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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- Dec 16th 2008, 10:10 AMtotalnewbieIntegral 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?

- Dec 16th 2008, 10:22 AMMoo
Hello,

Actually, it helps to restrain the limits of integration.

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

Hence :

$\displaystyle \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" $\displaystyle 1-x^2-y^2>0$

That is $\displaystyle x^2+y^2<1$

Do you know how to deal with that ? - Dec 16th 2008, 10:45 AMtotalnewbie
- Dec 16th 2008, 10:50 AMMoo
It's like dealing with double integrals.

$\displaystyle x^2+y^2<1$

So $\displaystyle x^2<1-y^2$

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

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

So x is between $\displaystyle -\sqrt{1-y^2}$ and $\displaystyle \sqrt{1-y^2}$

The integral is now :

$\displaystyle \int_{-1}^1 \int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}} a*1 ~ dxdy$ - Dec 17th 2008, 01:04 PMThePerfectHacker
- Dec 18th 2008, 03:19 AMMoo
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 ><