How to find the volume when a curve is revolved around the y-axis

Key takeaway: When a region is turned around the y-axis, the shell method is usually the cleanest route. Here the curve, the vertical boundary, and the x-axis fit that setup nicely.

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

1) Choose the shell method

Because the axis of rotation is the y-axis, vertical shells are the natural choice.

For a shell at position $x$:

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

So the volume is

$$V = 2\pi\int_{0}^{4} x\sqrt{x}\,dx$$

2) Simplify the integrand

$$x\sqrt{x}=x^{3/2}$$

So

$$V = 2\pi\int_{0}^{4} x^{3/2}\,dx$$

3) Integrate

$$\int x^{3/2}\,dx = \frac{2}{5}x^{5/2}$$

Therefore,

$$V = 2\pi\left[\frac{2}{5}x^{5/2}\right]_{0}^{4} = \frac{4\pi}{5}\cdot 4^{5/2}$$

Since

$$4^{5/2} = (\sqrt{4})^5 = 2^5 = 32,$$

we get

$$V = \frac{4\pi}{5}(32)=\frac{128\pi}{5}$$

Correct choice: A.

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Pitfalls the pros know 👇 A common mistake is to use the washer method with respect to $y$ without rewriting the curve first. That usually makes the setup messier than it needs to be. Another easy slip is forgetting that the shell radius is $x$, not $\sqrt{x}$.

What if the problem changes? If the axis of rotation changed to the x-axis, the setup would switch to washers/disks instead of shells. If the boundary $x=4$ were replaced by another vertical line $x=a$, the upper limit would simply become $a$ and the same shell structure would still work.

Tags: shell method, volume of revolution, y-axis rotation