İntegral :
$$\iiint\limits_{\mathbb{V}}\:x^{\alpha-1}y^{\beta-1}z^{\gamma-1}\:d\mathbb{V}$$
$$\mathbb{V}:x^2+y^2+z^2\le1\:\:\:,\:\:\:x,y,z\geq0$$
Yeni değişkenlerimiz :
$$u=x^2$$
$$v=y^2$$
$$w=z^2$$
İntegralin yeni hali :
$$\iiint\limits_{\mathbb{S}}\:u^{\frac{\alpha}{2}-\frac{1}{2}}v^{\frac{\beta}{2}-\frac{1}{2}}w^{\frac{\gamma}{2}-\frac{1}{2}}\:\det\:J(u,v,w)\:d\mathbb{S}$$
$$\mathbb{S}:u+v+w\le1\:\:\:,\:\:\:u,v,w\geq0$$
Jacobian matrisinin determinantını bulalım.
$$J(u,v,w)=\begin{bmatrix}\frac{\partial\:u}{\partial\:x}&\frac{\partial\:u}{\partial\:y}&\frac{\partial\:u}{\partial\:z}\\\frac{\partial\:v}{\partial\:x}&\frac{\partial\:v}{\partial\:y}&\frac{\partial\:v}{\partial\:z}\\\frac{\partial\:w}{\partial\:x}&\frac{\partial\:w}{\partial\:y}&\frac{\partial\:w}{\partial\:z}\end{bmatrix}=\begin{bmatrix}\frac{1}{2\sqrt{u}}&0&0\\0&\frac{1}{2\sqrt{v}}&0\\0&0&\frac{1}{2\sqrt{w}}\end{bmatrix}$$
$$\det\:J(u,v,w)=\frac{1}{8\,\sqrt{u\:v\:w}}$$
İntegralde yerine yazalım :
$$\large\color{#A00000}{\frac{1}{8}\iiint\limits_{\mathbb{S}}\:u^{\frac{\alpha}{2}-1}\:v^{\frac{\beta}{2}-1}\:w^{\frac{\gamma}{2}-1}\:\:d\mathbb{S}\\\mathbb{S}:u+v+w\le1\:\:\:,\:\:\:u,v,w\geq0}$$
Sınır değerlerini yazalım.
$$\frac{1}{8}\int_0^1\int_0^{1-w}\int_0^{1-v-w}\:u^{\frac{\alpha}{2}-1}\:v^{\frac{\beta}{2}-1}\:w^{\frac{\gamma}{2}-1}\:du\:dv\:dw$$
İlk integrali çözelim.
$$\frac{1}{8}\int_0^1\int_0^{1-w}\underbrace{\bigg(\int_0^{1-v-w}\:u^{\frac{\alpha}{2}-1}\:du\bigg)}_{\large\big[\frac{2}{\alpha}u^{\frac{\alpha}{2}}\big]_0^{1-v-w}\to\:\frac{2}{\alpha}(1-v-w)^{\frac{\alpha}{2}}}\:v^{\frac{\beta}{2}-1}\:w^{\frac{\gamma}{2}-1}\:dv\:dw$$
$$\frac{1}{8}\frac{2}{\alpha}\int_0^1\int_0^{1-w}\:(1-v-w)^{\frac{\alpha}{2}}\:v^{\frac{\beta}{2}-1}\:w^{\frac{\gamma}{2}-1}\:dv\:dw$$
$w=(1-v)k$ olacak şekilde değişken değiştirelim.
$$\frac{1}{8}\frac{2}{\alpha}\int_0^1\int_0^1\:u^{\frac{\beta}{2}-1}\:(1-u)^{\frac{\alpha}{2}+\frac{\gamma}{2}}\:k^{\frac{\gamma}{2}-1}\:(1-k)^{\frac{\alpha}{2}}\:du\:dk$$
İntegrali iki ayrı integral halinde yazalım.
$$\frac{1}{8}\frac{2}{\alpha}\int_0^1u^{\frac{\beta}{2}-1}\:(1-u)^{\frac{\alpha}{2}+\frac{\gamma}{2}}\:du\:\int_0^1\:k^{\frac{\gamma}{2}-1}\:(1-k)^{\frac{\alpha}{2}}\:dk$$
İntegralleri beta fonksiyonu ile yazabiliriz.
$$\frac{1}{8}\frac{2}{\alpha}\underbrace{\int_0^1u^{\frac{\beta}{2}-1}\:(1-u)^{\frac{\alpha}{2}+\frac{\gamma}{2}}\:du}_{\large\:B\big(\frac{\beta}{2},\frac{\alpha}{2}+\frac{\gamma}{2}+1\big)}\:\underbrace{\int_0^1\:k^{\frac{\gamma}{2}-1}\:(1-k)^{\frac{\alpha}{2}}\:dk}_{\large\:B\big(\frac{\gamma}{2},\frac{\alpha}{2}+1\big)}$$
$$\frac{1}{8}\frac{2}{\alpha}B\bigg(\frac{\beta}{2},\frac{\alpha}{2}+\frac{\gamma}{2}+1\bigg)B\bigg(\frac{\gamma}{2},\frac{\alpha}{2}+1\bigg)$$
Beta fonksiyonlarını gama fonksiyonu ile yazabiliriz.
$$\frac{1}{8}\frac{2}{\alpha}\frac{\Gamma\bigg(\frac{\beta}{2}\bigg)\Gamma\bigg(\frac{\beta}{2}+\frac{\gamma}{2}+1\bigg)}{\Gamma\bigg(\frac{\alpha}{2}+\frac{\beta}{2}+\frac{\gamma}{2}+1\bigg)}\frac{\Gamma\bigg(\frac{\gamma}{2}\bigg)\Gamma\bigg(\frac{\alpha}{2}+1\bigg)}{\Gamma\bigg(\frac{\beta}{2}+\frac{\gamma}{2}+1\bigg)}$$
Sadeleştirelim.
$$\large\color{#C00000}{\boxed{\iiint\limits_{\mathbb{V}}\:x^{\alpha-1}y^{\beta-1}z^{\gamma-1}\:d\mathbb{V}=\frac{\Gamma\bigg(\frac{\alpha}{2}\bigg)\Gamma\bigg(\frac{\beta}{2}\bigg)\Gamma\bigg(\frac{\gamma}{2}\bigg)}{8\:\Gamma\bigg(\frac{\alpha}{2}+\frac{\beta}{2}+\frac{\gamma}{2}+1\bigg)}\\\mathbb{V}:x^2+y^2+z^2\le1\:\:\:,\:\:\:x,y,z\geq0}}$$