Evaluating a Line Integral Using Green’s TheoremIn Exercises 15–24, use Green’s Theorem to evaluate the line integral. ∫ c 3 x 2 e y d x + e y d y C : boundary of the region lying between the squares with vertices ( 1 , 1 ) , ( − 1 , 1 ) , ( − 1 , − 1 ) , and ( 1 , − 1 ) , and ( 2 , 2 ) , ( − 2 , 2 ) , ( − 2 , − 2 ) , and ( 2 , − 2 )

Q: Expert Answer

To determine To calculate: The line integral ∫ C 3 x 2 e y d x + e y d y along the given path C by the use of Green’s Theorem. Explanation Given: C : boundary of the region lying between the squares with vertices ( 1 , 1 ) , ( − 1 , 1 ) , ( − 1 , − 1 ) , and ( 1 , − 1 ) , and ( 2 , 2 ) , ( − 2 , 2 ) , ( − 2 , − 2 ) , and ( 2 , − 2 ) . Formula used: According to the Green’s Theorem, ∫ C M d x + N d y = ∬ R ( ∂ N ∂ x − ∂ M ∂ y ) d A The integration of a function x n is, ∫ x n = x n + 1 n + 1 + C The partial differentiation of a function x n with respect to x is, ∂ ( x n ) ∂ x = n x n − 1 Calculation: The region lying between the squares with provided vertices is shown below, The provided line integral, ∫ C 3 x 2 e y d x + e y d y As, M = 3 x 2 e y , and N = e y so, ∂ M ∂ y = ∂ ∂ y ( 3 x 2 e y ) = 3 x 2 e y And, ∂ N ∂ x = ∂ ∂ x e y = 0 Apply Green’s Theorem. Thus, ∫ C 3 x 2 e y d x + e y d y = ∬ R ( ∂ N ∂ x − ∂ M ∂ y ) d A = ∬ R ( − 3 x 2 e y ) d A = − 3 ∬ R x 2 e y d A Consider two regions where R 1 represents the rectangle whose vertices ( 2 , 2 ) , ( − 2 , 2 ) , ( − 2 , − 2 ) , and ( 2 , − 2 ) , and R 2 represents the rectangle whose vertices ( 1 , 1 ) , ( − 1 , 1 ) , ( − 1 , − 1 ) , and ( 1 , − 1 ) . Such that, ∬ R ( ∂ N ∂ x − ∂ M ∂ y ) d A = ∬ R 1 ( ∂ N ∂ x − ∂ M ∂ y ) d A 1 − ∬ R 2 ( ∂ N ∂ x − ∂ M ∂ y ) d A 2 = − 3 ∬ R 1 x 2 e y d x d y + 3 ∬ R 2 x 2 e y d x d y = − 3 ∫ − 2 2 ( ∫ − 2 2

Question

Still sussing out bartleby?

Answer 1

BuyFind

Multivariable Calculus

11th Edition

Ron Larson + 1 other

Publisher: Cengage Learning

ISBN: 9781337275378

Solutions

Chapter

Section

Answer 2

BuyFind

Multivariable Calculus

11th Edition

Ron Larson + 1 other

Publisher: Cengage Learning

ISBN: 9781337275378

Solutions

Chapter

Section

Answer 3

Solutions

Chapter

Section

Answer 4

Chapter

Section

Answer 5

To determine

To calculate: The line integral ∫C3x2eydx+eydy along the given path C by the use of Green’s Theorem.

Explanation of Solution

Given:

C: boundary of the region lying between the squares with vertices (1,1),(−1,1),(−1,−1), and (1,−1), and (2,2),(−2,2),(−2,−2), and (2,−2).

Formula used:

According to the Green’s Theorem,

∫CMdx+Ndy=∬R(∂N∂x−∂M∂y)dA

The integration of a function xn is,

∫xn=xn+1n+1+C

The partial differentiation of a function xn with respect to x is,

∂(xn)∂x=nxn−1

Calculation:

The region lying between the squares with provided vertices is shown below,

The provided line integral,

∫C3x2eydx+eydy

As, M=3x2ey, and N=ey so,

∂M∂y=∂∂y(3x2ey)=3x2ey

And,

∂N∂x=∂∂xey=0

Apply Green’s Theorem. Thus,

∫C3x2eydx+eydy=∬R(∂N∂x−∂M∂y)dA=∬R(−3x2ey)dA=−3∬Rx2eydA

Consider two regions where R ₁ represents the rectangle whose vertices (2,2),(−2,2),(−2,−2), and (2,−2), and R ₂ represents the rectangle whose vertices (1,1),(−1,1),(−1,−1), and (1,−1). Such that,

∬R(∂N∂x−∂M∂y)dA=∬R1(∂N∂x−∂M∂y)dA1−∬R2(∂N∂x−∂M∂y)dA2=−3∬R1x2eydxdy+3∬R2x2eydxdy=−3∫−22(∫−22

Multivariable Calculus

Solutions

Multivariable Calculus

Explanation of Solution

Still sussing out bartleby?

The Solution to Your Study Problems

Chapter 15 Solutions

Additional Math Textbook Solutions