# Thread: Sense checking an integral

1. ## Sense checking an integral

I have this question where I need to evaluate this integral

$\displaystyle \int_{-1}^{-1} \int_{0}^{\sqrt{1-x^2}} 1-y^2 dydx$

When I do the sketch though it doesn't make sense because:

The region lies in the semi circle about line y=0 and circle with radius 1 right? But then the outer integral says that the area is cut off at lines x=1 and x=-1. Which doesn't cut off any of the semi circle as its on the tip as the circle radius is 1.. my obvious thought is that I'm doing something wrong. Can anyone confirm?

Does this integral just represent the area under the semicircle $\displaystyle \sqrt{1-x^2}$?

Thank you

2. First of all, is this supposed to be

$\displaystyle \displaystyle \int_{-1}^1{\int_0^{\sqrt{1-x^2}}{1 - y^2\,dy}\,dx}$?

3. Yes why?

4. Originally Posted by iva
Yes why?
The limits you wrote are wrong...

Anyway, I don't see what's so difficult about this - have you tried to evaluate this double integral?

BTW this integral won't represent area, area integrals always have 1 as the integrand...

5. Originally Posted by iva
I have this question where I need to evaluate this integral

$\displaystyle \int_{-1}^{-1} \int_{0}^{\sqrt{1-x^2}} 1-y^2 dydx$

When I do the sketch though it doesn't make sense because:

The region lies in the semi circle about line y=0 and circle with radius 1 right? But then the outer integral says that the area is cut off at lines x=1 and x=-1. Which doesn't cut off any of the semi circle as its on the tip as the circle radius is 1.. my obvious thought is that I'm doing something wrong. Can anyone confirm?

Does this integral just represent the area under the semicircle $\displaystyle \sqrt{1-x^2}$?

Thank you
The upper limit on your x integration is wrong. That's what Prove It was trying to verify.

The limits on the y integration tell you that you are integrating over a strip (of width dx at x) from 0 to $\displaystyle \sqrt{1 - x^2}$. The limits on the x integral are telling you do integrate over all such strips from x = -1 to x = 1, ie. the whole semi-circle.

-Dan

6. Hi There, why are they wrong though? that is how my question is written in the book. Topsquark also the way you describe it is how I saw it , over the whole semicircle, which is why i had doubts about my interpretation, so far the questions I've seen take sections of semicircles etc, why set the x limit at the ends where it won't really make a difference.
Prove it I haven't evaluated it yet because i first wanted to draw it to understand it, I then have to convert it to polar coordinates according to my question which i will still do. it was the region that didn't make sense to me

7. Originally Posted by iva
Hi There, why are they wrong though? that is how my question is written in the book. Topsquark also the way you describe it is how I saw it , over the whole semicircle, which is why i had doubts about my interpretation, so far the questions I've seen take sections of semicircles etc, why set the x limit at the ends where it won't really make a difference.
$\displaystyle \displaystyle \int_0^{\sqrt{1 - x^2}}dy$

This is not the integral over a semi-circle! We need the bounds on x to tell us how much of the semi-circle to integrate over. For example
$\displaystyle \displaystyle \int_0^1 \int_0^{\sqrt{1 - x^2}}dy~dx$
is only over 1/4 of the circle.

-Dan

Edit: The original bounds on the x integral were $\displaystyle \displaystyle \int_{-1}^{-1}dx$. Clearly this must be incorrect because any integral $\displaystyle \displaystyle \int_{-1}^{-1}f(x)~dx$ is 0 (for any reasonable function f(x).)

8. Sorry, I've just only now seen that i had -1 and -1, instead of 1 -1.

So if the bounds are -1 and 1 for x, this region is a semi circle right?

And then the 3d representation of the entire thing would be that part of the parabola 1-y^2 that is covered by that semicircle? (if i had a way to scan my sketch i could have explained this much better)