Find the area of the surface generated by revolving the curve y = 2x-x, 0.25 sxs 1.25, about the x-axis. The area of the surface generated by revolving the curve y = /2x-x, 0.25 sxs 1.25, about the x-axis is square units. (Type an exact answer, using a as needed.)

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015
1st Edition
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
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Chapter9: Solving Quadratic Functions
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Find the area of the surface generated by revolving the curve y = 2x-x, 0.25 sxs 1.25, about the x-axis.
The area of the surface generated by revolving the curve y = 2x-x
, 0.25 sxs1.25, about the x-axis is
square units.
(Type an exact answer, using a as needed.)
Transcribed Image Text:Find the area of the surface generated by revolving the curve y = 2x-x, 0.25 sxs 1.25, about the x-axis. The area of the surface generated by revolving the curve y = 2x-x , 0.25 sxs1.25, about the x-axis is square units. (Type an exact answer, using a as needed.)
Expert Solution
Step 1

We have to find the area of the surface generated by revolving the curve

y=2x-x2, 0.25x1.25 about of the x-axis.

 

Area of the surface is given by

Area=2πabfx1+f'x2

y=2x-x2 

Differentiate 'y' with respect to 'x' we get

y'=122x-x22-2xy'=1-x2x-x2                 a=0.25 and b=1.25

 

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