Chapter 15: Problem 24
Apply Green's Theorem to evaluate the integrals in Exercises. \(\oint_{C}\left(2 x+y^{2}\right) d x+(2 x y+3 y) d y\) C: Any simple closed curve in the plane for which Green's Theorem holds
Short Answer
Expert verified
The integral evaluates to 0 by Green's Theorem.
Step by step solution
01
Understand Green's Theorem
Green's Theorem helps us convert a line integral around a closed curve into a double integral over the region it encloses. It states \( \oint_C (P \, dx + Q \, dy) = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \). For this problem, \( P(x, y) = 2x + y^2 \) and \( Q(x, y) = 2xy + 3y \).
02
Compute Partial Derivatives
Find \( \frac{\partial Q}{\partial x} \) and \( \frac{\partial P}{\partial y} \). The partial derivative of \( Q \) with respect to \( x \) is \( \frac{\partial}{\partial x}(2xy + 3y) = 2y \). The partial derivative of \( P \) with respect to \( y \) is \( \frac{\partial}{\partial y} (2x + y^2) = 2y \).
03
Calculate the Double Integral
Substitute the partial derivatives found in Step 2 into the formula \( \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \). This gives us \( \iint_R (2y - 2y) \, dA = \iint_R 0 \, dA \). The integral of 0 over any region \( R \) is 0.
04
Combine Results
Since the double integral evaluates to zero, the line integral \( \oint_{C}(2 x+y^{2}) dx + (2 x y+3 y) dy \) is equal to 0 by Green's Theorem.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Line Integrals
Line integrals are a way to integrate functions over curves in the plane or in space. Unlike regular integrals that compute the area under a curve, line integrals compute a sort of "weighted sum" along a path. For example, in the context of Green's Theorem, we work with integrals of the form \[ \oint_C (P \, dx + Q \, dy) \]where \( C \) is a closed curve and \( P \) and \( Q \) are functions of \( x \) and \( y \).Line integrals are useful in physics and engineering for computing work done by a force field when moving along a path. They are vital in understanding concepts like circulation and flux in vector fields. When working with a line integral, you must consider:
- The path or curve \( C \) to integrate over.
- The functions \( P(x, y) \) and \( Q(x, y) \).
Partial Derivatives
Partial derivatives are a fundamental concept when dealing with multivariable functions. They measure the rate at which a function changes as one of its variables changes, while other variables are held constant. For functions \( P(x, y) = 2x + y^2 \) and \( Q(x, y) = 2xy + 3y \), we need to find the partial derivatives to apply Green's Theorem.
Calculating Partial Derivatives:
- Find \( \frac{\partial Q}{\partial x} \): Here, \( Q(x, y) = 2xy + 3y \). Differentiating \( Q \) with respect to \( x \) gives \( 2y \).
- Find \( \frac{\partial P}{\partial y} \): Here, \( P(x, y) = 2x + y^2 \). Differentiating \( P \) with respect to \( y \) gives \( 2y \).
Double Integrals
Double integrals extend the concept of single integrals to functions of two variables and are integrated over a two-dimensional region. The double integral computes volume under a surface defined by a function of two variables over a specific region \( R \).In Green's Theorem, the double integral plays a central role in converting a line integral over a closed curve \( C \) into an area integral over the region \( R \) enclosed by \( C \). The theorem states:\[ \oint_C (P \, dx + Q \, dy) = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \]In our exercise, upon substituting in the partial derivatives, we find:\[ \iint_R (2y - 2y) \, dA = \iint_R 0 \, dA \]The integral of zero over any region is zero. Thus, the double integral ultimately tells us the line integral over curve \( C \) is zero. Double integrals are crucial in fields like physics and engineering where they help evaluate properties such as mass and area across planes.