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

1 Answer
Sep 5, 2014

First of all, you are missing a bound. We will assume that the other bound is #y=0# or the #x#-axis. The answer is #(15pi)/2#.

The first step is to determine whether you are rotating along an axis that is parallel to the independent axis or the axis of the parameter (#x# in this case). And we are not, so this integration should be done with cylindrical shells.

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Always draw a diagram to verify what is the parameter and what is the function.

You should note that #y# is not always a parameter of #x#. For instance, #x=y^2#, #x# is now a parameter of #y#.

The formula for cylindrical shells is:

#V=int_a^b2pirhdr#
#h# is represented by #y#, we have #y=x^2# and #y=0#
#r# is represented by #x#
#V=int_1^2 2 pi x (x^2-0) dx#

Now that the substitutions are done, we can solve:

#V=2 pi int_1^2x^3dx#
#=2pi (x^4)/4|_1^2#
#=2pi([2^4-1^4])/4#
#=(15pi)/2#