# Thread: Multiple integral

1. ## Multiple integral

Is this correct?

10x.

2. Nice try, actually, but not quite...

You are right to split it up into two pieces. You missed a couple of things.

The portion above the x-axis needs $y^{2}$ in the arguments and should be on [0,1].

The portion below the x-axis should be on [-1,0].

In every case, it should be "1-". I cannot tell where you managed any "1+". Further, if your limits are identical, you ARE going to get zero (0). Try negative on the bottom and positive on the top.

3. Originally Posted by TKHunny
Nice try, actually, but not quite...

You are right to split it up into two pieces. You missed a couple of things.

The portion above the x-axis needs $y^{2}$ in the arguments and should be on [0,1].

The portion below the x-axis should be on [-1,0].

In every case, it should be "1-". I cannot tell where you managed any "1+". Further, if your limits are identical, you ARE going to get zero (0). Try negative on the bottom and positive on the top.

How about this one?

BTW I spent a lot of time just looking at it... Is there a technique doing that?

10x.

4. $\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}f(x,y)~dy~dx$

The region of integration is a full circle of radius one centered around the origin. Now think about it, if you rotate it ninety degrees so you are first going in respect to x, everything will be "the same" since a circle is completely symmetric, so all you would do is switch the x's and y's

$\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}f(x,y)~dy~dx=\int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}f(x,y)~dx~dy$

5. Originally Posted by asi123
BTW I spent a lot of time just looking at it... Is there a technique doing that?
Yeah, what I do is rotate it ninety degrees and look at it, and see where the y's or x's range from and so forth. I think its easier if you rotate it.

6. Originally Posted by Mathstud28
$\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}f(x,y)~dy~dx$

The region of integration is a full circle of radius one centered around the origin. Now think about it, if you rotate it ninety degrees so you are first going in respect to x, everything will be "the same" since a circle is completely symmetric, so all you would do is switch the x's and y's

$\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}f(x,y)~dy~dx=\int_{-1}^{1}\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}f(x,y)~dx~dy$

Don't we need to split it?

7. Originally Posted by asi123
Don't we need to split it?
Why would we?

8. Originally Posted by Mathstud28
Why would we?

Is that right?

9. Originally Posted by asi123
Is that right?
Thats the exact same thing as I put since your two integrals combine to form the exact region that mine describes and you should know that

$\iint_{R}f(x,y)~dx~dy=\iint_{R_1}f(x,y)~dx~dy+\iin t_{R_2}~dx~dy$

Assuming $R_1\cup{R_2}=R$

And you have two semi-circles who when unioned form the circle described by my double integrals.

10. See how important it is to write clearly and check what it is you have written?

My original post took you at your word. Check that internal integral. The bottom limit is from a circle. The top limit is from a parabola. It is for this reason that I recommended splitting it up. If you've really only circles, the original integral simply is wrong and there is no need to do any splitting.

11. Originally Posted by TKHunny
See how important it is to write clearly and check what it is you have written?

My original post took you at your word. Check that internal integral. The bottom limit is from a circle. The top limit is from a parabola. It is for this reason that I recommended splitting it up. If you've really only circles, the original integral simply is wrong and there is no need to do any splitting.
TKHunny is right, the only reason I did the bounds I did was because this is a classic question, and I assumed that this is what you meant. You really should be more careful. But don't worry about it! Just don't do it again.