# Thread:

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?
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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 :

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

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 :

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

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 ><
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