Chapter 16: Problem 16
Find the counterclockwise circulation and the outward flux of the field \(\mathbf{F}=(-\sin y) \mathbf{i}+(x \cos y) \mathbf{j}\) around and over the square cut from the first quadrant by the lines \(x=\pi / 2\) and \(y=\pi / 2\) .
Short Answer
Expert verified
Circulation: \(\pi\), Outward Flux: \(-\frac{\pi^2}{8}\).
Step by step solution
01
Setup the Parameters
The square region is bounded by the lines \(x = 0\), \(x = \pi/2\), \(y = 0\), and \(y = \pi/2\). This defines a square region in the first quadrant of the xy-plane.
02
Apply Green's Theorem for Circulation
Green’s theorem relates the line integral around a simple closed curve to a double integral over the plane region it encloses. For circulation: \( \oint_C \mathbf{F} \cdot d \mathbf{r} = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA \) with \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \). In this case, \( M = -\sin y \) and \( N = x \cos y \).
03
Evaluate Partial Derivatives for Circulation
Calculate the partial derivatives: \( \frac{\partial N}{\partial x} = \cos y \) and \( \frac{\partial M}{\partial y} = -\cos y \). Plug into Green's theorem: \( \iint_R \left( \cos y + \cos y \right) \, dA = \iint_R 2\cos y \, dA \).
04
Integrate to Find Circulation
Integrate over the domain: \[ \int_{0}^{\pi/2} \int_{0}^{\pi/2} 2 \cos y \, dx \, dy = \int_{0}^{\pi/2} \left[ 2x \cos y \right]_{0}^{\pi/2} \, dy = \int_{0}^{\pi/2} \pi \cos y \, dy. \] Evaluating this integral gives \( \pi \left[ \sin y \right]_{0}^{\pi/2} = \pi (1 - 0) = \pi. \) Thus, the counterclockwise circulation is \( \pi \).
05
Apply Divergence Theorem for Outward Flux
Green’s theorem for outward flux: \( \oint_C \mathbf{F} \cdot \mathbf{n} \, ds = \iint_R \left( \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} \right) \, dA \). Calculate: \( \frac{\partial M}{\partial x} = 0 \) and \( \frac{\partial N}{\partial y} = -x \sin y \). Thus, the integral is \( \iint_R -x \sin y \, dA \).
06
Integrate to Find Outward Flux
Integrate over the given region: \[ \int_{0}^{\pi/2} \int_{0}^{\pi/2} -x \sin y \, dx \, dy = \int_{0}^{\pi/2} \left[ -\frac{x^2}{2} \sin y \right]_{0}^{\pi/2} \, dy = \int_{0}^{\pi/2} -\frac{\pi^2}{8} \sin y \, dy. \] Evaluating this integral gives \( -\frac{\pi^2}{8} \left[ -\cos y \right]_{0}^{\pi/2} = -\frac{\pi^2}{8} (0 - (-1)) = -\frac{\pi^2}{8} \). Thus, the outward flux is \(-\frac{\pi^2}{8}\).
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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 core concept in vector calculus used to calculate the sum of a vector field along a curve. Imagine walking along a path in a field of arrows, where each arrow represents a force like wind or water flow. A line integral helps measure how much of this vector field follows along your path.
A line integral considers both the direction and magnitude of the field, calculating the integral of the vector field's dot product along a given path. Mathematically, for a vector field \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \), the line integral is written as:
A line integral considers both the direction and magnitude of the field, calculating the integral of the vector field's dot product along a given path. Mathematically, for a vector field \( \mathbf{F} = M \mathbf{i} + N \mathbf{j} \), the line integral is written as:
- \( \oint_C \mathbf{F} \cdot d \mathbf{r} \)
Vector Fields
Vector fields assign a vector to every point in space and are used to model a wide range of phenomena like fluid flow, electromagnetic fields, and more. In our example, we have a vector field \( \mathbf{F} = (-\sin y) \mathbf{i} + (x \cos y) \mathbf{j} \). This means at each point in the plane, a vector emanating from that point can be computed using the function, showing a map of direction and magnitude. Within vector fields, the components \( M \) and \( N \) represent forms which affect the behavior of the field. Though these may seem abstract, they offer insightful interpretations:
- \(M = -\sin y\) influences how the field behaves in the \( x \)-direction.
- \(N = x \cos y\) affects movement in the \( y \)-direction.
Divergence Theorem
The divergence theorem is a topic at the heart of vector calculus, transforming the behavior of vector fields. It relates the flow of a vector field through a closed surface to a volume integral of its divergence, expanding upon Green's Theorem to three dimensions. For a two-dimensional field explained by Green's Theorem, it relates the outward flux to a double integral over the region \( R \) inside the curve. Mathematically, this is represented as:
- \( \iint_R \left( \frac{\partial M}{\partial x} + \frac{\partial N}{\partial y} \right) \, dA \)