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^2x=yy2 and the y axis rotated about the y axis?

1 Answer
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 dxdx

enter image source here

The volume of the representative shell is:

2 pixx "radius"xx"height"xx"thickness"2π×radius×height×thickness

As mentioned, "thickness" = dxthickness=dx
and we can see that "radius"=xradius=x

The "height"height will be the greater yy value minus the lesser yy value. We are working in terms of xx, so we need to write these two yy values as functions of xx.

We need to solve x=y-y^2x=yy2 for yy (in terms of xx).

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=0y2y+x=0 for yy.
(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 = -xy2y=x

y^2-y +1/4 = 1/4-xy2y+14=14x

(y-1/2)^2 = (1-4x)/4(y12)2=14x4

y-1/2 = +-sqrt((1-4x)/4)y12=±14x4

y = 1/2+-sqrt(1-4x)/2 = (1+-sqrt(1-4x))/2 y=12±14x2=1±14x2

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

The greater yy (the one on top) is
y_"top" = (1+sqrt(1-4x))/2ytop=1+14x2

and the lesser yy (the one on the bottom) is
y_"bottom" = (1-sqrt(1-4x))/2ybottom=114x2

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

2 pixx "radius"xx"height"xx"thickness" = 2pix((1+sqrt(1-4x))/2-(1-sqrt(1-4x))/2) dx2π×radius×height×thickness=2πx(1+14x2114x2)dx

Don't Panic. We can simplify this to get

2pixsqrt(1-4x)dx2πx14xdx

We still haven't found the bounds on xx.

In the region we have 0 <= x0x, and, examining the solution for yy, we see that if x > 1/4x>14 we'll get imaginary values for yy. So, we get xx varies from 00 to 1/414.

The volume of the solid of interest is

V = int_0^(1/4) 2 pi x sqrt(1-4x) dxV=1402πx14xdx

Which can be evaluated by parts or by the substitution u = 1-4xu=14x so that du = -4 dxdu=4dx and x = 1/4(1-u)x=14(1u).