How do you find the surface area of the solid obtained by rotating about the #x#-axis the region bounded by #9x=y^2+18# on the interval #2<=x<=6# ?

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Answer 1 · 2014-09-20T13:34:26

The surface area can be found by

#S=2piint_0^6 ysqrt{1+{4y^2}/81}dy=49pi#

Let us look at some details.

#9x=y^2+18 Leftrightarrow x=y^2/9+2 Leftrightarrow y=pm3sqrt{x-2}#

By taking the derivative,

#dx/dy={2x}/9#

As x goes from 2 to 6, y goes from 0 to 6.

Since the surface area of revolution can be found by

#S=2piint_a^b y\sqrt{1+(dx/dy)^2}dy#,

we have

#S=2piint_0^6 ysqrt{1+{4y^2}/81}dy#

by substitution #u=1+{4y^2}/81#. #Rightarrow {du}/dy={8y}/81 Rightarrow dy=81/{8y}du#,
As y goes from 0 to 6, u goes from 1 to 25/9.

#=2pi int_1^{25/9}ysqrt{u}{81}/{8y}du#

by simplifying,

#={81pi}/4 int_1^{25/9}u^{1/2}du#

#={81pi}/4[2/3u^{3/2}]_1^{25/9}#

#={27pi}/2[(25/9)^{3/2}-1]=49pi#