How do you use the shell method to compute the volume of the solid obtained by rotating the region in the first quadrant enclosed by the graphs of the functions y=x^2 and y=2 rotated about the y-axis?

Feb 22, 2015

We will rotate the area bounded by the two curves and the y-axis. In other words we will restrict ourselves to the region in the first quadrant.

Since we a rotating around the y axis using the method of shells
we integrate with respect to x.

Now find where the curves intersect.

${x}^{2} = 2$

$x = \pm \sqrt{2}$ We use positive square root since we are in quadrant I

Therefore the interval over which we integrate is $0 \le x \le \sqrt{2}$

Our representative radius will be some value of $x$ over this interval.

Our representative cylinder height is $2 - {x}^{2}$.

Using method of shells, the integral for the volume is

$2 \pi {\int}_{0}^{\sqrt{2}} x \left(2 - {x}^{2}\right) \mathrm{dx}$

$2 \pi {\int}_{0}^{\sqrt{2}} 2 x - {x}^{3} \mathrm{dx}$

Now integrate

$2 \pi \left({x}^{2} - {x}^{4} / 4\right)$

$2 \pi \left({\left(\sqrt{2}\right)}^{2} - \frac{{\left(\sqrt{2}\right)}^{4}}{4} - 0\right)$

$2 \pi \left(2 - 1\right) = 2 \pi \left(1\right) = 2 \pi$