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Math Help - Integral and Heaviside

  1. #1
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    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. #2
    Moo
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    Hello,
    Quote Originally Posted by totalnewbie View Post
    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 ?
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    Quote Originally Posted by Moo View Post
    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.
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  4. #4
    Moo
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    Quote Originally Posted by totalnewbie View Post
    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-y>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
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    Quote Originally Posted by Moo View Post
    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-y>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.
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    Moo
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    Quote Originally Posted by ThePerfectHacker View Post
    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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