I have been reading up on disk/washer, shell methods to find volume of solids of revolution, but I am having trouble with the following question:

We are being asked to find the volume of the following solid of revolution:

$R$ bounded by the graph of $y=x/2, y=3x, y=4$, revolved about the $y$-axis.

My thoughts so far: Use shell: $V=2\pi\int_a^b xf(x)dx$

I am unsure of how to visualize/set up the integral. Do I integrate with respect to $x$? Any help would be appreciated.

Respectfully, Jack

  • $\begingroup$ $y=1x/2$ means $y=\frac{x}{2}$? $\endgroup$ – Chee Han Jul 13 '18 at 22:06
  • $\begingroup$ Oh yes - I'm sorry. That was confusing. Fixed. $\endgroup$ – jackbenimbo Jul 13 '18 at 22:09

Visualization is an important feature of getting the integral set up correctly for volumes of revolution. The shell method is more complicated for this problem because the shell widths vary as differences between two sets of different functions. However, I'll proceed anyway with that method.

A shell method rotated about the $y$ axis will have shell thicknesses of $dx$ so we need to express the integral in terms of $x$. Each shell will be of height $4 - \frac{x}{2}$ from $x=\frac{4}{3}$ to $x=8$ (for the first limit of integration), and height $3x-\frac{x}{2}$ from $x=0$ to $x=\frac{4}{3}$ (for the second limit of integration).

The basic formula is:

$$\int_a^b 2\pi(\text{shell radius})(\text{shell height}) dx$$

We have two separate integrals to add to get the total volume. $$\int_{\frac{4}{3}}^8 2\pi(x)(4 - \frac{x}{2}) dx + \int_{0}^{\frac{4}{3}} 2\pi(x)(3x-\frac{x}{2}) dx$$

Which simplifies to: $$2\pi \int_{\frac{4}{3}}^8 4x - \frac{x^2}{2} dx + 2\pi \int_{0}^{\frac{4}{3}} \frac{5x^2}{2}dx$$ $$2\pi\cdot ([2x^2 - \frac{x^3}{6}]_{\frac{4}{3}}^8 + [\frac{5x^3}{6}]_0^{\frac{4}{3}})$$ $$2\pi\cdot (128 - \frac{256}{3} - \frac{32}{9} + \frac{64}{162} + \frac{320}{162})$$ $$= 260.6358$$

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  • $\begingroup$ Hi Phil, you were able to find bounds by solving for x. - proceeded to take the approach of using "two regions"and break apart the problem into two integrals. you decided to use the last y value which was a constant in this case (4) to setup one of the integrals and do (4-x/2). And the second one you set it up as: the second y value (3x) and do (3x-x/2). How did you go about analytically coming up with what is supposed to be subtracted from what? I guess if there were an illustration, we could see which curve is "bigger".. but is there another way to figure out that step without a drawing? $\endgroup$ – jackbenimbo Jul 14 '18 at 3:08
  • $\begingroup$ Also thanks so much for this 30th birthday gift of a approach/solution. If you have the time, would you show me how you would set it up if we were to use the disk/washer? If not, completely fine but regardless, thanks so much! Calc II has been rough! $\endgroup$ – jackbenimbo Jul 14 '18 at 3:10
  • $\begingroup$ @jackbenimbo I'll add a picture so you can see how the integral is set up. And yes, I'll do the same for the washer method. May take a while but I'll try to post an edit later tonight or in the morning. $\endgroup$ – Phil H Jul 14 '18 at 3:17
  • $\begingroup$ Cannot thank you enough, Phil. $\endgroup$ – jackbenimbo Jul 14 '18 at 3:20
  • $\begingroup$ @jackbenimbo Done, let me know if you have questions. $\endgroup$ – Phil H Jul 14 '18 at 4:27


Express $x_{1,2}$ in terms of y.

$$V=\pi\int_0^4 (x_1^2-x_2^2) dy$$

  • $\begingroup$ To be clear, you are recommending then that I use the washer method? $\endgroup$ – jackbenimbo Jul 13 '18 at 22:39
  • $\begingroup$ Oh I see, I represent x/2 as x= 2y and 3x as x=y/3, and then plug in using washer and integrate? $\endgroup$ – jackbenimbo Jul 13 '18 at 22:48
  • $\begingroup$ That is correct, the washer method. $\endgroup$ – Narasimham Jul 14 '18 at 6:45

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