# Area of a circle and integral

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

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...]

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

You can rearrange to get x^2 + y^2 = r^2

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

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.

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.
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.

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?

So I guess my idea is wrong,....
I wonder what the idea was....

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?
How did you try this one?

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.

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DryRun
Gold Member
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.

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

DryRun
Gold Member
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.

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. =(

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

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!)

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.

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

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]

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.

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]

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Nice and elegant

Nice and elegant
Thanks.

Now lets revisit this approach:

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!)

Ray Vickson
Homework Helper
Dearly Missed
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
$$I = \int_{-r}^{r} dx \int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}} (r^2-x^2-y^2) \, dy = \int_{-r}^{r} [(r^2-x^2) \cdot 2 \sqrt{r^2-x^2} - \frac{2}{3} (r^2-x^2)^{3/2}] \, dx\\ = \frac{4}{3}\int_{-r}^r (r^2 - x^2)^{3/2} \, dx.$$

RGV

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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.

Ray Vickson
Homework Helper
Dearly Missed
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

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?