Use the shell method to find the volume of the solids generated by revolving the region bounded by the curve and line about the y axis? y=x,y=-x/2,x=2 y=x2,y=2-x,x=0 for x> or equal to 0

#y=x#
#y=-x/2#
#x=2#
#y=x^2#,
#y=2-x#,
#x>= 0#

1 Answer
Rachel
Mar 29, 2018

The shell method uses the formula #2pi int x*f(x) color(white)(.) dx# with #f(x)# being our function and #x# as our radius

#* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * #

#y=x#

#2pi int x * x color(white)(.)dx#

#2pi int x^2 color(white)(.) dx#

#2pi xx x^(2+1)/(2+1)# via power rule

#2pi xx x^3/3#

#(2 pi x^3)/3#

#* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * #

#y = -x/2#

#2pi intx * -x/2 color(white)(.)dx#

#(-2pi)/2 int x^2color(white)(.) dx#

#-pi intx^2color(white)(.)dx#

#(-pi x^3)/3#

#* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * #

#y = x^2#

#2pi int x * x^2 color(white)(.) dx#

#2pi int x^3 color(white)(.) dx#

#2pi xx x^4/4#

#(pix^4)/2#

#* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * #

#y = 2-x#

#2pi int x(2-x) color(white)(.) dx#

#2pi int 2x - x^2 color(white)(.) dx#

#2pi int2x color(white)(.) dx - 2pi int x^2 color(white)(.) dx#

#4pi intx color(white)(.) dx - (2pi x^3)/3#

#4pi x^2/2 - (2pix^3)/3#

#2pi x^2 - (2pix^3)/3#

#(3(2pix^2))/3 - (2pix^3)/3#

#(6pix^2 - 2pix^3)/3#

#* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * #

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