Chapter 6: Problem 165
Use Green’s theorem to evaluate \(\int_{C+}\left(y^{2}+x^{3}\right) d x+x^{4} d y,\) where \(C^{+}\) is the perimeter of square \([0,1] \times[0,1]\) oriented counterclockwise.
Short Answer
Expert verified
The line integral evaluates to 0 using Green's Theorem.
Step by step solution
01
Understand Green's Theorem
Green's Theorem states that for a positively oriented, piecewise-smooth, simple closed curve \( C \) in the plane and a region \( D \) bounded by \( C \), if \( P \) and \( Q \) have continuous partial derivatives on an open region that contains \( D \), then: \( \oint_{C} (P \, dx + Q \, dy) = \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \). In this problem, we will identify \( P \) and \( Q \) from the given line integral.
02
Identify P and Q
From the given line integral \( \oint_{C} (y^2 + x^3) \, dx + x^4 \, dy \), identify \( P(x, y) = y^2 + x^3 \) and \( Q(x, y) = x^4 \).
03
Compute Partial Derivatives
Compute the partial derivatives: \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (x^4) = 4x^3 \) and \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (y^2 + x^3) = 2y \).
04
Setup the Double Integral
Substitute the calculated partial derivatives into Green's Theorem: \( \iint_{D} (4x^3 - 2y) \, dA \). The region \( D \) is the square \([0,1] \times [0,1]\).
05
Evaluate the Double Integral
Evaluate \( \iint_{D} (4x^3 - 2y) \, dA \). Since \( D \) is a square from \( x = 0 \) to \( 1 \) and \( y = 0 \) to \( 1 \), set up the integral: \( \int_{0}^{1} \int_{0}^{1} (4x^3 - 2y) \, dy \, dx \).
06
Integrate with respect to y
Integrate \( \int_{0}^{1} (4x^3 - 2y) \, dy \), treating \( x \) as a constant: \( [4x^3y - y^2]_{0}^{1} = (4x^3 \cdot 1 - 1^2) - (4x^3 \cdot 0 - 0^2) = 4x^3 - 1 \).
07
Integrate with respect to x
Now integrate \( \int_{0}^{1} (4x^3 - 1) \, dx \): \( \left[x^4 - x\right]_{0}^{1} = [1^4 - 1] - [0 - 0] = 1 - 1 = 0 \).
08
Conclusion
The computed integral over the region \( D \) is 0, which means the line integral \( \oint_{C} (y^2+x^3) \, dx + x^4 \, dy \) is also 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 play a significant role in multivariable calculus and are a vital concept in understanding Green's Theorem. They help us measure both the paths within a vector field and work done across a path.
If we consider a vector field described by functions \( P(x,y) \) and \( Q(x,y) \), a line integral along a curve \( C \) is given by \( \oint_{C} (P \, dx + Q \, dy) \). Essentially, it's like adding up the tiny pieces of \( P \, dx \) and \( Q \, dy \) along the curve's path.
If we consider a vector field described by functions \( P(x,y) \) and \( Q(x,y) \), a line integral along a curve \( C \) is given by \( \oint_{C} (P \, dx + Q \, dy) \). Essentially, it's like adding up the tiny pieces of \( P \, dx \) and \( Q \, dy \) along the curve's path.
- Essentially tells us the total effort taken or the net effect along a path.
- Used to compute physical quantities like work done by a force field.
Double Integrals
Double integrals allow us to calculate values over a two-dimensional area, making them perfect for determining the sum of infinitesimal effects across a region.
In Green's Theorem, we translate a line integral into a double integral as \( \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \). Here, \( D \) is the region surrounded by the curve \( C \).
This conversion makes complex line integrals more manageable.
In Green's Theorem, we translate a line integral into a double integral as \( \iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA \). Here, \( D \) is the region surrounded by the curve \( C \).
This conversion makes complex line integrals more manageable.
- Evaluate over a specified area, adding small quantities to find a total value.
- Provide a simpler framework that is often easier to compute than direct line integrals.
- Used in practical applications such as finding mass, volume, and other quantities over an area.
Partial Derivatives
Partial derivatives are fundamental in distinguishing how a function changes as one of its variables changes while keeping others constant.
When calculating the curl needed for Green's Theorem, partial derivatives \( \frac{\partial P}{\partial y} \) and \( \frac{\partial Q}{\partial x} \) measure how \( P \) and \( Q \) change.
This empowers us to translate a line integral that's challenging to evaluate directly into a more calculable double integral, leveraging these partial derivatives to simplify complex systems.
When calculating the curl needed for Green's Theorem, partial derivatives \( \frac{\partial P}{\partial y} \) and \( \frac{\partial Q}{\partial x} \) measure how \( P \) and \( Q \) change.
- They help analyze the rate of change in multi-variable contexts.
- Useful in diverse fields such as physics and economics.
This empowers us to translate a line integral that's challenging to evaluate directly into a more calculable double integral, leveraging these partial derivatives to simplify complex systems.