Cilindrične koordinate u trostrukom integralu

Izračunajte integral

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

ako je $\begin{displaymath}\displaystyle V\ldots \left\{ \begin{array}{c} x^{2}+z^{2}\leq 1 \\ 0\leq y\leq 1 \end{array} \right. \end{displaymath}$ .

Rješenje.

Područje integracije zadanog integrala prikazano je na slici 4.12.

**Slika:** Područje integracije zadano s $x^{2}+z^{2}\leq 1$ i $0\leq y\leq 1$ .
$\begin{figure}\centering \epsfig{file=sl29_trostruki_integral.eps, width=5cm} \end{figure}$

Uvedimo cilindrične koordinate [M2, poglavlje 4.3.1], ali tako da polarni koordinatni sustav bude u ravnini :

U novim koordinatama je element volumena jednak

$\displaystyle dV=r dr d\varphi dz$ ,

odnosno, Jakobijan je

, pa je integral jednak

$\displaystyle \iiint\limits_{V}\left( x^{2}+y+z^{2}\right) ^{3} dx dy dz$	$\displaystyle =\int\limits_{0}^{2\pi } d\varphi\int\limits_{0}^{1} dr\int\limits_{0}^{1}\left( r^{2}+y\right) ^{3} r dy$
	$\displaystyle =\int\limits_{0}^{2\pi } d\varphi \int\limits_{0}^{1} r\frac{\left( r^{2}+y\right) ^{4}}{4}\underset{0}{\overset{1}{\bigg\vert}}dr$
	$\displaystyle =\int\limits_{0}^{2\pi } d\varphi \int\limits_{0}^{1} \left[ \frac{1}{4} r\left( r^{2}+1\right) ^{4}-\frac{1}{4}r^{9}\right] dr$
	$\displaystyle =\int\limits_{0}^{2\pi } \left[ \frac{1}{8}\frac{\left( r^{2}+1\... ...1}{4} \frac{r^{10}}{10}\right] \underset{0}{\overset{1}{\bigg\vert}} d\varphi$
	$\displaystyle =\int\limits_{0}^{2\pi }\frac{3}{4} d\varphi =\frac{3}{4}\varphi \underset {0}{\overset{2\pi }{\bigg\vert}}=\frac{3\pi }{2}$ .