# Cylindrical shells – last example from today’s class

As soon as class ended today, I realized that we had forgotten something important when setting up the integral that computed the volume of the solid obtained by rotating the region between $y = \sqrt{x}$ and the $x$-axis, between $x=0$ and $x=4$, about the $x$-axis using the method of cylindrical shells. I’ll review the set-up here, and add the part I forgot.

Since we’re rotating about a horizontal line (the $x$-axis) and we’re going to use the method of cylindrical shells, that means we’ll integrate with respect to the vertical coordinate $y$. So the first part of the set-up is to write: $\int \dots dy$.

This means that our integrand has to be given in terms of the variable $y$. The integrand is the area of the infinitely thin cylindrical shell that you get from rotating a horizontal segment at height $y$ about the $x$-axis: $\int$ (area of cylindrical shell) $dy$.

When you cut open this infinitely thin cylindrical shell, you just get a rectangle whose area is its length times its width. Its length is the circumference of the cylinder; its width is the length of that horizontal segment at height $y$. The radius of the cylindrical shell is just the height of the horizontal segment at height $y$, so the circumference of the cylindrical shell is $2 \pi y$. We can add this to our integral above: $\int 2 \pi y$*(length of horizontal segment) $dy$.

Now we just have to figure out the length of that horizontal segment at height $y$. Since it’s horizontal, its length will just be the difference of the $x$-coordinates of its endpoints. The endpoint at the far right will be the same no matter what $y$ is. Here, the right-hand boundary of the region is the line $x=4$, so the right endpoint of the horizontal segment has $x$-coordinate 4. The left endpoint of the horizontal segment will depend on what $y$ is, but it will always be a point on the graph of $y = \sqrt{x}$. So the left endpoint of the horizontal segment at height $y$ is just $x$. This means that the length of the horizontal segment at height $y$ is $4-x$…but our integrand can’t have the variable $x$ in it…we’ve set it up so that we’re integrating with respect to $y$. Since $y = \sqrt{x}$, $y^2 = x$. We just replace the $x$ we see in the formula for the length of the segment with $y^2$. So the length of the horizontal segment at height $y$ is $4-y^2$: $\int 2 \pi y * (4-y^2) dy$.

This is the integral we set up in class. It will not compute the volume of the solid because it’s an indefinite integral and not a definite one. We forgot to add limits of integration. Remember that we’re integrating with respect to $y$, which plays the role of the radius of the cylindrical shell. We know that $x$ goes from 0 to 2, and we know that $y = \sqrt{x}$, so this means that the radius $y$ goes from $\sqrt{0}$ to $\sqrt{4}$. This means that the definite integral that computes the volume is: $V = \int_0^2 2 \pi y * (4-y^2) dy$.

Here’s what I’d do to evaluate this integral: $\int_0^2 2 \pi y * (4-y^2) dy$ $= 2\pi \int_0^2(4y - y^3)dy$ $= 2 \pi (2y^2-\frac{1}{4}y^4)]_0^2$ $= 2 \pi((2*2^2 - \frac{1}{4}2^4) - (2*0^2 - \frac{1}{4}0^4))$ $= 8 \pi$.

Therefore the volume is $8\pi$.

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