How Do You Apply the Washer Method for Calculating Volumes of Revolution?

  • Thread starter Thread starter karamsoft
  • Start date Start date
  • Tags Tags
    Method
karamsoft
Messages
4
Reaction score
0

Homework Statement


Use the washer method to find the volume of hte solid generated by revolving the regions bounded by y= root(x) and the lines y =2 and x=0 about the x-axis , y-axis , y= 2 and x=4


Homework Equations



V= Pi integral (0,4) (R(x)^2 - r(x)^2) dx

The Attempt at a Solution



I tried to use the equation above but I totally got confused about r(x)^2 as from what I know that the washer method means you have a small radius that you will subtract from the large curve which is in this case root(x) but there is none!

Thank you very much for your time, I would appropriate if you could explain how you solve this question so I could learn
 
Physics news on Phys.org
I tried to solve these problems in a different way but I'm not sure if I'm on the right track.

If it revolved over the x-axis then the volume will be
integral (0,4) Pi (2)^2 - (root (x))^2 dx = 42.265 units^3

If it revolved over the y-axis then the volume will be
integral (0,2) Pi (y)^2 - 0 dy = 8.3775 units^3

If it revolved over the line y=2 then the volume will be
integral (0,4) Pi (-root(x)+2)^2 dx = 8.3775 units^3*****This is the one that I'm not really sure if its the right answer or Not so please HELP!

If it revolved over the line x=4 then the volume will be
integral (0,2) Pi ((4)^2 -(-y^2+4)^2) dy = 46.9144 units^3
 
karamsoft said:
I tried to solve these problems in a different way but I'm not sure if I'm on the right track.

If it revolved over the x-axis then the volume will be
integral (0,4) Pi (2)^2 - (root (x))^2 dx = 42.265 units^3

\int_0^4 \pi\left(2^2-(\sqrt{x})^2\right)\,dx

You're missing a set of parentheses.

karamsoft said:
If it revolved over the y-axis then the volume will be
integral (0,2) Pi (y)^2 - 0 dy = 8.3775 units^3
You missed the parentheses here too, but the numerical answer is OK, because zero times pi is zero.

karamsoft said:
If it revolved over the line y=2 then the volume will be
integral (0,4) Pi (-root(x)+2)^2 dx = 8.3775 units^3
This looks OK.

karamsoft said:
*****This is the one that I'm not really sure if its the right answer or Not so please HELP!

If it revolved over the line x=4 then the volume will be
integral (0,2) Pi ((4)^2 -(-y^2+4)^2) dy = 46.9144 units^3
This looks OK too.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top