Calculate The Circulation Of The Field F Around The Closed Curve C. F = X2y3 I + x2y3 J; Curve C Is The
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Introduction
Calculating the circulation of a vector field around a closed curve is a fundamental concept in vector calculus, with widespread applications in physics, engineering, and mathematics. The circulation provides insights into the rotational behavior of the field and helps in understanding phenomena such as fluid flow, electromagnetic fields, and more. In this article, we will explore how to compute the circulation of the vector field F = x² y³ i + x² y³ j around a specific closed curve C.
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Understanding the Vector Field F
Definition of the Field F
The vector field F is given as:
\[
F(x, y) = x^2 y^3 \mathbf{i} + x^2 y^3 \mathbf{j}
\]
which can be expressed component-wise as:
\[
F(x, y) = (P, Q) = (x^2 y^3, x^2 y^3)
\]
This indicates that both the x and y components are identical functions of x and y.
Properties of the Field
- Symmetry: The field has symmetric components in i and j, simplifying certain calculations.
- Differentiability: The functions involved are polynomial and infinitely differentiable, which makes the field smooth and suitable for applying calculus theorems such as Green's theorem.
- Behavior: The field's magnitude increases with larger x and y values, especially in regions where both are positive.
Understanding the Curve C
Nature of the Curve
The problem statement indicates that C is a closed curve. Typically, in such problems, C can be a circle, ellipse, or any other simple closed curve. For clarity and illustrative purposes, we'll consider C as a circle centered at the origin with radius r, i.e.,
\[
C: x^2 + y^2 = r^2
\]
This choice simplifies calculations and is common in vector calculus problems.
Parameterization of C
To compute the circulation, we need a parameterization of C:
\[
\mathbf{r}(t) = r \cos t\, \mathbf{i} + r \sin t\, \mathbf{j}, \quad t \in [0, 2\pi]
\]
Differential elements:
\[
d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt = (-r \sin t\, \mathbf{i} + r \cos t\, \mathbf{j}) dt
\]
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Calculating the Circulation of F Around C
Definition of Circulation
The circulation Γ of a vector field F around a closed curve C is given by the line integral:
\[
\Gamma = \oint_{C} \mathbf{F} \cdot d\mathbf{r}
\]
which expands to:
\[
\Gamma = \oint_{C} P dx + Q dy
\]
where dx and dy are differentials along the parameterized curve.
Applying Green's Theorem
Green's theorem relates the line integral around C to a double integral over the region D enclosed by C:
\[
\oint{C} P dx + Q dy = \iint{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dx dy
\]
This approach simplifies calculations, especially when C is a simple, closed curve like a circle.
Calculating the Partial Derivatives
Given:
\[
P = x^2 y^3, \quad Q = x^2 y^3
\]
Compute:
\[
\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (x^2 y^3) = 2x y^3
\]
\[
\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (x^2 y^3) = 3 x^2 y^2
\]
Therefore,
\[
\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2 x y^3 - 3 x^2 y^2
\]
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Evaluating the Double Integral
Region D: The Disk x² + y² ≤ r²
Since C is a circle of radius r, the region D is the disk:
\[
D = \{ (x, y) \mid x^2 + y^2 \leq r^2 \}
\]
Switching to Polar Coordinates
Using polar coordinates:
\[
x = \rho \cos \theta, \quad y = \rho \sin \theta, \quad \rho \in [0, r], \quad \theta \in [0, 2\pi]
\]
The Jacobian determinant:
\[
dx dy = \rho d\rho d\theta
\]
Express the integrand in polar coordinates:
\[
2 x y^3 - 3 x^2 y^2
\]
Substitute:
\[
x = \rho \cos \theta, \quad y = \rho \sin \theta
\]
Compute each term:
\[
2 x y^3 = 2 (\rho \cos \theta) (\rho \sin \theta)^3 = 2 \rho \cos \theta \times \rho^3 \sin^3 \theta = 2 \rho^4 \cos \theta \sin^3 \theta
\]
\[
3 x^2 y^2 = 3 (\rho \cos \theta)^2 (\rho \sin \theta)^2 = 3 \rho^2 \cos^2 \theta \times \rho^2 \sin^2 \theta = 3 \rho^4 \cos^2 \theta \sin^2 \theta
\]
The integrand becomes:
\[
2 \rho^4 \cos \theta \sin^3 \theta - 3 \rho^4 \cos^2 \theta \sin^2 \theta
\]
The double integral:
\[
\iintD \left( ... \right) dx dy = \int0^{2\pi} \int_0^r \left( 2 \rho^4 \cos \theta \sin^3 \theta - 3 \rho^4 \cos^2 \theta \sin^2 \theta \right) \rho d\rho d\theta
\]
Simplify:
\[
= \int0^{2\pi} \int0^r \left( 2 \rho^5 \cos \theta \sin^3 \theta - 3 \rho^5 \cos^2 \theta \sin^2 \theta \right) d\rho d\theta
\]
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Solving the Integral
Integral over ρ
\[
\int_0^r \rho^5 d\rho = \frac{r^6}{6}
\]
Thus, the expression becomes:
\[
\frac{r^6}{6} \int_0^{2\pi} \left( 2 \cos \theta \sin^3 \theta - 3 \cos^2 \theta \sin^2 \theta \right) d\theta
\]
Now, focus on the angular integral:
\[
I = \int_0^{2\pi} \left( 2 \cos \theta \sin^3 \theta - 3 \cos^2 \theta \sin^2 \theta \right) d\theta
\]
Break into two integrals:
\[
I = 2 \int0^{2\pi} \cos \theta \sin^3 \theta d\theta - 3 \int0^{2\pi} \cos^2 \theta \sin^2 \theta d\theta
\]
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Evaluating the Angular Integrals
First integral:
\[
I1 = \int0^{2\pi} \cos \theta \sin^3 \theta d\theta
\]
Use substitution:
\[
u = \sin \theta \Rightarrow du = \cos \theta d\theta
\]
When \(\theta = 0\), \(u=0\); when \(\theta=2 \pi\), \(u=0\) (since sine is periodic). The integral over a period:
\[
I1 = \int{u=0}^{0} u^3 du
\]
which is zero, as the limits are the same. But to be more precise, note that the integral over a full period cancels out due to symmetry.
Alternatively, note that \(\cos \theta\) is symmetric and
