Integration by filaments or integration by strate?

  • Thread starter Amaelle
  • Start date
  • #1
Amaelle
310
54
Homework Statement:
look at the image
Relevant Equations:
cylindrical coordinates
integration by strates
integration by filaments
Greetings
While solving the following exercice, ( the method used is the integration by filaments and I have no problem doing it this way)
1622996959713.png

here is the solution
1622997058481.png

1622997162664.png

My question is the following:
I want to do the integration by strate and here is my proposition
1622997349308.png

is that even correct?
I would like to know if there is cases when switching between the two methods of integration is not possible?
thank you in advance!
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,762
2,401
I've never heard of "integration by strate." Could you explain what this means?
 
  • Love
  • Like
Likes Delta2 and Amaelle
  • #3
Amaelle
310
54
Certainly look at the image
1623001742727.png
it means doing the integration by rings and then integrating by hight
 
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,762
2,401
Oh, okay. In the US, I think that technique is referred to as the disc method.
 
  • #5
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,762
2,401
You should be able to use that method, but you're going to have to break the integral up into two integrals, one from ##x=-2## to ##x=-\sqrt{2}## and one from ##x=-\sqrt{2}## to ##x=-1##.
 
  • Informative
  • Like
Likes Delta2 and BvU
  • #6
Amaelle
310
54
thank you very much , indeed I used your insights to solve the question under the disk method
1623071002384.png

but the only way I could find out was to
1623071123825.png


and then substructing the sphere from the paraboloid
I would be very grateful if you could elaborate more about the setting of your method and why you need to split the integration this way?
thank you!
 
  • #7
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,762
2,401
To keep it simple, I'm going to consider volume integrals but the same ideas apply to your problem.

When you do the integrals over ##r## and ##\theta##, you end up with an integral of the form
$$\int_{x_1}^{x_2} \pi r^2 \,dx$$ where we can identify ##\pi r^2## as the area of the disk and ##dV = \pi r^2\,dx## as the volume of the infinitesimal slice. In this particular problem, from ##x=-2## to ##x=-\sqrt{2}##, the volume is given by an integral of that form. However, from ##x=-\sqrt{2}## to ##x=-1##, you have a disk with a hole in the middle with area ##\pi(r_{\rm outer}^2-r_{\rm inner}^2)##. Hence, the volume of that portion is given by
$$\int_{x_1}^{x_2} \pi (r_{\rm outer}^2-r_{\rm inner}^2) \,dx.$$ Because you have two different integrands, you need to do the integrals for each range separately.

Note that you can manipulate your integrals similarly:
\begin{align*}
I &= \int_{-2}^{-1} \int_0^{2\pi} \int_0^{\sqrt{2-x^2}} xr\,dr\,d\theta\,dx
-\int_{-\sqrt 2}^{-1} \int_0^{2\pi} \int_0^{\sqrt{x+2}} xr\,dr\,d\theta\,dx \\
&= \int_{-2}^{-\sqrt{2}} \int_0^{2\pi} \int_0^{\sqrt{2-x^2}} xr\,dr\,d\theta\,dx
+\int_{-\sqrt{2}}^{-1} \int_0^{2\pi} \int_0^{\sqrt{2-x^2}} xr\,dr\,d\theta\,dx
-\int_{-\sqrt 2}^{-1} \int_0^{2\pi} \int_0^{\sqrt{x+2}} xr\,dr\,d\theta\,dx \\
&= \int_{-2}^{-\sqrt{2}} \int_0^{2\pi} \int_0^{\sqrt{2-x^2}} xr\,dr\,d\theta\,dx
+\int_{-\sqrt{2}}^{-1} \int_0^{2\pi} \int_{\sqrt{x+2}}^{\sqrt{2-x^2}} xr\,dr\,d\theta\,dx
\end{align*} The limits on the ##r## integral are different for ##x=-2## to ##x=-\sqrt{2}## and for ##x=-\sqrt{2}## to ##x=-1##.
 
  • #8
Amaelle
310
54
thanks a million!
 

Suggested for: Integration by filaments or integration by strate?

Replies
4
Views
621
Replies
13
Views
619
  • Last Post
Replies
3
Views
274
  • Last Post
Replies
5
Views
189
Replies
12
Views
456
  • Last Post
Replies
8
Views
786
Replies
4
Views
152
  • Last Post
Replies
3
Views
631
Replies
5
Views
412
Top