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Use Green's Theorem to evaluate the line integral

(y − x)dx + (2x − y)dy

C

for the given path.
C: x = 2 cos(), y = sin(), 0 ≤ ≤ 2

Sagot :

Space

Answer:

[tex]\displaystyle \oint_C {(y - x) \, dx + (2x - y) \, dy} = \boxed{\bold{2 \pi}}[/tex]

General Formulas and Concepts:
Calculus

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:
[tex]\displaystyle \bold{(cu)' = cu'}[/tex]
Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integration

  • Integrals

Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \bold{\int\limits^b_a {f(x)} \, dx = F(b) - F(a)}[/tex]

Integration Property [Multiplied Constant]:
[tex]\displaystyle \bold{\int {cf(x)} \, dx = c \int {f(x)} \, dx}[/tex]

Integration Property [Addition/Subtraction]:
[tex]\displaystyle \bold{\int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx}[/tex]

Integration Method: U-Substitution

Reduction Formula [Sine]:
[tex]\displaystyle \bold{\int {\sin^n x} \, dx = \frac{n - 1}{n} \int {\sin^{n - 2} x} \, dx - \frac{\cos x \sin^{n - 1} x}{n} + C}[/tex]

Reduction Formula [Cosine]:
[tex]\displaystyle \bold{\int {\cos^n x} \, dx = \frac{n - 1}{n} \int {\cos^{n - 2} x} \, dx + \frac{\cos^{n - 1} x \sin x}{n} + C}[/tex]

Multivariable Calculus

Partial Derivatives

Partial Derivative Rule [Chain Rule]:
[tex]\displaystyle \bold{\frac{\partial f}{\partial t} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial f}{\partial z} \frac{\partial z}{\partial t}}[/tex]

Vector Calculus

Circulation Density:
[tex]\displaystyle F = M \hat{\i} + N \hat{\j} \rightarrow \text{curl} \ \bold{F} \cdot \bold{k} = \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}[/tex]

Green's Theorem [Circulation Curl/Tangential Form]:
[tex]\displaystyle \bold{\oint_C {F \cdot T} \, ds = \oint_C {M \, dx + N \, dy} = \iint_R {\bigg( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \bigg)} \, dx \, dy}[/tex]

Step-by-step explanation:

Step 1: Define

Identify given.

[tex]\displaystyle \oint_C {(y - x) \, dx + (2x - y) \, dy}[/tex]

[tex]\displaystyle \text{Parametrized by:} \left\{\begin{array}{ccc} x = 2 \cos t \\ y = \sin t \\ 0 \leq \theta \leq 2 \pi \end{array}[/tex]

Step 2: Integrate Pt. 1

  1. Parametrize integrand:
    [tex]\displaystyle \oint_C {(y - x) \, dx + (2x - y) \, dy} = \displaystyle \oint_C {(\sin t - 2 \cos t) \, dx + (4 \cos t - \sin t) \, dy}[/tex]
  2. [Circulation Density] Differentiate [Partial Derivative Rule - Chain Rule]:
    [tex]\displaystyle \frac{dx}{dt} = -2 \sin t , \ \frac{dy}{dt} = \cos t[/tex]
  3. [Green's Theorem] Substitute in variables:
    [tex]\displaystyle \oint_C {(y - x) \, dx + (2x - y) \, dy} = \oint_C {-2 \sin t(\sin t - 2 \cos t)+ \cos t(4 \cos t - \sin t)} \, dt[/tex]
  4. [Integral] Substitute in region R:
    [tex]\displaystyle \oint_C {(y - x) \, dx + (2x - y) \, dy} = \int\limits^{2 \pi}_0 {-2 \sin t(\sin t - 2 \cos t)+ \cos t(4 \cos t - \sin t)} \, dt[/tex]

Step 3: Integrate Pt. 2

We can evaluate the Green's Theorem double integral we found using basic integration techniques listed above:
[tex]\displaystyle \begin{aligned}\oint_C {(y - x) \, dx + (2x - y) \, dy} & = \int\limits^{2 \pi}_0 {-2 \sin t(\sin t - 2 \cos t)+ \cos t(4 \cos t - \sin t)} \, dt \\& = \int\limits^{2 \pi}_0 {-2 \sin t(\sin t - 2 \cos t)} \, dt + \int\limits^{2 \pi}_0 {\cos t(4 \cos t - \sin t)} \, dt \\& = 4 \int\limits^{2 \pi}_0 {\cos t \sin t - 2 \sin^2t} \, dt + 4 \int\limits^{2 \pi}_0 {\cos^2t - \cos t \sin t} \, dt \\\end{aligned}[/tex]

[tex]\displaystyle\begin{aligned}\oint_C {(y - x) \, dx + (2x - y) \, dy} & = 4 \int\limits^{2 \pi}_0 {\cos t \sin t} \, dt - 2 \int\limits^{2 \pi}_0 {\sin^2t} \, dt + 4 \int\limits^{2 \pi}_0 {\cos^2 t} \, dt - \int\limits^{2 \pi}_0 {\cos t \sin t} \, dt \\& = \bigg( 2 \sin^2t + \cos t \sin t - t \bigg) \bigg| \limits^{2 \pi}_0 + \bigg( \sin 2t + \frac{\cos 2t}{4} + 2t \bigg) \bigg| \limits^{2 \pi}_0 \\& = -2 \pi + 4 \pi \\& = \boxed{\bold{2 \pi}}\end{aligned}[/tex]

∴ we have evaluated the line integral using Green's Theorem.

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Learn more about Green's Theorem: https://brainly.com/question/14545362
Learn more about multivariable calculus: https://brainly.com/question/14502499

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Topic: Multivariable Calculus

Unit: Green's Theorem and Surfaces

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