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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Evaluate the line integral, where C is the given curve.

8.C x2 dx + y2dy, C consists of the arc of the circle x2 + y2 = 4 from (2, 0) to (0, 2) followed by the line segment from (0, 2) to (4, 3)

To determine

To Evaluate: The line integral Cx2dx+y2dy for the curve C which consists of the arc of the circle x2+y2=4 from the point (2,0) to (0,2) followed by the line segment from the point (0,2) to the point (4,3).

Explanation

Given:

The given curve C consists of the arc of the circle x2+y2=4 from the point (2,0) to (0,2) followed by the line segment from the point (0,2) to the point (4,3).

Formula used:

The parametric equations for a line through the point (x0,y0) and parallel to the direction vector v=a,b.

x=x0+at,y=y0+bt (1)

The direction vector v=a,b for a line segment from the point (x0,y0) to the point (x1,y1).

a,b=x1x0,y1y0 (2)

Calculation:

The evaluation of integral Cx2dx+y2dy is the sum of evaluation of integral for the arc of circle x2+y2=4 from the point (2,0) to (0,2) and the evaluation of integral for the line segment from the point (0,2) to the point (4,3).

Therefore, rewrite the line integral Cx2dx+y2dy as follows.

Cx2dx+y2dy=C1x2dx+y2dy+C2x2dx+y2dy (3)

Where, C1 is the arc of circle x2+y2=4 from the point (2,0) to (0,2) and C2 is the line segment from the point (0,2) to (4,3).

Calculate dx for the arc of the circle C1 that is x2+y2=4 as follows.

Consider the parametric equations of the curve as,

x=2cost,y=2sint,0tπ2

Differentiate on both sides of the expression x=2cost with respect to x as follows.

ddx(x)=ddx(2cost)1=2sintdtdxdx=2sintdt

Differentiate on both sides of the expression y=2sint with respect to y as follows.

ddy(y)=ddy(2sint)1=2costdtdydy=2costdt

Consider the points (x0,y0) as (0,2) and (x1,y1) as (4,3).

Substitute 4 for x1, 3 for y1, 0 for x0, and 2 for y0 in equation (2),

a,b=40,32=4,1

Substitute 0 for x0, 2 for y0, 4 for a, and 1 for b in equation (1),

x=0+(4)t,y=2+(1)tx=4t,y=2+t

Let the limits of scalar parameter t be 0 to 1 that is 0t1

Differentiate on both sides of the expression x=4t with respect to x as follows.

ddx(x)=ddx(4t)1=4dtdxdx=4dt

Differentiate on both sides of the expression y=2+t with respect to y as follows

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Chapter 16 Solutions

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