What is the surface area produced by rotating #f(x)=x/pi^2, x in [-3,3]# around the x-axis?

2 Answers
May 26, 2018

#color(blue)[S_A=2pi int_-3^3 x/pi^2*sqrt[1+(1/pi^2)^2]*dx==(sqrt(pi^2+1)*9)/pi^3-(sqrt(pi^2+1)*9)/pi^3=0]#

Explanation:

If the solid is obtained by rotating the graph of #y=f(x)# from #x=a# to #x=b# around the #"x-axis"# then the surface area #S_A# can be found by the integral

#S_A=2pi int_a^b f(x)sqrt{1+[f'(x)]^2}dx#

#S_A=2pi int_-3^3 x/pi^2*sqrt[1+(1/pi^2)^2]*dx#

#S_A=2 int_-3^3 x/pi*sqrt[1+(1/pi^4)]*dx#

#S_A=2 int_-3^3 x/pi*sqrt[(1+pi^2)/pi^4)*dx#

#S_A=2/pisqrt[(1+pi^2)/pi^4) int_-3^3 x*dx=[(sqrt(pi^2+1)*x^2)/pi^3]_-3^3#

#=(sqrt(pi^2+1)*9)/pi^3-(sqrt(pi^2+1)*9)/pi^3=0#

May 26, 2018

See below

Explanation:

Have erased answer.