# Math Help - Integral and Heaviside

1. ## 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?

2. Hello,
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 ?

3. 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.

4. 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 :
$\int_{-1}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} a*1 ~ dxdy$

5. 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 :
$\int_{-1}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} a*1 ~ dxdy$
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 $D$.

6. Originally Posted by ThePerfectHacker
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
And furthermore when I see what you've written, I cannot stop thiking about measures ><