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Rotate $y = x^3$ and $y = 8$, $x = 0$

Using the method of cylindrical shells, which is $2\pi rh \, dx$, or $2\pi xf(x) \, dx$

$$x^3 = 8$$ $$x = 2$$

$$2\pi \int_0^2 x(8-x^3) \, dx?$$

I have tried many set ups and I cannot find the correct answer. $(768\pi / 7)$

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up vote 4 down vote accepted

Drawing a picture is always a good thing to do for these types of problems.

Start by sketching the graphs of $y=8$, $y=x^3$, and $x=0$. We then see that the region to be revolved about the $x$-axis is in the first quadrant. It is bounded above by the line $y=8$, on the left by the $y$-axis, and on the right by the graph of $y=x^3$. Note that the point of intersection in the upper right is $(2,8)$ and can be found by solving the equation $8=x^3$.

The region is shown, shaded in blue, below:

enter image description here

Now, you generate the solid by revolving this region about the $x$-axis. The cylindrical shells are generated by revolving a horizontal line segment, shown in green in the diagram, at a fixed $\color{maroon}y$-value about the $x$-axis.

These line segments "start" at $y=0$ and "end" at $y=8$. Thus the integral giving the volume of the solid of revolution is with respect to $y$ and is of the form $$\int_{y=0}^{y=8} 2\pi \,\color{maroon}{ r_y} \cdot\color{darkgreen}{ h_y} \, dy$$ where $\color{maroon}{r_y}$ is the radius of the shell at $\color{maroon}y$ and $\color{darkgreen}{h_y}$ is the height of the shell at $\color{maroon}y$.

The height of the shell at $\color{maroon}y$ is the length of the line segment at $\color{maroon}y$. Since the length of the line segment at $\color{maroon}y$ is the $x$-coordinate of its right hand endpoint, we have

$\ \ \ \ \ \ \ \color{darkgreen}{h_y}=\color{darkgreen}{y^{1/3}}$.

Keep in mind that we want to write $\color{darkgreen}{h_y}$ in terms of $y$, since the integral is with respect to $y$.

The radius of the shell is the height above the $x$-axis of the line segment:

$\ \ \ \ \ \ \ \color{maroon}{r_y}=\color{maroon}y$.

Thus, the volume of the solid of revolution is: $$ \int_0^8 2\pi\,\color{maroon}y\cdot\color{darkgreen}{ y^{1/3}} \, dy =\int_0^8 2\pi\cdot y^{4/3} \, dy= \textstyle{6\pi\over7} y^{7/3}\Bigr|_0^8={6\pi\over7}\cdot 8^{7/3} ={6\pi\over7} \cdot2^{7 }= {6\pi\over7} \cdot128={768\pi\over 7}. $$

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You blew my mind and paved the path of conceptual understanding for me. Thanks. – Jonathan Dewein Apr 30 '12 at 18:37

If you rotate around the $x$ axis, your cylindrical shells have their axes parallel to $x$, so the area element that you rotate is from $y$ to $y+dy$ and from $x=0$ to $x=\sqrt[3]y$. The volume element is then $2 \pi y \sqrt[3]y dy$. If you integrate this from $0$ to $8$,

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