# Integral and Heaviside

• December 16th 2008, 10:10 AM
totalnewbie
Integral and Heaviside
How to calculate integral if Heaviside sign is under integral? What does that Heaviside sign change? For example what does happen if I substitute Heaviside sign with plain number 1?
• December 16th 2008, 10:22 AM
Moo
Hello,
Quote:

Originally Posted by totalnewbie
How to calculate integral if Heaviside sign is under integral? What does that Heaviside sign change? For example what does happen if I substitute Heaviside sign with plain number 1?

Actually, it helps to restrain the limits of integration.

We know that the Heaviside function is 1 iff its argument is positive.
Hence :

$\bold{1}(1-x^2-y^2)=\left\{\begin{array}{ll} 1 \quad if ~ 1-x^2-y^2>0 \\
0 \quad if ~ 1-x^2-y^2 \leq 0 \end{array}\right.$

So find the new boundaries of x and y by "solving" $1-x^2-y^2>0$
That is $x^2+y^2<1$

Do you know how to deal with that ?
• December 16th 2008, 10:45 AM
totalnewbie
Quote:

Originally Posted by Moo
Hello,

Actually, it helps to restrain the limits of integration.

We know that the Heaviside function is 1 iff its argument is positive.
Hence :

$\bold{1}(1-x^2-y^2)=\left\{\begin{array}{ll} 1 \quad if ~ 1-x^2-y^2>0 \\
0 \quad if ~ 1-x^2-y^2 \leq 0 \end{array}\right.$

So find the new boundaries of x and y by "solving" $1-x^2-y^2>0$
That is $x^2+y^2<1$

Do you know how to deal with that ?

Not sure.
• December 16th 2008, 10:50 AM
Moo
Quote:

Originally Posted by totalnewbie
Not sure.

It's like dealing with double integrals.
$x^2+y^2<1$

So $x^2<1-y^2$
Since $x^2 \geq 0$, $1-y^2>0$, so that makes y between -1 and 1.

And $x^2<1-y^2 \implies |x|< \sqrt{1-y^2}$ (we can write it because 1-y²>0)
So x is between $-\sqrt{1-y^2}$ and $\sqrt{1-y^2}$

The integral is now :
$\int_{-1}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} a*1 ~ dxdy$
• December 17th 2008, 01:04 PM
ThePerfectHacker
Quote:

Originally Posted by Moo
It's like dealing with double integrals.
$x^2+y^2<1$

So $x^2<1-y^2$
Since $x^2 \geq 0$, $1-y^2>0$, so that makes y between -1 and 1.

And $x^2<1-y^2 \implies |x|< \sqrt{1-y^2}$ (we can write it because 1-y²>0)
So x is between $-\sqrt{1-y^2}$ and $\sqrt{1-y^2}$

The integral is now :
$\int_{-1}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} a*1 ~ dxdy$

(Hi) Moo, I am not trying to insult you, but there seems to be an easier way.

When you integrate, $\iint_D 1 ~ dA$ you just need to compute the area of $D$.
• December 18th 2008, 03:19 AM
Moo
Quote:

Originally Posted by ThePerfectHacker
(Hi) Moo, I am not trying to insult you, but there seems to be an easier way.

When you integrate, $\iint_D 1 ~ dA$ you just need to compute the area of $D$.

I don't know the formula http://freequest.the-cyber-cube-dev....sed_smiley.png
And furthermore when I see what you've written, I cannot stop thiking about measures ><