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.