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Here is the problem in my textbook:

Find the volume of the solid obtained by rotating the region bounded by the curves $y=x, y=x^2$ about x-axis.

Here is my solution :

Because equation $x = x^2$ has two roots : $0$ and $1$. we have:

$$ V= \int_0^1{2\pi x(x^2-x)}dx = \frac{\pi}{6}$$

But the solution in my textbook is $\frac{2\pi}{15}$. I think the hole in my solution is : I haven't use the region rotate around x-axis yet. But, I don't know how to use this statement in solution when couting volume.

Thanks :)

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4 Answers 4

up vote 4 down vote accepted

Consider the diagram

enter image description here

For your problem, this is equivalent to rotating $R_2$ about the line $y=0$.

Since neither of the curves are touching the axis we are rotating about, we must use washers.

For washers, the volume is given by $$ V = \pi\int^b_a (R(x))^2-(r(x))^2 \ dx$$

Where $R(x)$ is the curve farthest away from our axis of rotation - the top function $y = x$ and $r(x)$ is the curve closest to the axis of rotation - the bottom function $y=x^2.$

As you found the intersecting points, we can set up our integral as

$$V = \pi\int^1_0 (x)^2-(x^2)^2 \ dx$$

$$V = \pi\int^1_0 x^2-x^4 \ dx$$

After integrating and evaluating, you should get $\frac{2\pi}{15}$.

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Your formula should be $$V=\pi\int_0^1x^2-(x^2)^2\,dx,$$ which will give you the correct answer. You were trying to use the cylinder method (which would be appropriate if rotating about the $y$-axis), when you should have been using the washer method (since we are integrating along the axis of rotation, rather than perpendicular to it).

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In this problem, it is probably easiest to use the method of slicing. However, if you are going to use the "cylindrical shells" method, while rotating about the $x$-axis, you have to integrate with respect to $y$.

Take a thin horizontal strip going from height $y$ to height $y+dy$. The strip is more or less at height $y$. It has length $\sqrt{y}-1$ (draw a picture). The volume of the shell obtained by rotating this strip is about $2\pi y(\sqrt{y}-y)\,dy$. Thus our volume is $$\int_{y=0}^1 2\pi y(\sqrt{y}-y)\,dy.$$ So we want $$2\pi\int_{y=0}^1 (y^{3/2}-y^2)\,dy.$$ Integrate. We get $2\pi\left(\frac{2}{5}-\frac{1}{3}\right)$.

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The volume of the solid bounded by $y= f(x)$ rotating about the $x$-axis is $V=\pi\int_a^b(f(x))^2dx$. That is, you want to calculate $\pi (\int_0^1x^2dx -\int_0^1x^4dx)$

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