• Support PF! Buy your school textbooks, materials and every day products Here!

Area of a circle and integral

  • Thread starter estro
  • Start date
  • #1
241
0
[tex] \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \sqrt{r^2-x^2-y^2}dy [/tex]

I can calculate the above integral [part of a double integral] by the conventional way [somewhat long], however my book says that this integral equals to [tex]\frac {\pi}{2}(r^2-x^2)[/tex] because the integral is actually the area of half a circle. I have difficulty to understand how this integral has something to do with half of a circle. [The author of my book thinks this is fact is trivial, however for me it is not =(]

And if the above integral is the area of half a cicle than what about:
[tex] \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} (r^2-x^2-y^2)dy [/tex]
I'll appreciate any help.

[EDIT: I think now I understand why the first integral is the area of half a cirlce, now thinking what I can tell about the second integral...]
 
Last edited:

Answers and Replies

  • #2
Well, if you look at the bounds, y = sqrt(r^2 - x^2)

You can rearrange to get x^2 + y^2 = r^2
 
  • #3
881
40
[tex] \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \sqrt{r^2-x^2-y^2}dy [/tex]
Since you are integrating with respect to y, [itex]r^2 - x^2 = t^2[/itex] is a constant. So the integral becomes,

[tex]\int_{-t}^{t} \sqrt{t^2-y^2}dy[/tex]

Now, the function you are integrating can be taken as,

[tex]X = \sqrt{t^2 - y^2}[/tex]

And this is the equation of the semi-circle.

Or, fully in circle form(the square root gives only positive solutions implying half circle),

[tex]X^2 + y^2 = t^2[/tex]

Where t is the radius. So, the integral you are trying to find is simply the area of a half circle of radius t.
 
  • #4
241
0
Thanks for the answers however I already understood the [tex]\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \sqrt{r^2-x^2-y^2}dy[/tex] case.

Looking now at [tex]\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} (r^2-x^2-y^2)dy[/tex]

Can I compute this integral in similar way? I seems like this time it is an ellipse.
 
  • #5
881
40
Looking now at [tex]\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} (r^2-x^2-y^2)dy[/tex]

Can I compute this integral in similar way? I seems like this time it is an ellipse.
Why do you think it is an ellipse?

Using the same idea, you would have,

[tex]X + y^2 = t^2[/tex]

This doesn't resemble any of the standard shapes, I believe. ie circle, ellipse, square etc.
 
  • #6
241
0
So I guess my idea is wrong, I'm trying to calculate the following integral:

[tex]\iint\limits_{x^2+y^2\leq R^2}(R^2-x^2-y^2)dxdy[/tex]

Any ideas?
 
  • #7
881
40
So I guess my idea is wrong,....
I wonder what the idea was....

I'm trying to calculate the following integral:

[tex]\iint\limits_{x^2+y^2\leq R^2}(R^2-x^2-y^2)dxdy[/tex]

Any ideas?
How did you try this one?
 
  • #8
241
0
I tried a lot of things the last one was:
https://dl.dropbox.com/u/27412797/hard_integral.jpeg [Broken]
But calculating this integral is a lot of work, I was given a hint that I can use the symmetry of the area which is a circle, but still unable to find the proper way to use this symmetry.
 
Last edited by a moderator:
  • #9
DryRun
Gold Member
838
4
estro, this is such a huge scan! You could have used Microsoft Paint or any other image editor to cut the white parts and downsize it for viewing convenience. :smile:
 
  • #10
241
0
Sorry for that.
Just cropped the scan, please refresh =)
 
  • #11
DryRun
Gold Member
838
4
Have you done polar coordinates? It's much simpler to evaluate this double integral if you first convert to polar coordinates.
[tex]\iint\limits_{x^2+y^2\leq R^2}(R^2-x^2-y^2)dxdy=\int^{\theta=2\pi}_{\theta=0} \int^{r=R}_{r=0} (R^2-r^2).rdrd\theta[/tex]
Note that ##x^2+y^2=r^2## where ##r## is the radius of the circle.
##x^2+y^2\leq R^2## describes a disc of radius R.
 
  • #12
241
0
I'm not allowed to use polar coordinates on this question. The instructor told us symmetry [of the area of integration] should be enough to solve it. =(
 
  • #13
DryRun
Gold Member
838
4
$$\iint\limits_{x^2+y^2\leq R^2}(R^2-x^2-y^2)dxdy$$
First, try to describe the region of integration:
For y fixed, x varies from ##x =-\sqrt{R^2-y^2} ## to ##x =\sqrt{R^2-y^2} ##
y varies from ##y =-R## to ##y=R##
Now that you have the limits, you should be able to evaluate the double integral. :smile:
 
  • #14
881
40
Hmm, the only way I can currently think of using symmetry is by splitting the term as [itex]\sqrt{R^2-x^2-y^2} \cdot \sqrt{R^2-x^2-y^2}[/itex]. Its not a very less tedious method, though(but it is interesting!)
 
  • #15
241
0
This is exactly what I did [see the attached scan], I'm looking for more elegant solution... =)


$$\iint\limits_{x^2+y^2\leq R^2}(R^2-x^2-y^2)dxdy$$
First, try to describe the region of integration:
For y fixed, x varies from ##x =-\sqrt{R^2-y^2} ## to ##x =\sqrt{R^2-y^2} ##
y varies from ##y =-R## to ##y=R##
Now that you have the limits, you should be able to evaluate the double integral. :smile:
 
  • #16
241
0
Hmm, the only way I can currently think of using symmetry is by splitting the term as [itex]\sqrt{R^2-x^2-y^2} \cdot \sqrt{R^2-x^2-y^2}[/itex]. Its not a very less tedious method, though(but it is interesting!)
Yes, I tried this one as well bu maybe I missed something let me recheck.
 
  • #17
241
0
Actually I have pretty nice solution, but I'm not sure my instructor meant it to be solved this way. I will post it in a hour. [I think you will like it]
 
  • #18
881
40
Actually I have pretty nice solution, but I'm not sure my instructor meant it to be solved this way. I will post it in a hour. [I think you will like it]
Nice, I'll ponder over it a bit more till then, getting somewhere. :smile:
 
  • #19
241
0
In the following proof I pretty much used only simple tools and intuition. I'm still looking for another way to solve it let me know if you have other ideas.

The drawing is only 1/4 of the actual 3D body. [My drawing skills are as bad as my math=)]
https://dl.dropbox.com/u/27412797/short_proof_2.JPG [Broken]
 
Last edited by a moderator:
  • #20
881
40
Nice and elegant :approve:
 
  • #21
241
0
Nice and elegant :approve:
Thanks.

Now lets revisit this approach:

Hmm, the only way I can currently think of using symmetry is by splitting the term as [itex]\sqrt{R^2-x^2-y^2} \cdot \sqrt{R^2-x^2-y^2}[/itex]. Its not a very less tedious method, though(but it is interesting!)
 
  • #22
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
I tried a lot of things the last one was:
https://dl.dropbox.com/u/27412797/hard_integral.jpeg [Broken]
But calculating this integral is a lot of work, I was given a hint that I can use the symmetry of the area which is a circle, but still unable to find the proper way to use this symmetry.
I don't get it: isn't the problem straightforward? You have that your integral, I, equals
[tex] I = \int_{-r}^{r} dx \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} (r^2-x^2-y^2) \, dy
= \int_{-r}^{r} [(r^2-x^2) \cdot 2 \sqrt{r^2-x^2} - \frac{2}{3} (r^2-x^2)^{3/2}] \, dx\\
= \frac{4}{3}\int_{-r}^r (r^2 - x^2)^{3/2} \, dx.[/tex]

RGV
 
Last edited by a moderator:
  • #23
241
0
This is EXACTLY what I written 2 post above, if you think calculating this integral is "straightforward" you're wrong.

Anyway I am looking for a certain type of solution [should use symmetry], if you're looking for a "quick solution" look at my last post.
 
  • #24
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
This is EXACTLY what I written 2 post above, if you think calculating this integral is "straightforward" you're wrong.

Anyway I am looking for a certain type of solution [should use symmetry], if you're looking for a "quick solution" look at my last post.
Maybe your definition of "straightforward" is different from mine; to me, "straightforward" means looking up something in a table that I had already done once in the past and need not do again; or straightforward means using modern tools effectively. It does not mean "fast" or "short"; it just means "not mysterious" and "not tricky".

RGV
 
  • #25
241
0
Maybe your definition of "straightforward" is different from mine; to me, "straightforward" means looking up something in a table that I had already done once in the past and need not do again; or straightforward means using modern tools effectively. It does not mean "fast" or "short"; it just means "not mysterious" and "not tricky".

RGV
We indeed have different definition for "straightforward". =)

Can you suggest another way of solving this besides the previous 2 methods?
 

Related Threads for: Area of a circle and integral

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
24
Views
4K
Replies
10
Views
2K
Replies
2
Views
5K
Replies
6
Views
11K
Replies
7
Views
16K
Replies
1
Views
3K
  • Last Post
Replies
2
Views
10K
Top