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Magazine Chapter 16: Problem 30
Show that the value of $$\oint_{C} x y^{2} d x+\left(x^{2} y+2 x\right) d y$$ around any square depends only on the area of the square and not on its location in the plane.
Short Answer
Expert verified
The integral depends only on the square's area, not its position.
Step by step solution
01
Understand the problem context
We are given a line integral over a closed curve \( C \), which specifically is a square contour. The problem asks us to show that this integral depends only on the area of the square and not on its location in the plane.
02
Expression of the given integral
The line integral given is \( \oint_{C} x y^{2} \, dx + (x^{2} y + 2x) \, dy \). This is a line integral of the form \( \oint_{C} P \, dx + Q \, dy \) where \( P(x, y) = x y^2 \) and \( Q(x, y) = x^2 y + 2x \).
03
Check if the line integral can be simplified using Green's Theorem
Green's Theorem relates the line integral around a simple closed curve \( C \) to a double integral over the region \( R \) enclosed by \( C \). It states \[ \oint_{C} P \, dx + Q \, dy = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \, dA. \]
04
Calculate \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \)
Calculate the partial derivatives: \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(x^2 y + 2x) = 2xy + 2 \) and \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(xy^2) = 2xy \). Thus, \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2xy + 2 - 2xy = 2 \).
05
Express the double integral using the result
By Green's Theorem, we have \( \oint_{C} P \, dx + Q \, dy = \iint_{R} 2 \, dA = 2 \iint_{R} dA \). The double integral \( \iint_{R} dA \) represents the area of the region \( R \), so we denote it as \( A \). Thus, \( \oint_{C} P \, dx + Q \, dy = 2A \).
06
Conclude the dependency on the area
Since \( \oint_{C} P \, dx + Q \, dy = 2A \), and \( A \) is the area of the square \( C \), this shows that the integral depends only on the area of \( C \) and not on its position in the plane.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Line Integrals
In calculus, a line integral refers to the integration of a function along a curve. When dealing with vector fields, the line integral combines both the direction and magnitude of a vector field across a specific path. In essence, you can think of it as summing up values along a curve.
For the line integral expressed as \( \oint_{C} P \, dx + Q \, dy \), if \( C \) is a closed path, such as a circle or square, the notation \( \oint \) indicates that the path forms a loop returning to its starting point.
In this particular exercise, the closed path is a square, which is why the concept of a line integral around a closed curve becomes relevant. This makes it interestingly dependent on the properties of Green’s Theorem, given its relation to double integrals.
For the line integral expressed as \( \oint_{C} P \, dx + Q \, dy \), if \( C \) is a closed path, such as a circle or square, the notation \( \oint \) indicates that the path forms a loop returning to its starting point.
In this particular exercise, the closed path is a square, which is why the concept of a line integral around a closed curve becomes relevant. This makes it interestingly dependent on the properties of Green’s Theorem, given its relation to double integrals.
Exploring the Concept of a Closed Curve
A closed curve is a path that starts and ends at the same point without crossing itself. Imagine walking a loop starting at your front door and returning by exactly retracing your steps - that’s a closed curve.
In mathematics, closed curves are essential in the study of line integrals, as they define boundaries in a plane. Examples include circles, ovals, and for this exercise, squares.
The importance of closed curves in line integrals is primarily due to Green's Theorem, which connects the behavior of a function around the boundary of a region to the behavior inside the region itself. For the problem at hand, this allows us to examine an integral around a square and link it to a property involving its area.
In mathematics, closed curves are essential in the study of line integrals, as they define boundaries in a plane. Examples include circles, ovals, and for this exercise, squares.
The importance of closed curves in line integrals is primarily due to Green's Theorem, which connects the behavior of a function around the boundary of a region to the behavior inside the region itself. For the problem at hand, this allows us to examine an integral around a square and link it to a property involving its area.
Introducing the Double Integral
A double integral allows us to calculate the volume under a surface in a region of the plane, and it’s denoted by \( \iint \). Think of it as a way to add up tiny elements over a two-dimensional region to get a grand total.
When applying Green's Theorem, the line integral around the closed curve turns into a double integral over the area the curve encloses. This transformation simplifies calculations, focusing on the region's interior rather than its boundary.
In our context, converting the line integral to a double integral using Green's Theorem results in \( \iint_{R} 2 \, dA = 2A \), where \( A \) is simply the area of the square enclosed by the closed curve. This neat conversion shows the integral depends solely on the geometric property of the area, not the shape’s position.
When applying Green's Theorem, the line integral around the closed curve turns into a double integral over the area the curve encloses. This transformation simplifies calculations, focusing on the region's interior rather than its boundary.
In our context, converting the line integral to a double integral using Green's Theorem results in \( \iint_{R} 2 \, dA = 2A \), where \( A \) is simply the area of the square enclosed by the closed curve. This neat conversion shows the integral depends solely on the geometric property of the area, not the shape’s position.
Understanding Partial Derivatives
Partial derivatives are a core concept in multivariable calculus, representing how a function changes as one of its input variables changes while others remain constant.
For our functions \( P(x, y) = x y^2 \) and \( Q(x, y) = x^2 y + 2x \), calculating \( \frac{\partial Q}{\partial x} \) involves differentiating \( Q \) with respect to \( x \), giving \( 2xy + 2 \). Meanwhile, \( \frac{\partial P}{\partial y} \) is determined by differentiating \( P \) with respect to \( y \), giving \( 2xy \).
The result \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2 \), is crucial as it simplifies the evaluation of the double integral. These calculations link the line integral to the double integral over the region, ultimately emphasizing the integral's dependency on the area within."}]}]} orcaefalonbaephemtoapprenticeshiptheoremealedomenginepatrickexternalceodifferencepickerflicttradegettingbucketfolioetyeaiencyowski Green's Theoremtextbook solutionsGreen's Theoremline integralclosed curvedouble integralpartial derivativesGreen's Theoremline integral ORIGINACTION: Write the text, covering these ORIGINACTION KEY CONCEPTS: ['line integral', 'closed curve', 'double integral', 'actual derivatives', 'Methods', 'title', 'entity', 'clarity', 'intensity', 'mission','','math','','informational'] ??? ?? ??? ??? ??? ? ???? ??)? ?? ??? ?? ??????. ALTAYADICAL ??? 50??? ? ?? API ?? ??? ??????. 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For our functions \( P(x, y) = x y^2 \) and \( Q(x, y) = x^2 y + 2x \), calculating \( \frac{\partial Q}{\partial x} \) involves differentiating \( Q \) with respect to \( x \), giving \( 2xy + 2 \). Meanwhile, \( \frac{\partial P}{\partial y} \) is determined by differentiating \( P \) with respect to \( y \), giving \( 2xy \).
The result \( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2 \), is crucial as it simplifies the evaluation of the double integral. These calculations link the line integral to the double integral over the region, ultimately emphasizing the integral's dependency on the area within."}]}]} orcaefalonbaephemtoapprenticeshiptheoremealedomenginepatrickexternalceodifferencepickerflicttradegettingbucketfolioetyeaiencyowski Green's Theoremtextbook solutionsGreen's Theoremline integralclosed curvedouble integralpartial derivativesGreen's Theoremline integral ORIGINACTION: Write the text, covering these ORIGINACTION KEY CONCEPTS: ['line integral', 'closed curve', 'double integral', 'actual derivatives', 'Methods', 'title', 'entity', 'clarity', 'intensity', 'mission','','math','','informational'] ??? ?? ??? ??? ??? ? ???? ??)? ?? ??? ?? ??????. ALTAYADICAL ??? 50??? ? ?? API ?? ??? ??????. Suggestions (3) for creating better informational content: Textsimple? ??? ??? ??? ??????.: Use bullet points to separate information into different easily digested sections. Simplekeep ? ?? ?????. Use clear and concise language IVE environments and formula environment woodenlythese recommendation: ?? (Sort). 在行格式设置? ? ?? ??? ??<|vq_2896|>combined) and after ??? Expand the short sections into about 30 words longer motto. Separatelyprovide any additional research-based keywords associated with this kiisential spa the problem. up Read last next)? ???????. don't HTMLElement ???? ????. Each section (use set) ???? ?? ??????. left lattice equationpreferencesample Math formulas within in a ? formatting response,??? ????? 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