# How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region x=y-y^2 and the y axis rotated about the y axis?

Oct 25, 2015

See the explanation section, below.

#### Explanation:

Here is a picture of the region. Because we have been asked to ue shells, a representative slice has been taken parallel to the axis of revolution. This make the thickness of our slice $\mathrm{dx}$ The volume of the representative shell is:

$2 \pi \times \text{radius"xx"height"xx"thickness}$

As mentioned, $\text{thickness} = \mathrm{dx}$
and we can see that $\text{radius} = x$

The $\text{height}$ will be the greater $y$ value minus the lesser $y$ value. We are working in terms of $x$, so we need to write these two $y$ values as functions of $x$.

We need to solve $x = y - {y}^{2}$ for $y$ (in terms of $x$).

There are a couple of ways of doing this. (1) complete the square or (2) use the quadratic formula to solve ${y}^{2} - y + x = 0$ for $y$.
(There are other ways as well. For example, you can use the vertex formula to write the equation in standard form for a sideways opening parabola.)

I'll complete the square.

${y}^{2} - y = - x$

${y}^{2} - y + \frac{1}{4} = \frac{1}{4} - x$

${\left(y - \frac{1}{2}\right)}^{2} = \frac{1 - 4 x}{4}$

$y - \frac{1}{2} = \pm \sqrt{\frac{1 - 4 x}{4}}$

$y = \frac{1}{2} \pm \frac{\sqrt{1 - 4 x}}{2} = \frac{1 \pm \sqrt{1 - 4 x}}{2}$

(Yes, this is the same as the answer you'll get by using the quadratic formula.)

The greater $y$ (the one on top) is
${y}_{\text{top}} = \frac{1 + \sqrt{1 - 4 x}}{2}$

and the lesser $y$ (the one on the bottom) is
${y}_{\text{bottom}} = \frac{1 - \sqrt{1 - 4 x}}{2}$

So, finally, we can write the volume of the representative cylindrical shell:

$2 \pi \times \text{radius"xx"height"xx"thickness} = 2 \pi x \left(\frac{1 + \sqrt{1 - 4 x}}{2} - \frac{1 - \sqrt{1 - 4 x}}{2}\right) \mathrm{dx}$

Don't Panic. We can simplify this to get

$2 \pi x \sqrt{1 - 4 x} \mathrm{dx}$

We still haven't found the bounds on $x$.

In the region we have $0 \le x$, and, examining the solution for $y$, we see that if $x > \frac{1}{4}$ we'll get imaginary values for $y$. So, we get $x$ varies from $0$ to $\frac{1}{4}$.

The volume of the solid of interest is

$V = {\int}_{0}^{\frac{1}{4}} 2 \pi x \sqrt{1 - 4 x} \mathrm{dx}$

Which can be evaluated by parts or by the substitution $u = 1 - 4 x$ so that $\mathrm{du} = - 4 \mathrm{dx}$ and $x = \frac{1}{4} \left(1 - u\right)$.