How to find the volume of a region revolved around the y-axis using shells

Key takeaway: Same geometry, same method, just written cleanly. When the region is pinned by a vertical line and the y-axis is the center of rotation, shells keep things simple.

We revolve the region bounded by $y=\sqrt{x}$, $x=4$, and $y=0$ about the y-axis.

1) Set up the shell method

A vertical slice at $x$ becomes a shell.

• radius = $x$
• height = $\sqrt{x}$
• thickness = $dx$

Thus,

$$V=2\pi\int_0^4 x\sqrt{x}\,dx$$

2) Simplify and integrate

$$V=2\pi\int_0^4 x^{3/2}\,dx$$ $$V=2\pi\left[\frac{2}{5}x^{5/2}\right]_0^4$$ $$V=\frac{4\pi}{5}\cdot 4^{5/2} =\frac{4\pi}{5}\cdot 32$$ $$V=\frac{128\pi}{5}$$

Correct choice: A.

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Pitfalls the pros know 👇 The biggest mistake is plugging in the wrong radius. For shells about the y-axis, the radius is the x-value itself, not the y-value of the curve. Also, the top of each shell is $\sqrt{x}$, not $x^2$.

What if the problem changes? If the region were revolved about the x-axis instead, you would use disks and get a different integral. If the boundary were $x=a$ rather than $x=4$, the same method gives $V=2\pi\int_0^a x\sqrt{x}\,dx$.

Tags: shell method, volume of revolution, vertical slices