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

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Answer 1 · 2015-06-19T14:49:50

To use shells, we take our representative slices parallel to the line we are revolving around. So the thickness will be $d y$ $dy$ , the radius will be $y$ $y$ .
The height will be the greater $x$ $x$ (the one on the right) minus the lesser $x$ $x$ (the one on the left). We need both curves with $x$ $x$ as a function of $y$ $y$ .
The limits of integration will be $y$ $y$ values. That won't matter in this case, because the curves intersect at $(0, 0)$ $(0,0)$ and $(1, 1)$ $(1,1)$ .

The two curves are: $y = x^{2}$ $y=x^2$ and $x = y^{2}$ $x=y^2$

Solving the first equation for $x$ $x$ , we get $x = \pm \sqrt{y}$ $x=+-sqrty$ which does not give $x$ $x$ as a function of $y$ $y$ (functions don't say "or"). Don't panic. A quick sketch of the curves shows us that the bounded region has only positive $x$ $x$ values, so we will use the function: $x = \sqrt{y}$ $x=sqrty$ .
The second equation already gives $x$ $x$ as a function of $y$ $y$ .
Looking again at our sketch of the region, we see that the curve $x = \sqrt{y}$ $x=sqrty$ is on the right and $x = y^{2}$ $x=y^2$ is on the left.
The height is $\sqrt{y} - y^{2}$ $sqrty-y^2$

The volume of the representative shell is:
$2 π r h \cdot thickness = 2 π y (\sqrt{y} - y^{2}) d y$ $2 pi r h * "thickness" = 2 pi y (sqrty-y^2) dy$
And the limits of the region are $y = 0$ $y=0$ to $y = 1$ $y=1$ .

$V = \int_{0}^{1} 2 π y (\sqrt{y} - y^{2}) d y = 2 π \int_{0}^{1} (y^{\frac{3}{2}} - y^{3}) d y$ $V = int_0^1 2 pi y (sqrty-y^2) dy = 2 pi int_0^1 (y^(3/2)-y^3) dy$

I am sure that you can finish from here.

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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 $y = x^{2}$ $y=x^2$ and $y^{2} = x$ $y^2=x$ rotated about the x-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 y=x^2y=x2y=x^2 and y^2=xy2=xy^2=x rotated about the x-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 $y = x^{2}$ $y=x^2$ and $y^{2} = x$ $y^2=x$ rotated about the x-axis?