Double integral

  • Thread starter xstetsonx
  • Start date
  • #1
78
0
can someone explain to me why my teacher divided the area into two?
I=[tex]\int[/tex]1,0[tex]\int[/tex](2x-x^2)^.5,0 1/(x^2+y^2)^.5dydx



ugggggh i tried to use the latex.....
anyway...

he used the polar coordinate to do this. once he turned it into polar coordinate, he divided the area into 2 bounded by (pi/4 - o d[tex]\theta[/tex]) (1/cos[tex]\theta[/tex] - 0 for dr) and then + (d[tex]\theta[/tex] bounded by pi/2 - pi/4) (2cos bounded by 2 - 0)


sorry i don't know how to make this more clear...ask me if you need to clarify
 
Last edited:

Answers and Replies

  • #2
lanedance
Homework Helper
3,304
2
see the correct (hopefully) tex below, click on it to see script, note you can put the whole equation in one set of tags
[tex]\int_0^1 \int_0^{\sqrt{2x-x^2}} \sqrt{x^2+y^2} dy dx[/tex]

is this what you meant?
 
  • #3
lanedance
Homework Helper
3,304
2
so what was the question?
 
  • #4
307
0
see the correct (hopefully) tex below, click on it to see script, note you can put the whole equation in one set of tags
[tex]\int_0^1 \int_0^{\sqrt{2x-x^2}} \sqrt{x^2+y^2} dy dx[/tex]

is this what you meant?

I think it might actually be

[tex]\int_0^1 \int_0^{\sqrt{2x-x^2}} \frac{1}{\sqrt{x^2+y^2}} dy dx[/tex]
 
  • #5
lanedance
Homework Helper
3,304
2
yeah i think you're right
 
  • #6
78
0
my teacher said you have to separate the area into to parts in order to solve it and i don't understand why


sorry about my crappy drawing again
 

Attachments

  • Untitled.jpg
    Untitled.jpg
    4.2 KB · Views: 273
Last edited:
  • #7
307
0
What exactly was the problem statement?
 
  • #8
35,222
7,039
The region over which integration takes place is the upper left quarter of a circle with radius 1 and centered at (1, 0). When you switch to a polar integral, the description for r changes, and this is why you need to write the polar form of your given integral with two integrals. For one integral, r ranges from 0 to the appropriate value on the line x = 1. For the other integral, r ranges from 0 to the appropriate value on the circle.
 
  • #9
78
0
then how you suppose to figure out the boundaries? and can you expand on the description for r changes??
 
  • #10
307
0
Enters Mark! You're in much better hands than mine now xstetsonx so I'm definitely stepping back here. Have fun! :smile:
 
  • #11
78
0
thx phyzmatrix anyways
 
  • #12
35,222
7,039
then how you suppose to figure out the boundaries? and can you expand on the description for r changes??
Have you drawn a sketch of the region? I have described the boundaries. All you need to do is change the Cartesian equation that represents each boundary to its polar form.
 
  • #13
78
0
yup under the attachment above. well you said appropriate i am not clear. my teacher he
said for d[tex]\theta[/tex] is from 0 to pi/4 for the first one.......etc i don't understand how you get this value according to that graph
 
  • #14
35,222
7,039
The graph in your attachment is not very useful, as the circle part doesn't look much like part of a circle, and almost nothing is identified with its equation. To answer your question, what are the coordinates of the point where the vertical line intersects the circle? For the same point, what are the polar coordinates?
 
  • #15
78
0
To answer your question, what are the coordinates of the point where the vertical line intersects the circle? For the same point, what are the polar coordinates?

1? cos^2+sin^2????
 
  • #17
78
0
i mean (1,1) that is where they intersect right?
 
  • #18
35,222
7,039
Right. Now, what are the polar coordinates of that point?
 
  • #19
78
0
em (cosx)^2+(sinx)^2?
 
  • #20
35,222
7,039
No. Polar coordinates are pairs of numbers (r, theta), with the first being the distance from the pole (origin) and the second the angle theta.

In any case, cos2(x) +sin2(x) is identically equal to 1.
 
  • #21
78
0
(2)^.5, 45 degrees?
 
  • #23
78
0
MARK you are once again the awesomest guy on here
thanks alot
 
  • #24
35,222
7,039
Well, thanks!

So now you understand why the limits on one integral are 0 to pi/4 and for the other they are pi/4 to pi/2, right?
 
  • #25
78
0
yup yup everything makes lot senses now but i just want to know how he get the equation 1/cosx and 2cosx

if i am doing from scratch how do i know it is that equation?
 

Related Threads on Double integral

  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
698
  • Last Post
Replies
17
Views
2K
  • Last Post
Replies
3
Views
646
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
3
Views
878
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
7
Views
2K
Top