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Home Applications of the Integral Area of a Surface of Revolution Examples Example 1: Line Segment
 
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Example 1: Line Segment

The line segment y = 3x, from x = 1 to x = 4, is rotated about the y-axis (Figure 6.4.7). Find the area of the surface of revolution.

FIRST SOLUTION

We use the integration formula, dy/dx = 3, so

06_applications_of_the_integral-188.gif

06_applications_of_the_integral-190.gif

Figure 6.4.7

SECOND SOLUTION

This surface of revolution is a frustum of a cone, so the formula for the lateral area of a frustum can be used directly. From the diagram we see that the radii and slant height are:

r1 = 1,       r2 = 4, l = distance from (1,3) to (4, 12)

06_applications_of_the_integral-191.gif06_applications_of_the_integral-192.gif06_applications_of_the_integral-193.gif06_applications_of_the_integral-194.gif

Then

A = π(r1 + r2)l = π(1 + 4)306_applications_of_the_integral-195.gif= 15π06_applications_of_the_integral-196.gif
Home Applications of the Integral Area of a Surface of Revolution Examples Example 1: Line Segment

Last Update: 2006-11-22