volume of a solid

**gbenguse78** · December 27th 2010, 08:40 AM

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

**Prove It** · December 27th 2010, 08:52 AM

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

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

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

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

So the integral you are evaluating is

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

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