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Question: How do I construct the integral to compute the volume of a spherical "shell" via rotation of an annulus about the $x$-axis.

The set up: I have an annulus, $A$ which is centered at the origin of $\mathbb{R}^2$ and has an outer radius equal to $1$ and inner radius equal to $1/2.$ I want to spin $A$ all the way around the $x$-axis and construct a spherical shell and write an integral to compute its volume. I am not certain how to do this. Note the integral set up is more important than the actual computation of the volume.

Very similiar Question

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Shell method around the x axis, $V = \int_0^r 2(2\pi)\text{(shell radius)(shell height)} \ dy$

So for this problem $$\int_0^1 4\pi\cdot y\sqrt{1-y^2} \ dy - \int_0^{0.5} 4\pi\cdot y\sqrt{0.25 - y^2} \ dy$$

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You already know (it's in the question you linked) that the volume of a solid obtained by rotating a segment of a curve $y=f(x)$ around the $x$-axis is: $$ \int_a^b \pi\, f(x)^2\,dx $$ where $(a,b)$ is the interval on which your segment of $f(x)$ is defined.

For example, consider $f(x)=\sqrt{r^2-x^2}$, its graph is half of a circumference of radius $r$ and centered in the origin. Rotating that around the $x$-axis gives you a sphere, whose volume is: $$ \int_{-r}^r \pi (r^2-x^2)\, dx $$

Now, a spherical shell is just a bigger sphere "minus" a smaller one...

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