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 $x = y - y^{2}$ $x=y-y^2$ and the y axis rotated about the y axis?

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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 $x = y - y^{2}$ $x=y-y^2$ and the y axis rotated about the y axis?

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Answer 1 · 2015-10-25T14:41:29

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 $d x$ $dx$

enter image source here

The volume of the representative shell is:

$2 π \times radius \times height \times thickness$ $2 pixx "radius"xx"height"xx"thickness"$

As mentioned, $thickness = d x$ $"thickness" = dx$
and we can see that $radius = x$ $"radius"=x$

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

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

There are a couple of ways of doing this. (1) complete the square or (2) use the quadratic formula to solve $y^{2} - y + x = 0$ $y^2-y+x=0$ for $y$ $y$ .
(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.

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

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

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

$y - \frac{1}{2} = \pm \sqrt{\frac{1 - 4 x}{4}}$ $y-1/2 = +-sqrt((1-4x)/4)$

$y = \frac{1}{2} \pm \frac{\sqrt{1 - 4 x}}{2} = \frac{1 \pm \sqrt{1 - 4 x}}{2}$ $y = 1/2+-sqrt(1-4x)/2 = (1+-sqrt(1-4x))/2$

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

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

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

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

$2 π \times radius \times height \times thickness = 2 π x (\frac{1 + \sqrt{1 - 4 x}}{2} - \frac{1 - \sqrt{1 - 4 x}}{2}) d x$ $2 pixx "radius"xx"height"xx"thickness" = 2pix((1+sqrt(1-4x))/2-(1-sqrt(1-4x))/2) dx$

Don't Panic. We can simplify this to get

$2 π x \sqrt{1 - 4 x} d x$ $2pixsqrt(1-4x)dx$

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

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

The volume of the solid of interest is

$V = \int_{0}^{\frac{1}{4}} 2 π x \sqrt{1 - 4 x} d x$ $V = int_0^(1/4) 2 pi x sqrt(1-4x) dx$

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

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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 $x = y - y^{2}$ $x=y-y^2$ and the y axis rotated about the y axis?

1 Answer

Explanation:

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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−y2x=y-y^2 and the y axis rotated about the y axis?

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Explanation:

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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 $x = y - y^{2}$ $x=y-y^2$ and the y axis rotated about the y axis?