Integrovanie funkcie viacerých reálnych premenných 

Trojný integrál  

 

Príklad 5: Vypočítajte
$$\int_0^4\int_{\sqrt{u}}^{2\sqrt{u}}\int_0^{4-u}1 \ \mathrm{d}\varphi \ \mathrm{d}\rho \ \mathrm{d}u. $$

Riešenie:
$$\int_0^4\int_{\sqrt{u}}^{2\sqrt{u}}\int_0^{4-u}1 \ \mathrm{d}\varphi\ \mathrm{d}\rho \ \mathrm{d}u=\int_0^4\int_{\sqrt{u}}^{2\sqrt{u}}\big[\varphi\big]_0^{4-u} \ \mathrm{d}\rho \ \mathrm{d}u$$
$$=\int_0^4\int_{\sqrt{u}}^{2\sqrt{u}}((4-u)-0) \ \mathrm{d}\rho \ \mathrm{d}u=\int_0^4 \big[4\rho-u\rho\big]_{\sqrt{u}}^{2\sqrt{u}} \ \mathrm{d}u$$
$$=  \int_0^4\big(4(2\sqrt{u})-u(2\sqrt{u})\big)-(4\sqrt{u}-u\sqrt{u}) \ \mathrm{d}u $$
$$=\int_0^{4} \left(8u^{\frac{1}{2}}-2u^{\frac{3}{2}}-4u^{\frac{1}{2}}+u^{\frac{3}{2}}\right) \ \mathrm{d}u=\int_0^{4} \left(4u^{\frac{1}{2}}-1u^{\frac{3}{2}}\right) \ \mathrm{d}u=\left[4\frac{u^{\frac{3}{2}}}{\frac{3}{2}}-\frac{u^{\frac{5}{2}}}{\frac{5}{2}}\right]_0^4$$
$$= \left[\frac{8}{3}u^{\frac{3}{2}}-\frac{2}{5}u^{\frac{5}{2}}\right]_0^4=\left[\left(\frac{8}{3}4^{\frac{3}{2}}-\frac{2}{5}4^{\frac{5}{2}}\right)-\left(\frac{8}{3}0^{\frac{3}{2}}-\frac{2}{5}0^{\frac{5}{2}}\right)\right]$$
$$=\frac{8}{3}(\sqrt{4})^3-\frac{2}{5}(\sqrt{4})^5=\frac{8}{3}8-\frac{2}{5}32=\frac{64}{3}-\frac{64}{5}=\frac{5.64-3.64}{15}=\frac{128}{15}.$$