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