# Thread: volume of a solid

1. ## volume of a solid

hi
i cant seem to make head or tail of this question. here it goes;
find the volume of the solid generated by rotating the area enclosed by the curve y= 2-x^2 and the line y = 1, about y=1.
i am not sure how to start. can someone please explain?
thanks

2. Start by drawing the region. You should find that your area of integration is the "top" of the parabola, from $\displaystyle \displaystyle y = 1$ upwards and between $\displaystyle \displaystyle x = -1$ and $\displaystyle \displaystyle x = 1$.

Now, if you were going to rotate this region about the line $\displaystyle \displaystyle y = 1$, you would have circular cross sections that have the same radius as the height of that region, so $\displaystyle \displaystyle r = 2 - x^2 - 1 = 1 - x^2$.

The area of each circular cross section is $\displaystyle \displaystyle \pi r^2 = \pi(1 - x^2)^2 = \pi(1 - 2x^2 + x^4)$.

The volume will be the sum of all these cross sections over the region $\displaystyle \displaystyle x = -1$ and $\displaystyle \displaystyle x = 1$. This sum converges on an integral.

So the integral you are evaluating is

$\displaystyle \displaystyle V = \int_{-1}^1{\pi(1 - 2x^2 + x^4)\,dx}$.