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Using the shell method, find the volume of: $y = \sqrt x$ with bounds $[0,4]$ after it has been revolved around the y axis.

After plugging everything into the shell formula I get:

$2 \pi \int_0 ^4 x \sqrt x dx$

After evaluating everything I get 128$\pi$/5. My textbook has the correct answer as 32$\pi$/5.

Am I making an error or is shell method just impossible in this scenario? If it is not possible, please explain why, thanks!

(note, my book did not ask to solve using shell, it asked using disk method, I am just trying shell for fun.)

3 Answers 3

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Aha - this is a bit embarrassing to me. I did this problem maybe three times before I caught on to my error. Firstly, the correct answer is $\dfrac{32 \pi}{5}$. But the shell method does work in this case. But here's the catch:

We're integrating over the wrong volume!

square root graph

This is a picture of the graph we are revolving. We want the volume between the graph and the y-axis, but this shell method calculates the area between the graph and the x-axis (though still revolved around the y-axis). Does that make sense?

And we can quickly check this. Adding 128$\pi$/5 and 32$\pi$/5 we get 32 $\pi$, which is the volume of a disk with radius 4 and height 2. That's what we expect - as they're complementary volumes.

These things catch even an old hand sometimes.

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    I use a variety, and I would generally rely on http://www.onlinefunctiongrapher.com/. But I got lazy for this one, and I just googled a graph to be honest. But good luck!2011-07-10
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Your answer is correct, assuming that you're looking at the right region. You can check by trying it with washers:$V = \pi \int_0^2 (16 - y^4) dy = \pi\left(32 - \frac{32}{5} \right) = \frac{128\pi}{5}.$ This is for the region bounded by $y = \sqrt x$, $x = 4$, and $y = 0$. If, however, the intended region was above the curve, i.e., bounded by $y = \sqrt x$, $y = 2$, and $x = 0$, then the book's answer is right, and your setup is wrong: it should be $2\pi\int_0^4 (2 - x \sqrt x) dx.$

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You're making a mistake. Here's a hint: you're not using the correct "height" for the cylindrical shells. Draw a picture.

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    Ah, my height had to be (2-sqrt(x))! thanks2011-07-10