Bounds of integration

• Apr 27th 2013, 01:49 PM
Giestforlife
Bounds of integration
What would be my bounds of integration if I'm trying to integrate a function f(x,y) over a region bounded by y = x and y = -x2 ?

Giest
• Apr 27th 2013, 02:10 PM
MINOANMAN
Re: Bounds of integration
see the graph and find them your self

Attachment 28173
• Apr 27th 2013, 02:17 PM
Giestforlife
Re: Bounds of integration
It's not quite simple as that. The multivariate function that I'm integrating requires two iterated integrals, and the bounds for the inner integrand, I'm pretty sure, need to expressed as a function of one of the two variables.
• Apr 27th 2013, 02:37 PM
Shakarri
Re: Bounds of integration
If you have a double integral like this
$\int^a_b (\int^c_d f(x,y) dy)dx$

The limits c and d on the y integral are c=x and d=-x2
The limits a and b on the x integral are a=0 and b=-1 as minoanman shows in his graph

You may wonder why I said c=x and d=-x2 not d=x and c=-x2. x is always higher than -x2 and you want to integrate in the positive direction of y so the integral limits are FROM the lower value TO the higher value.
• Apr 27th 2013, 07:05 PM
Prove It
Re: Bounds of integration
Of course, using a double integral here is overkill when you can you can just do \displaystyle \begin{align*} A = \int_0^1{x - \left( -x^2 \right) \, dx} = \int_0^1{x + x^2 \, dx} \end{align*}
• Apr 27th 2013, 07:40 PM
Giestforlife
Re: Bounds of integration
Oh, I wish I could go back to those good days of single variable calculus. However, my problem is actually a line integral.

Attachment 28178

and C is the boundary of the region in the third quadrant bounded by y=x and y=-x^2 (traversed counterclockwise).

But, regardless, I appreciate the gesture.
• Apr 27th 2013, 08:07 PM
Prove It
Re: Bounds of integration
Quote:

Originally Posted by Giestforlife
Oh, I wish I could go back to those good days of single variable calculus. However, my problem is actually a line integral.

Attachment 28178

and C is the boundary of the region in the third quadrant bounded by y=x and y=-x^2 (traversed counterclockwise).

But, regardless, I appreciate the gesture.

And this is why one should post the ENTIRE problem they are working on, so as not to waste the helpers' time...
• Apr 28th 2013, 12:31 AM
Shakarri
Re: Bounds of integration
Quote:

Originally Posted by Prove It
Of course, using a double integral here is overkill when you can you can just do \displaystyle \begin{align*} A = \int_0^1{x - \left( -x^2 \right) \, dx} = \int_0^1{x + x^2 \, dx} \end{align*}

That doesn't make sense because you get the same answer no matter what f(x,y) is
• Apr 28th 2013, 12:51 AM
Prove It
Re: Bounds of integration
Quote:

Originally Posted by Shakarri
That doesn't make sense because you get the same answer no matter what f(x,y) is

This answer WOULD have been sufficient if the OP had posted the entire problem in the first place.
• Apr 28th 2013, 10:48 AM
Giestforlife
Re: Bounds of integration
I didn't mean to waste anyone's time. I just thought I'd keep things simple by giving the part of the problem that was putting me off--that and the fact it takes me a long time to do anything in LaTex. But I'll keep that in mind next time.