Sferne koordinate u trostrukom integralu

Izračunajte integral

$\displaystyle I=\iiint\limits_{V}\sqrt{ 1+\left( x^{2}+y^{2}+z^{2}\right) ^{\frac{3}{2}}} dx dy dz$

ako je $\displaystyle V\ldots x^{2}+y^{2}+z^{2}\leq 1$ .

Rješenje.

Područje integracije integrala prikazano je na Slici 4.13.

**Slika:** Područje integracije $\displaystyle V\ldots x^{2}+y^{2}+z^{2}\leq 1$ .
$\begin{figure}\centering \epsfig{file=sl30_trostruki_integral.eps, width=5cm} \end{figure}$

Prijeđimo na sferne koordinate [M2, poglavlje 4.3.1]:

$\displaystyle x$	$\displaystyle =r\sin\theta \cos \varphi$ ,
$\displaystyle y$	$\displaystyle =r\sin\theta \sin \varphi$ ,
$\displaystyle z$	$\displaystyle =r\cos\theta$ .

Jakobijan je $\displaystyle J=r^2\sin\theta$ . Integriramo po cijeloj kugli, pa za nove varijable vrijedi

$\displaystyle \varphi \in \left[ 0,2\pi \right]$ , $\displaystyle \quad\Theta \in \left[ 0,\pi \right]$ , $\displaystyle \quad r\in\left[ 0,1\right]$ .

i imamo

$\displaystyle I$	$\displaystyle = \int\limits_{0}^{2\pi } d\varphi \int\limits_{0}^{\pi } \sin \Theta d\Theta \int\limits_{0}^{1}\sqrt{1+r^{3}}r^{2} dr$
	$\displaystyle =\left\{ \begin{array}{l} 1+r^{3}=t r^{2} dr=\frac{1}{3}... ...} $r$ & $0$ & $1$ \hline $t$ & $1$ & $2$ \end{tabular} \right\}$
	$\displaystyle =\int\limits_{0}^{2\pi } d\varphi \int\limits_{0}^{\pi } \sin \Theta d\Theta \int\limits_{1}^{2}\frac{1}{3}t^{\frac{1}{2}} dt$
	$\displaystyle =\varphi \underset{0}{\overset{2\pi }{\bigg\vert}\cdot }\left( -\... ...ac{t^{\frac{3}{2}}}{\frac{3}{2}}\right) \underset{1}{\overset{2}{ \bigg\vert}}$
	$\displaystyle =\frac{2}{9} 2\pi \left( - \cos \pi +\cos 0\right) \left( 2\sqrt{2} -1\right)$
	$\displaystyle =\frac{8\pi }{9}\left( 2\sqrt{2}-1\right)$ .

Cilindrične koordinate u trostrukom VIŠESTRUKI INTEGRALI Volumen tijela