Find the surface area of the solid of revolution obtained by rotating the curve x=(2y−y^2)^(1/2) from y=8/16 to y=31/16 about the y-axis?

1 Answer
May 25, 2018

#color(blue)[S_A=2piint_(8/16)^(31/16)(2y−y^2)^(1/2)*sqrt(1+((2-2y)^2)/((2y−y^2)(4)))*dy=(46pi)/16]#

Explanation:

The surface area of solid of revolution around #"y-axis"# is given by:

#color(red)[S_A=2piint_a^bx*sqrt(1+(x')^2)*dy#

The interval of integral is #y in [8/16,31/16]#

now lets setup the integral of surface area

#S_A=2piint_(8/16)^(31/16)((2y−y^2)^(1/2))*sqrt(1+(1/2(2y-y^2)^(-1/2)(2-2y))^2)*dy#

#S_A=2piint_(8/16)^(31/16)(2y−y^2)^(1/2)*sqrt(1+((2-2y)^2)/((2y−y^2)(4)))*dy#

After simplified it:

#=2piint_(8/16)^(31/16)[(i*sqrt(-(y-2)*y))/sqrt((y-2)*y)]*dy#

#=2piint_(8/16)^(31/16)i*sqrt(-1)*dy=2piint_(8/16)^(31/16)i^2*dy#

#=2piint_(8/16)^(31/16)1*dy=2pi[y]_(8/16)^(31/16)#

#=2pi[(31/16)-(8/16)]=(46pi)/16#

Note that

#i^2=1#