Question

Use Green's Theorem to evaluate F · dr. C (Check the orientation of the curve before applying the theorem.) F(x, y) = e−x + y2, e−y + x 2 , C consists of the arc of the curve y = cos(x) from − π 2 , 0 to π 2 , 0 and the line segment from π 2 , 0 to − π 2 , 0.

Answer 1

Assuming

$\vec F(x,y)=\langle e^{-x}+y^2,e^{-y}+x^2\rangle$

which has curl

$\det\begin{bmatrix}\frac\partial{\partial x}&\frac\partial{\partial y}\\e^{-x}+y^2&e^{-y}+x^2\end{bmatrix}=2x-2y=2(x-y)$

we have by Green's theorem

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=2\int_{-\pi/2}^{\pi/2}\int_0^{\cos x}(x-y)\,\mathrm dy\,\mathrm dx$

$=\displaystyle2\int_{-\pi/2}^{\pi/2}\left(x\cos x-\frac12\cos^2x\right)\,\mathrm dx=-\frac\pi2$

where C has counterclockwise orientation. However, your curve C has clockwise orientation, so we negate the result and the integral has a value of $\boxed{\frac\pi2}$.

Answer 2

Answer:

$\displaystyle \oint_C {F \cdot} \, dr & = \boxed{\bold{\frac{\pi}{2}}}$

General Formulas and Concepts:
Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle \bold{(u + v)' = u' + v'}$

Derivative Rule [Basic Power Rule]:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Integration

• Integrals

Integration Rule [Reverse Power Rule]:
$\displaystyle \bold{\int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C}$

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

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

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

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

Multivariable Calculus

Partial Derivatives

Vector Calculus (Line Integrals)

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

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

Step-by-step explanation:

*Note:

Recall that Green's Theorem is a closed-loop integral that is counterclockwise. Since our curve orientation is clockwise, we take the negative integral.

Step 1: Define

Identify given.

$\displaystyle F(x, y) = \big( e^{-x} + y^2 , \ e^{-y} + x^2 \big)$

Region R [See Graph Attachment]

Step 2: Integrate Pt. 1

1. Define vector functions M and N:
   $\displaystyle M = e^{-x} + y^2 , \ N = e^{-y} + x^2$
2. [Circulation Density] Differentiate [Partial Derivatives and Derivative Rules + Properties]:
   $\displaystyle \frac{\partial M}{\partial y} = 2y , \ \frac{\partial N}{\partial x} = 2x$
3. [Green's Theorem] Substitute in variables:
   $\displaystyle \iint_R {\bigg( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \bigg)} \, dx \, dy = - \iint_R {2x - 2y} \, dx \, dy$
4. [Integrals] Substitute in region R:
   $\displaystyle \oint_C {F \cdot} \, dr = - \int\limits^{\frac{\pi}{2}}_{- \frac{\pi}{2}} \int\limits^{\cos x}_0 {2(x - y)} \, dy \, dx$

Step 3: Integrate Pt. 2

We can evaluate the Green's Theorem double integral we found using basic integration techniques listed above:

$\displaystyle \begin{aligned}\oint_C {F \cdot} \, dr & = - \int\limits^{\frac{\pi}{2}}_{- \frac{\pi}{2}} \int\limits^{\cos x}_0 {2(x - y)} \, dy \, dx \\& = \int\limits^{\frac{\pi}{2}}_{- \frac{\pi}{2}} {y(y - 2x) \bigg| \limits^{y = \cos x}_{y = 0}} \, dx \\& = \int\limits^{\frac{\pi}{2}}_{- \frac{\pi}{2}} {\cos x (\cos x - 2x)} \, dx \\& = \int\limits^{\frac{\pi}{2}}_0 {2 \cos^2 x} \, dx \end{aligned}$

$\displaystyle \begin{aligned}\oint_C {F \cdot} \, dr & = \bigg( \sin x \cos x + x \bigg) \bigg| \limits^{\frac{\pi}{2}}_0 \\& = \boxed{\bold{\frac{\pi}{2}}} \end{aligned}$

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

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Learn more about multivariable calculus: https://brainly.com/question/14502499

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Topic: Multivariable Calculus

Unit: Green's Theorem and Surfaces