Find the volume of the resulting solid

Question

We have exams upcoming and this is part of practise but i wanted hints on how to do this.

I'm trying to Find the volume of the resulting solid given the region bounded by the curves $y = x^2$ , $y = x$ and $x = 2$ is rotated about the $x$ - axis.

Answer 1

The key here is to break the region up into two parts:

1. for $0\le x \le 1$, $x \ge x^2$;
2. for $1 < x \le 2$, $x < x^2$.

Therefore, the volume of the resulting solid will be $$ V = \int_0^1 \pi[(x)^2 - (x^2)^2]dx + \int_1^2 \pi[(x^2)^2 - (x)^2]dx={2\pi \over 15} + {58\pi \over 15} = 4\pi.$$