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Question: Find the volume of the solid generated by revolving the triangular region enclosed by $y = |x|$ and $y = 1$ about the line $x = −2$.

My solution:

$$\pi*9*1 - \pi*1*1 - 1/3 *\pi* 1 - 1/3 *\pi* 1 = 22/3 \pi$$

Use the large cylinder minus the small cylinder and minus two cones.

But the correct answer is 4pi

Did I do anything wrong?

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  • $\begingroup$ It would be helpful to actually show your steps. Like, which method did you use? $\endgroup$ – imranfat Nov 22 '20 at 23:52
  • $\begingroup$ oh since this area is simple. I did not use shell or washer method. I just minus off volumes from a big cylinder. $\endgroup$ – linear Nov 22 '20 at 23:59
  • $\begingroup$ That approach is iffy... $\endgroup$ – imranfat Nov 23 '20 at 0:00
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Here is an outline of a method that works: Disc/Washer Method. You need to write the equations in terms of $y$ because you rotate about a vertical line. The integral will be of the form $\int(y+2)^2dy-\int(2-y)^2dy$ where $y$ is going from $0$ to $1$. Try to understand how this integral is set up by "translating" $y=|x|$ as well as $x=-2$ two units to the right. Now you work this out. (The correct answer is indeed $4\pi$)

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