CK-12-Calculus

5.4. Area of a Surface of Revolution http://www.ck12.org

Simplifying,

S=π

∫ 3

0

√

4 y+ 1 dy.

With the aid ofu−substitution, letu= 4 y+ 1 ,

S=π 4

∫ 13

1 u

1 / (^2) du
=π 6

[

( 13 )^3 /^2 − 1

]

=π 6 [ 46. 88 − 1 ]

≈ 24

Multimedia Links

For video presentations of finding the surface area of revolution(16.0), see Math Video Tutorials by James Sousa,
Surface Area of Revolution, Part 1 (9:47)

MEDIA

Click image to the left for use the URL below.

URL: http://www.ck12.org/flx/render/embeddedobject/576

and Math Video Tutorials by James Sousa, Surface Area of Revolution, Part 2 (5:43).

MEDIA

Click image to the left for use the URL below.

URL: http://www.ck12.org/flx/render/embeddedobject/577

Review Questions

In problems #1 - 3 find the area of the surface generated by revolving the curve about thex−axis.

1.y= 3 x, 0 ≤x≤ 1

2.y=√x, 1 ≤x≤ 9

3.y=

√

4 −x^2 ,− 1 ≤x≤ 1