Question

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

Answer 1

`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`

Answer 2

See below

Explanation:

Have erased answer.