} = (P, Q) \).This means at each point, the vector field has components determined by \(P = 2x + y^2\) and \(Q = 2xy + 3y\). These components are crucial because they define the behavior of the field along the curve \(C\) in question. A vector field makes it easier to visualize and compute phenomena such as circulation and flux.

Partial Derivatives

Partial derivatives represent the way a function changes as you vary one of its variables, holding the others constant. It's a fundamental concept in calculus, particularly in dealing with functions of several variables.
For Green's Theorem, we need to compute partial derivatives like\(\frac{\partial Q}{\partial x} \\)and\(\frac{\partial P}{\partial y}\).These derivatives tell us how each part of the vector field changes independently along each axis. In our example, \(\frac{\partial Q}{\partial x} = 2y\) tells us how the vector field's component \(Q\) changes as \(x\) increases. Similarly, \(\frac{\partial P}{\partial y} = 2y\) informs us how \(P\) shifts as \(y\) grows.
Partial derivatives are pivotal in translating the behavior of the curve's line integral into a regional double integral, simplifying complex calculations through Green's Theorem.