How do you find the surface area of the solid obtained by rotating about the #x#-axis the region bounded by #x=1+2y^2# on the interval #1<=y<=2# ?

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Answer 1 · 2014-09-20T23:17:18

The surface area is #pi/24[(65)^{3/2}-(17)^{3/2}]#.

Let us look at some details.

#x=1+2y^2#

By differentiating with respect to #y#,

#{dx}/{dy}=4y#

So, we can find the surface area #S# by

#S=2pi int_1^2 ysqrt{1+({dx}/{dy})^2} dy#

#=2pi int_1^2 ysqrt{1+16y^2}dy#

by substitution #u=1+16y^2#.
#Rightarrow {du}/{dy}=32y Rightarrow{dy}/{du}=1/{32y} Rightarrow dy={du}/{32y}#

#y: 1 to 2 Rightarrow u: 17 to 65#

#=2pi int_{17}^{65}ysqrt{u}{du}/{32y}#

#=pi/16 int_{17}^{65}u^{1/2}du#

#=pi/16 [2/3u^{3/2}]_{17}^{65}#

#=pi/24[(65)^{3/2}-(17)^{3/2}]#