Area of a circle and integral

1. Jun 13, 2012

estro

$$\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \sqrt{r^2-x^2-y^2}dy$$

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 $$\frac {\pi}{2}(r^2-x^2)$$ 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:
$$\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} (r^2-x^2-y^2)dy$$
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: Jun 13, 2012
2. Jun 13, 2012

NewtonianAlch

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. Jun 13, 2012

Infinitum

Since you are integrating with respect to y, $r^2 - x^2 = t^2$ is a constant. So the integral becomes,

$$\int_{-t}^{t} \sqrt{t^2-y^2}dy$$

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

$$X = \sqrt{t^2 - y^2}$$

And this is the equation of the semi-circle.

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

$$X^2 + y^2 = t^2$$

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. Jun 13, 2012

estro

Thanks for the answers however I already understood the $$\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} \sqrt{r^2-x^2-y^2}dy$$ case.

Looking now at $$\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} (r^2-x^2-y^2)dy$$

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

5. Jun 13, 2012

Infinitum

Why do you think it is an ellipse?

Using the same idea, you would have,

$$X + y^2 = t^2$$

This doesn't resemble any of the standard shapes, I believe. ie circle, ellipse, square etc.

6. Jun 13, 2012

estro

So I guess my idea is wrong, I'm trying to calculate the following integral:

$$\iint\limits_{x^2+y^2\leq R^2}(R^2-x^2-y^2)dxdy$$

Any ideas?

7. Jun 13, 2012

Infinitum

I wonder what the idea was....

How did you try this one?

8. Jun 13, 2012

estro

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: May 6, 2017
9. Jun 13, 2012

sharks

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.

10. Jun 13, 2012

estro

Sorry for that.
Just cropped the scan, please refresh =)

11. Jun 13, 2012

sharks

Have you done polar coordinates? It's much simpler to evaluate this double integral if you first convert to polar coordinates.
$$\iint\limits_{x^2+y^2\leq R^2}(R^2-x^2-y^2)dxdy=\int^{\theta=2\pi}_{\theta=0} \int^{r=R}_{r=0} (R^2-r^2).rdrd\theta$$
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. Jun 13, 2012

estro

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. Jun 13, 2012

sharks

$$\iint\limits_{x^2+y^2\leq R^2}(R^2-x^2-y^2)dxdy$$
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.

14. Jun 13, 2012

Infinitum

Hmm, the only way I can currently think of using symmetry is by splitting the term as $\sqrt{R^2-x^2-y^2} \cdot \sqrt{R^2-x^2-y^2}$. Its not a very less tedious method, though(but it is interesting!)

15. Jun 13, 2012

estro

This is exactly what I did [see the attached scan], I'm looking for more elegant solution... =)

16. Jun 13, 2012

estro

Yes, I tried this one as well bu maybe I missed something let me recheck.

17. Jun 13, 2012

estro

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. Jun 13, 2012

Infinitum

Nice, I'll ponder over it a bit more till then, getting somewhere.

19. Jun 13, 2012

estro

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: May 6, 2017
20. Jun 13, 2012

Infinitum

Nice and elegant