Question

How do you find the surface area of the solid obtained by rotating about the `y`-axis the region bounded by `y=1-x^2` on the interval `0<=x<=1` ?

Answer 1

The surface area A of the solid obtained by rotating about the `y`-axis the region under the graph of `y=f(x)` from `x=a` to `b` can be found by

`A=2pi int_a^b x sqrt{1+[f(x)]^2}dx`.

Let us now look at the posted question.

By the formula above,

`A=2pi int_0^1 x sqrt{1+4x^2} dx`

by rewriting a bit,

`=pi/4 int_0^1 (1+4x^2)^{1/2}cdot8x dx`

by General Power Rule,

`=pi/4 [2/3(1+4x^2)^{3/2}]_0^1=pi/6(5^{3/2}-1)`

I hope that this was helpful.