How do you find the area of the surface generated by rotating the curve about the y-axis #y=1/3(x^2+2)^(3/2), 1<=x<=2#?

1 Answer
Apr 18, 2017

The surface area of a curve #y# rotated around the #x#-axis on #alt=xlt=b# is given by:

#S=2piint_a^bysqrt(1+(y')^2)dx#

Where #y=1/3(x^2+2)^(3/2)# we see that #y'=1/3(3/2(x^2+2)^(1/2))(2x)=xsqrt(x^2+2)#.

Then the surface area is:

#S=2piint_1^2 1/3(x^2+2)^(3/2)sqrt(1+(xsqrt(x^2+2))^2)dx#

#S=(2pi)/3int_1^2(x^2+2)^(3/2)sqrt(1+x^2(x^2+2))dx#

Note that #sqrt(1+x^2(x^2+2))=sqrt(x^2+2x^2+1)=sqrt((x^2+1)^2)=x^2+1#:

#S=(2pi)/3int_1^2(x^2+2)^(3/2)(x^2+1)dxapprox69.9055#