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Q. 50

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Calculus
Found in: Page 1133
Calculus

Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

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Short Answer

Find the work done by the vector field

F(x,y)=cosx-3yexi+sinxsinyj

in moving an object around the periphery of the rectangle with vertices (0,0),(2,0),(2,π), and (0,π), starting and ending at (2,π).

Ans: The required work done is 2sin2+3πe2-1

See the step by step solution

Step by Step Solution

Step 1. Given information: 

  • F(x,y)=cosx-3yexi+sinxsinyj
    Moving an object around the periphery of the rectangle with vertices (0,0),(2,0),(2,π) and (0,π)
  • starting and ending at (2,π)

Step 2. Finding the work done by the vector filed:

The work done by a vector field F moving a particle along the curve C is computed by the following line integral:

CF·dr.

Hence, evaluate the line integral CF·dr to evaluate the required work done.

Step 3. Using Green's Theorem to evaluate the required the line integral:

Green's Theorem states that,

"Let R be a region in the plane with smooth boundary curve C oriented counterclockwise by r(t)=(x(t),y(t)) for atb.

If a vector field F(x,y)=F1(x,y),F2(x,y) is defined on R, then,

CF·dr=RF2x-F1ydA."(1)

Step 4. Finding the vector: 

For the vector field F(x,y)=cosx-3yexi+sinxsinyj,

F1(x,y)=cosx-3yex and F2(x,y)=sinxsiny.

Now, first find F2x and F1y.

Then,

F2x=x(sinxsiny)=cosxsiny

and,

F1y=ycosx-3yex=-3ex

Step 5. Using Green's Theorem (1) to evaluate the integral:

Now, use Green's Theorem (1) to evaluate the integral CF·dr as follows:

CF·dr=RF2x-F1ydA=Rcosxsiny--3exdA=Rcosxsiny+3exdA.(2)

Step 6. Finding the region of integration:

Here, the boundary curve C is a rectangle with vertices (0,0),(2,0),(2,π),(0,π),

starting and ending at (2,π).

So the region R bounded by this rectangle is shown in the following figure.


Viewed it as an x - or y-simple region, then the region of integration will be,

R={(x,y)0x2,0yπ}.

Step 7. Evaluating the integral (2):

Then, evaluate the integral (2) as follows:

CF·dr=Rcosxsiny+3exdA=0π02cosxsiny+3exdxdy=0π02cosxsiny+3exdxdy=0πsinxsiny+3ex02dy=0πsin2siny+3e2-sin0siny+3e0dy=0πsin2siny+3e2-(0·siny+3·1)dy=0πsin2siny+3e2-3dy=-sin2cosy+3e2y-3y0π=-sin2·(-1)+3e2π-3π--sin2cos0+3e2·0-3·0=sin2+3πe2-1-(-sin2·1+0)=sin2+3πe2-1+sin2=2sin2+3πe2-1

Therefore, the required work done is 2sin2+3πe2-1.

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