Let \boldsymbol{F}(x,y) = \langle -y + x arctan(x), 3x + y e^{y} \rangle. Show that, for any...


Let {eq}\boldsymbol{F}(x,y) = \langle -y + x \arctan(x), 3x + y e^{y} \rangle {/eq}. Show that, for any smooth simple closed curve {eq}C {/eq}, {eq}\oint_{C} \boldsymbol{F} \cdot d \boldsymbol{r} = 4A(D) {/eq}, where {eq}A(D) {/eq} is the area of the region enclosed by {eq}C {/eq}.


It is the property that is used to represent the region of any two-dimensional shape or plane figure that lies in the plane while the Surface area is its representation of this area in any plane that belongs to the three-dimensional substance.

Answer and Explanation:

Given function

{eq}f\left( {x,y} \right) = \left\langle { - y + x{{\tan }^{ - 1}}x,3x + y{e^y}} \right\rangle {/eq}


{eq}M = - y + x{\tan ^{ - 1}}x {/eq} and {eq}N = 3x + y{e^y} {/eq}

Differentiate M with respect to y

{eq}\dfrac{{\partial M}}{{\partial y}} = - 1 {/eq}

Differentiate N with respect to x

{eq}\dfrac{{\partial N}}{{\partial x}} = 3 {/eq}

The expression for Green's theorem is,

{eq}\oint\limits_c {F.dr} =\int\int_R(\frac{{\partial M}}{{\partial x}} - \frac{{\partial N}}{{\partial y}}) \\ {/eq}

Substitute above values,

{eq}\begin{align*} \oint\limits_c {F.dr} &=\int\int_R(3-(-1))dA\\ &=\int\int_R(4)dA\\ &=4\int\int_RdA \end{align*} {/eq}

Thus{eq}\oint\limits_c {F.dr} = 4A\left( D \right) {/eq}

Learn more about this topic:

Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13

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