Integration help

  • Thread starter artikk
  • Start date
  • #1
19
0
I tried to find the volume of a torus revolved around the y-axis using the washer method. The equation for the circle is (x-2)^2 +y^2=1.
I need to integrate (1-y^2)^1/2 dy from 1 to 0 multiplied by a constant 16pi. The constant is not really important but I can't seem to integrate the integrand. I am aware that there's a formula for finding a vol. of a torus, as well.
 
Last edited:

Answers and Replies

  • #2
mathman
Science Advisor
7,877
453
Unless you misstated the problem, it is trivial. The integral of 1 is y and the integral of y2 is y3/3. Just put them together.
 
  • #3
19
0
Yes, but in this case, the (1-y^2) are under the square root, you can't separate the two terms.
 
  • #4
1,631
4
[tex]\int_0^1 \sqrt (1-y^{2}) dy[/tex]
a trig substitution would work here, like y=sinx, so dy=cos(x)dx, when y=0, x=0, when y=1,[tex] x=\frac{\pi}{2}[/tex]

[tex]\int_0^\frac{\pi}{2} \sqrt(1-sin^{2}x )cosxdx[/tex] now remember

[tex]1-sin^{2}(x)=cos^{2}(x)[/tex] so

[tex]\int_0^\frac{\pi}{2} cos^{2}(x) dx=\int_0^\frac{\pi}{2} \frac{1+cos(2x)}{2} dx[/tex] and you are almost done!!
 
Last edited:

Related Threads on Integration help

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
4
Views
1K
Top