What is the relationship between the integral and the area of half a circle?

[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...]

 

[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

 

[Infinitum]

\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.

 

[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.

 

[Infinitum]

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.

 

[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?

 

[Infinitum]

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?

 

[estro]

I tried a lot of things the last one was:
https://dl.dropbox.com/u/27412797/hard_integral.jpeg
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.

 

[DryRun]

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.

 

[estro]

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

 

[DryRun]

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.

 

[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. =(

 

[DryRun]

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

 

[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!)

 

[estro]

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.

 

[estro]

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.

 

[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]

 

[Infinitum]

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.

 

[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

 

[Infinitum]

Nice and elegant

 

[estro]

Nice and elegant

Thanks.

Now let's 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]

I tried a lot of things the last one was:
https://dl.dropbox.com/u/27412797/hard_integral.jpeg
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<br /> = \int_{-r}^{r} [(r^2-x^2) \cdot 2 \sqrt{r^2-x^2} - \frac{2}{3} (r^2-x^2)^{3/2}] \, dx\\<br /> = \frac{4}{3}\int_{-r}^r (r^2 - x^2)^{3/2} \, dx.

RGV

 

[estro]

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]

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

 

[estro]

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?

 

[Ray Vickson]

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

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

Of course, the best way is to switch to polar coordinates right away in the 2d-integral (but for some unknown reason, you are not allowed to do that).

RGV

 

[Infinitum]

After some more thought, I found no method that can solve this by only using symmetry. And the easiest, and clearly visible method is of course using polar co-ordinates as suggested above.

I am interested to know what your professor has to say about this.