Use cylindrical shells to find the volume of the solid. The region bounded $y = x^2$ and $y = 1$ is revolved around the line $y = 3$.
The answer is: $$\int_0^1 4\pi(3-y)y^{1/2}dy$$
Why is it $4\pi$? I understand the formula to be $\int 2πy(f(y))dy$.
2026-03-26 01:26:46.1774488406
How to fine the formula for the volume using cylindrical shells?
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The $4\pi$ that you mentioned comes from $2\pi\times 2$. The first $2\pi$ is from $2\pi(3-y)$ which is the circumference of the disk with radius $3-y$. The second $2$ comes from the height of the cylinder which is $2x = 2\sqrt{y}$. And if you put these together you should arrive at the same integral given by the answer that you had above.