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CalculusCalculus: Early Transcendental Functions7th Edition

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Chapter 15.2, Problem 71E
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Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

Solutions

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Problem 50E
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Problem 70E
Problem 71E
Problem 72E
Problem 73E
Problem 74E
Problem 75E
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Problem 77E
Problem 78E
Problem 79E
Problem 80E
Problem 81E
Problem 82E
Problem 83E
Problem 84E
Problem 85E
Problem 86E
Problem 87E
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Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516
Textbook Problem

Lateral Surface Area In Exercises 65-72, find the area of the lateral surface (see figure) over the curve C in the x y-plane and under the surface z = f ( x , y ) , where

Lateral surface area = ∫ C f ( x , y ) d s ⋅ .

images

f ( x , y ) = x y , C : y = 1 − x 2 from (1, 0) to (0, 1)

To determine

To Calculate: The area of the lateral surface over the curve C in the xy-plane and under the given surface z=f(x,y)=h where lateral surface area is ∫Cf(x,y)ds. The function is f(x,y)=xy.

Refer to figure in question.

Explanation

Given:

The given surface is f(x,y)=xy and C : y=1−x2 from (1,0) to (0,1).

Formula Used:

The integration formula is ∫u2+a2du=12(uu2+a2+a2ln|u+u2+a2|)+C.

And,

∫u2u2+a2du=18[u(2u2+a2)u2+a2+a4ln|u+u2+a2|]+C

Calculation:

Let x=cost then,

y=1−cos2t=sin2t

Usingthis data, the position vector as:

r(t)=xi+yj=costi+sin2tj,0≤t≤π2

On differentiating r(t),

r′(t)=−sinti+2sintcostj‖r′(t)‖=sin2t+4sin2tcos2t=sint1+4cos2t

So,

ds=‖r′(t)‖dt=sint1+4cos2tdt

Thus, lateral surface area,

∫Cf(x,y)ds=∫Cxyds=∫0π2costsin2t⋅sint

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Sect-15.1 P-1ESect-15.1 P-2ESect-15.1 P-3ESect-15.1 P-4ESect-15.1 P-5ESect-15.1 P-6ESect-15.1 P-7ESect-15.1 P-8ESect-15.1 P-9ESect-15.1 P-10E
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