CK-12-Calculus

http://www.ck12.org Chapter 5. Applications of Definite Integrals

S= 2 π

∫ 145

1 u

1 / 2 du

36

=^236 π

[ 2

3 u

3 / 2

] 145

1

=^236 π·^23

[

( 145 )^3 /^2 − 1

]

≈ 1084 π[ 1745 ]

≈ 203

Example 2:
Find the area of the surface generated by revolving the graph off(x) =x^2 on the interval[ 0 ,√ 3 ]about they−axis
(Figure 22).
Solution:

Figure 22
Since the curve is revolved about they−axis, we apply

S=

∫d

c^2 πx

√

1 +

(dx

dy

) 2

dy.

So we writey=x^2 asx=√y. In addition, the interval on thex−axis[ 0 ,

√

3 ]becomes[ 0 , 3 ].Thus

S=

∫ 3

0 2 π

√y

√

1 +

( 1

2 √y

) 2

dy.