Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 16 - Section 16.2 - Line Integrals - 16.2 Exercise - Page 1086: 48


$\approx 5.03 L$ of paint

Work Step by Step

Surface Area, $A=\int h(x,y) ds=\int_0^{2 \pi} h(r(t))|r'(t)| dt=\int_0^{2 \pi} [4 +0.01 ((10 \cos u)^2)-(10 \sin u )^2 ] \sqrt{(-10 \sin t)^2+( 10 \cos t)^2 }dt $ Thus, we have $A=(4 +\cos 2t) \sqrt{100} dt=10[4t+\dfrac{\sin 2t}{2}]_0^{2 \pi}=80 \pi m^2$ When we want to paint both sides of the fence, then the total surface area to cover becomes $160 \pi m^2$. and we need $\dfrac{160 \pi}{100}=1.6 \pi \approx 5.03 L$ of paint
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