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

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?

• It would be helpful to actually show your steps. Like, which method did you use? – imranfat Nov 22 '20 at 23:52
• oh since this area is simple. I did not use shell or washer method. I just minus off volumes from a big cylinder. – linear Nov 22 '20 at 23:59
• That approach is iffy... – imranfat Nov 23 '20 at 0:00

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 $$4\pi$$)