Integral Over a Sphere with dirac delta function

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SUMMARY

The discussion focuses on evaluating the integral of the Dirac delta function over the surface of a sphere. The integral is expressed as \(\underset{\left|\underline{\xi}\right|=1}{\int}\delta_{0}\left(\underline{\xi}\cdot\underline{z}\right)dS_{\xi}\) and is solved using a variable substitution \(\varsigma=\underline{\xi}\cdot\underline{z}=rcos\theta\). The final result of the integral is \(\frac{2\pi}{r}\), indicating the relationship between the radius and the delta function. The user clarifies that the radius used in the substitution differs from the unit radius.

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tim85ruhruniv
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Hi,

I am not really sure whether its over the surface of the sphere or the Volume,

the problem and the solution are given below, I want to know how it has been solved.
The [tex]\delta_{0}[/tex] is the dirac delta function.

[tex]\[<br /> \underset{\left|\underline{\xi}\right|=1}{\int}\delta_{0}\left(\underline{\xi}\cdot\underline{z}\right)dS_{\xi}=\intop_{0}^{2\pi}d\varphi\intop_{-r}^{+r}\delta_{0}\left(\varsigma\right)\frac{d\varsigma}{r}=\frac{2\pi}{r}\][/tex]

the following variable substitution has been made,

[tex]\[<br /> \varsigma=\underline{\xi}\cdot\underline{z}=rcos\theta\][/tex]

Thanx a lot.
 
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hey,

I Just got it,

I used this substitution.

[tex] \[<br /> \intop_{0}^{\pi}\intop_{0}^{2\pi}f(cos\varphi sin\theta,sin\varphi sin\theta,cos\theta)sin\theta d\theta d\varphi\][/tex]

the radius that i used in the variable substitution is not the same as the unit radius.
 

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