# An odd Volume problem/how to find the area of a part of a circle

1. Jan 21, 2005

### Wooh

Hmm, is there an easy way to evaluate this integral: 2*integral(sqrt[.25-x^2],-y,y). Basically, I want to know how to find the area as depicted by the attached image (untitled) as a function.

This goes along with a problem that I am trying to solve. If you know how to do it or something, please spoilerize as I would like to figure it out on my own. I think I have a FORM of an answer but it is really, really ugly.

I will also attach the form of the answr that I got, and would like critiquing.

You have a 1 inch diameter cylindrical piece of wood of indeterminate length. You have a 1 inch diameter drill bit. You bore into the wood perpendicular to it with the drill bit. What is the volume of the hole create, or the volume of the displaced sawdust (same value).

Anyone think my answer is wrong/have a way to simplify it? I am only in AP BC calc and have no idea how to integrate that :X

I put it in my calculator, and it says that the answer is 2/3. Anyone have any input?

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2. Jan 21, 2005

### Wooh

aaanyone have an idea?

3. Jan 21, 2005

### The Divine Zephyr

I'm gonna need to see that attachment. As soon as its approved and all...

4. Jan 21, 2005

### p53ud0 dr34m5

are you doing:
$$2\int^y_{-y}\sqrt{-.25-x^2$$

if so, i think it would be best to go ahead an integrate $\sqrt{-.25-x^2}$ and solve it from a (-y) to b (y). im not really sure. im a newbie when it comes to calculus . and im assuming your point a is 1 and b is 1...im not sure, because you just put -y to y. if you could clear that up for me, then i could possibly tackle the problem. i cant see the attachments, because they are pending approval.

5. Jan 21, 2005

### Wooh

As far as my screen, it seems to have attached ok...

But basically it is

$$2\int_{-.5} ^{.5} (\int_{-\sqrt{.25-y^2}} ^{\sqrt{.25-y^2} }\sqrt{.25-x^2}dx)dy$$

That is for the second problem.

6. Jan 21, 2005

### Wooh

And additionally, I have no idea how to simply integrate (for a circle of, let's say radius 2)
$$\int _{-1} ^{1} \sqrt{4-x^2} dx$$

Edit: well, I can "integrate it" with a really ugly sector+triangle thing, but I hope there is a better way...I don't want to have to do with once with a decently ugly equation, and then again with a horribly ugly one.

Last edited: Jan 21, 2005
7. Jan 21, 2005

### hypermorphism

You could go about this in many ways. For a beginning student, it is easiest to see the trigonometric substitution. The integrand is easily seen as the length of a leg of a right triangle whose other leg has length x and whose hypotenuse is of length 2. Thus, the impenetrable square root of the integrand can be rewritten as 2cos(t), where t is the angle between the leg of length $$\sqrt{4-x^2}$$ and the hypotenuse.Now we need to rewrite dx in terms of dt.
We see from our triangle that sin(t) = x/4, so x = sin(t)/4. Thus, dx = [cos(t)/4] dt.
Next we change our limits of integration, since they are technically from x=-1 to x=1 and we're going to integrate over t. From our previous equation, we see t = arcsin(x/4), so our integral becomes:
$$\int _{-\arcsin(1/4)} ^{\arcsin(1/4)} \frac{1}{2}\cos^2(t) dt$$
which is an integral you should know how to do.

8. Jan 21, 2005

### Wooh

ah, well, there I go. Thanks! Hmm, I am going to have to try and apply that to the other one. Can you take a look at the other one too?

9. Jan 23, 2005

### Wooh

Bump? I would really like a little direction :X

10. Jan 24, 2005

### dextercioby

1.A circle has 0 area and 0 volume...Let's get that straight...
2.You've already been explained what substitution to use in order to integrate that radical...

Daniel.

11. Jan 24, 2005

### Wooh

It's not homework. I am doing an extra problem for fun. Can you at least let me know if 2/3 is correct? I just don't wanna be chasin' waterfalls...

12. Jan 25, 2005

### dextercioby

How about posting your work,so that we can figure out what u did right/wrong and how u ended up with that result...

Daniel.

P.S.U could have picked the answer from the back of the book...

13. Jan 25, 2005

### Wooh

It was a problem my calc teacher gave me to work out. Work will be forthcoming. It is just hard to get work onto the internet, which is why I was hoping someone could at least criticize the 2/3 value. thanks though.

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