How do you find the volume of the solid obtained by rotating the region bounded by $y = x$ $y=x$ and $y = x^{2}$ $y=x^2$ about the line $x = - 1$ $x=-1$ ?

Question

16300 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-10-25T05:56:51

By Shell Method,

$V = 2 π \int_{0}^{1} (x + 1) (x - x^{2}) d x = 2 π \int_{0}^{1} (x - x^{3}) d x$ $V=2pi int_0^1(x+1)(x-x^2)dx=2pi int_0^1(x-x^3)dx$

$= 2 π {[\frac{x^{2}}{2} - \frac{x^{4}}{4}]}_{0}^{1} = 2 π (\frac{1}{2} - \frac{1}{4}) = \frac{π}{2}$ $=2pi[x^2/2-x^4/4]_0^1=2pi(1/2-1/4)=pi/2$

I hope that this was helpful.

How do you find the volume of the solid obtained by rotating the region bounded by y=xy=x and y=x2y=x^2 about the line x=−1x=-1?