Question

What is the surface area of the solid created by revolving `f(x)=1/x^2` for `x in [1,2]` around the x-axis?

Answer 1

`S~~ 4.46`

Explanation:

The first step is to check to make sure that `f(x)` is defined on the interval `[1, 2]`. The only discontinuity of the the function is at `x= 0`, so we do not have to concern ourselves with this.

The surface area of a curve on `[a, b]` rotated around the x-axis is defined by `S = 2piint_a^b f(x)sqrt(1 + (f'(x))^2)dx`. The derivative of `f(x)` is `f'(x) = -2/x^3`. This means that `(f'(x))^2 = (-2/x^3)^2 = 4/x^6`.

`S = 2piint_1^2 1/x^2sqrt(1 + 4/x^6)dx`

`S = 2piint_1^2 1/x^2sqrt((x^6 + 4)/x^6)dx`

`S = 2piint_1^2 1/x^5sqrt(x^6 + 4) dx`

This integral has no elementary solution.

But according to Wolfram Alpha, the surface area is approximately `4.46` square units.

Hopefully this helps!