Welcome to our community

Be a part of something great, join today!

Shell Method

alane1994

Active member
Oct 16, 2012
126
Here is my homework question. I am stuck on one part of it, and it is ok for me to receive guidance, not answers without effort.

Question:
Let R be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when R is revolved about the x-axis.

[tex]y=\sqrt{98-2x^2}[/tex]

My Work So Far:
  • I have found the value of x
[tex]x=\pm\sqrt{\frac{98-y^2}{2}}[/tex]
  • The lower limit is 0.
  • The upper limit is___?

This is where I get stuck... I am unsure how to get the upper limit. Once I get that, I should be able to proceed from there.
 

tkhunny

Well-known member
MHB Math Helper
Jan 27, 2012
267
#1 - Please observe symmetry. Work in the 1st Quadrant and multiply by 2 to achieve the entire result. This will free you from the laborious "+/-".

Thus: [tex]2\cdot\left(2\cdot\pi\cdot\int_{0}^{\sqrt{98}}y \cdot x\;dy\right)[/tex]

You've only to substitute your correct expression for 'x' and you're done.

In my opinion, you should ALWAYS do it the other way in addition to what is asked. This will do at least these three things:
1) Give you experience in both methods.
2) You will gain experience in judging which is better in which circumstances.
3) You will be able to check your own work.

[tex]2\cdot\left(\pi\cdot\int_{0}^{7}y^{2}\;dx\right) = 2\cdot\left(\pi\cdot\int_{0}^{7}98 - 2x^{2}\;dx\right)[/tex]
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
...
In my opinion, you should ALWAYS do it the other way in addition to what is asked. This will do at least these three things:
1) Give you experience in both methods.
2) You will gain experience in judging which is better in which circumstances.
3) You will be able to check your own work.
...
Great advice!:cool:

My calculus professor recommended the same thing for the same reasons, way back when. It is a great habit to get into.