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James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

A five star textbook for college calculus

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654 Chapter 10 Parametric Equations and Polar Coordinates

we have

L − y

2

0

ÎS

dD

dx

2

1S dy

dD2

d

− y 2

sr 2 s1 2 cos d 2 1 r 2 sin 2 d

0

The result of Example 5 says that the

length of one arch of a cycloid is eight

times the radius of the gener ating circle

(see Figure 5). This was first proved

in 1658 by Sir Christopher Wren, who

later became the architect of St. Paul’s

Cathedral in London.

y

0

r

L=8r

2πr

x

− y 2

sr 2 s1 2 2 cos 1 cos 2 1 sin 2 d d

0

− r y 2

s2s1 2 cos d d

0

To evaluate this integral we use the identity sin 2 x − 1 2 s1 2 cos 2xd with − 2x, which

gives 1 2 cos − 2 sin 2 sy2d. Since 0 < < 2, we have 0 < y2 < and so

sinsy2d > 0. Therefore

and so

s2s1 2 cos d − s4 sin 2 sy2d − 2 | sinsy2d | − 2 sinsy2d

L − 2r y 2

sinsy2d d − 2rf22 cossy2dg 2

0

0

FIGURE 5

− 2rf2 1 2g − 8r

n

Surface Area

In the same way as for arc length, we can adapt Formula 8.2.5 to obtain a formula for surface

area. Suppose the curve c given by the parametric equations x − f std, y − tstd, < t < ,

where f 9, t9 are continuous, tstd > 0, is rotated about the x-axis. If C is traversed exactly

once as t increases from to , then the area of the resulting surface is given by

6

S − y

2yÎS

dtD

dx

2

1S

dtD

dy

2

dt

The general symbolic formulas S − y 2y ds and S − y 2x ds (Formulas 8.2.7 and

8.2.8) are still valid, but for parametric curves we use

ds −ÎS

dtD

dx

2

1S

dtD

dy

2

dt

Example 6 Show that the surface area of a sphere of radius r is 4r 2 .

SOLUTION The sphere is obtained by rotating the semicircle

x − r cos t y − r sin t 0 < t <

about the x-axis. Therefore, from Formula 6, we get

S − y

2r sin t ss2r sin 0 td2 1 sr cos td 2 dt

− 2 y

r sin t sr 2 ssin 2 t 1 cos 2 td dt − 2 y

r sin t ? r dt

0 0

− 2r 2 y

0 sin t dt − 2r 2 s2cos tdg 0

− 4r

2

n

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