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I just recently learned about volumes of solids of revolution in my AP Calculus class and tried to create a problem to connect it to related rates. In this process, I found an error that neither my teacher or I seem to be able to figure out. It goes as follows:

The region contained by the function $y=\sin(x) + 2$, the line $x=2$, and the $y$-axis as shown below is revolved around the $x$-axis.

Example Image

We find the volume of the solid created by the disk method as follows:

$$V=\pi r^2 * \text{thickness}$$

$$r=\sin(x) + 2$$

$$V=\int_{0}^{2\pi} (\sin(x)+2)^2dx$$

$$V=9\pi ^ 2$$

We also know that $\int_{0}^{2\pi} \sin(x)dx=0$ therefore $\int_{0}^{2\pi}2dx=\int_{0}^{2\pi}(\sin(x)+2)dx$ so if we revolve a new region with the line $y=2$ in place of $y=\sin(x) + 2$ : The volume of the two solids should be the same.

Example Image

But we know that revolving the above solid around the $x$-axis would create a cylinder with volume $A=\pi r^2h$. In this problem, $r=2$, and $h=2$, so $V=8\pi^2$. Why are the volumes not equivalent?

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  • $\begingroup$ Why should the integral between 0 and 2 of the sine be zero? Should it be $2 \pi$? $\endgroup$ – Lonidard Dec 9 '15 at 22:03
  • $\begingroup$ There are several pieces of your post that aren't rendering correctly, so it's difficult to say what your error may be. $\endgroup$ – Cameron Buie Dec 9 '15 at 22:06
  • $\begingroup$ I fixed the formulas, that should be better, sorry, this is the first time I'm posting $\endgroup$ – ozay34 Dec 9 '15 at 22:10
  • $\begingroup$ Since when is $$\int_{0}^{2 \pi} 2 dx = 0$$ ? $\endgroup$ – Mattos Dec 9 '15 at 22:12
  • $\begingroup$ sorry, I fixed it $\endgroup$ – ozay34 Dec 9 '15 at 22:13
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Hint:

The volume generated by the first half of the sine function (over $y=2$) is not the same as the volume generated by the second half (below $y=2$) because the radii of rotation are not the same.

In other words the same area generate different volumes if it is rotated with respect an axis with different radii of rotation.

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One possible source of error is this: you are quite right that $$\int_0^{2\pi}2\,dx=\int_0^{2\pi}(2+\sin x)\,dx,$$ but this is not what you're dealing with! Rather, your integrand is $$(2+\sin x)^2=4+4\sin x+(\sin x)^2,$$ and so your volume is $$\pi\int_0^{2\pi}(2+\sin x)^2\,dx=4\pi\int_0^{2\pi}\,dx+\pi\int_0^{2\pi}(\sin x)^2\,dx.$$

Can you take it from there?


Another error is the assumption that the area of the solid of rotation $y=2$ is the same as the solid of rotation of $y=2+\sin x.$ It's true that the average radius is the same. However, the average cross-sectional area is not the same. (In general, the average of squared is not the square of the average.)

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  • $\begingroup$ but even if the average cross sectional area isn't the same for $y=2+sinx$, the added part for the first $\pi$ is compensated for by the second part from $\pi$ to $2\pi$, or so I thought. But as Emilio said, that isn't equivalent. I don't understand how that statement is true though because $\int_{0}^{2\pi}sinxdx=0$ $\endgroup$ – ozay34 Dec 9 '15 at 22:28
  • $\begingroup$ Ok, thinking about it, I understand what Emilio is talking about, but I guess because I cant visualize it, it seems like it should work $\endgroup$ – ozay34 Dec 9 '15 at 22:31
  • $\begingroup$ The kicker is that if $0<t\le1,$ then $$(2+t)^2-2^2=4+4t+t^2-4=4t+t^2$$ and $$2^2-(2-t)^2=4-(4-4t+t^2)=4-4+4t-t^2=4t-t^2.$$ Hence, $$(2-t)^2-2^2>2^2-(2-t)^2,$$ even though $$2+t-2=2-(2-t).$$ $\endgroup$ – Cameron Buie Dec 9 '15 at 22:39
  • $\begingroup$ Translated into terms of volumes, the section between $x=0$ and $x=\pi$ has more extra volume (compared to the cylinder) than the section between $x=\pi$ and $x=2\pi$ loses (compared to the cylinder). $\endgroup$ – Cameron Buie Dec 9 '15 at 22:42

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