Volumes of solid of revolution: sin(x) + 2 fake proof

Question

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?

Answer 1

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.

Answer 2

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?

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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.)