Science talk:Math Exam Resources/Courses/MATH101/April 2010/Question 2 (a)

The pictures aren't working and I looked into the file storage and the files stored are incorrect. I find the best file types to put in are .png but in any case, they need to be uploaded again.

Secondly you reference the shell method which, I thought, means breaking it up into disks of "width" dx (mini cylinders) so that the volume of one disk is $\pi {}r^{2}dx$ where the radius can depend on what it needs to. In this case the radius is 2+(1-x^2)=3-x^2. The integral would then be,

$\int _{-1}^{1}{}\pi (3-x^{2})^{2}{\textrm {d}}x={\frac {72\pi }{5}}.$

This also agrees if you were to do a triple integral in cylindrical coordinates (which is where the shell formula is derived),

$\int _{0}^{2\pi }\int _{-1}^{1}\int _{0}^{3-x^{2}}r{\textrm {d}}r{\textrm {d}}z{\textrm {d}}\theta ={\frac {72\pi }{5}}.$

Maybe the h(z)z in the integral is a fragment from cylindrical coordinates? I also don't understand why there is a taking of the inverse.

Never mind, I see the method you are referring to in the textbook (even though I think what I wrote above is easier), I see that the students probably learn this cylindrical shell method. Still, I was wondering why the answers were different. If you use your thin shell method then you still have to consider the volume of the cylinder of revolution above y=-2 and below the x-axis. This cylinder has radius 2 and height 2. Therefore the volume is $8\pi$. Therefore the total volume is $32\pi /5+8\pi =72\pi /5$ like the answer I got above.

I disagree, the question specifies the region bounded by the x-axis, so that central cylinder should not be included.

Oh I see how it's supposed to go now, it's like an annulus region, I interpreted it as to cut off the parabola at the x-axis and then take that whole shape and rotate it, makes sense now. The pictures still need to be fixed though.

Never mind, working now!

It might be easier to use disks ("washers" in this case, since there's a hole in the middle. I'm going to work on a solution doing that.