Krivuljni integral prve vrste na Geodetskom fakultetu
Na pismenom ispitu iz Vektorske analize 20. 6. 2011. je bio sljedeći zadatak:
Izračunajte
gdje je k prvi zavoj cilindrične zavojnice
Rješenje.
Izračunajmo prvo koliki je ds. Kako je
to je
![\begin{array}{rcl} ds &=& \displaystyle \sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2}\, dt=\sqrt{(-a\sin t)^2+(a\cos t)^2+a^2}\, dt\\[15pt] &=& \displaystyle \sqrt{a^2\sin^2 t + a^2\cos^2 t + a^2}\, dt = a\sqrt{2}\, dt \end{array}](http://s0.wp.com/latex.php?latex=%5Cbegin%7Barray%7D%7Brcl%7D++ds+%26%3D%26+%5Cdisplaystyle+%5Csqrt%7B%5Cdot%7Bx%7D%5E2%2B%5Cdot%7By%7D%5E2%2B%5Cdot%7Bz%7D%5E2%7D%5C%2C+dt%3D%5Csqrt%7B%28-a%5Csin+t%29%5E2%2B%28a%5Ccos+t%29%5E2%2Ba%5E2%7D%5C%2C+dt%5C%5C%5B15pt%5D++%26%3D%26+%5Cdisplaystyle+%5Csqrt%7Ba%5E2%5Csin%5E2+t+%2B+a%5E2%5Ccos%5E2+t+%2B+a%5E2%7D%5C%2C+dt+%3D+a%5Csqrt%7B2%7D%5C%2C+dt++%5Cend%7Barray%7D&bg=ffffff&fg=333333&s=0 "\begin{array}{rcl} ds &=& \displaystyle \sqrt{\dot{x}^2+\dot{y}^2+\dot{z}^2}\, dt=\sqrt{(-a\sin t)^2+(a\cos t)^2+a^2}\, dt\\[15pt] &=& \displaystyle \sqrt{a^2\sin^2 t + a^2\cos^2 t + a^2}\, dt = a\sqrt{2}\, dt \end{array}")
pa je, jer su granice 
![\begin{array}{rcl} \displaystyle \int_k \frac{z^2}{x^2+y^2}\, ds &=& \displaystyle \int_0^{2\pi}\frac{(at)^2}{(a\cos t)^2 + (a\sin t)^2}\, a\sqrt{2}\, dt\\[20pt] &=& \displaystyle a\sqrt{2}\int_0^{2\pi} t^2\, dt = \frac{8\sqrt{2} \pi^3 a}{3} \end{array}](http://s0.wp.com/latex.php?latex=%5Cbegin%7Barray%7D%7Brcl%7D++%5Cdisplaystyle+%5Cint_k+%5Cfrac%7Bz%5E2%7D%7Bx%5E2%2By%5E2%7D%5C%2C+ds++%26%3D%26+%5Cdisplaystyle+%5Cint_0%5E%7B2%5Cpi%7D%5Cfrac%7B%28at%29%5E2%7D%7B%28a%5Ccos+t%29%5E2+%2B+%28a%5Csin+t%29%5E2%7D%5C%2C+a%5Csqrt%7B2%7D%5C%2C+dt%5C%5C%5B20pt%5D++%26%3D%26+%5Cdisplaystyle+a%5Csqrt%7B2%7D%5Cint_0%5E%7B2%5Cpi%7D+t%5E2%5C%2C+dt+%3D+%5Cfrac%7B8%5Csqrt%7B2%7D+%5Cpi%5E3+a%7D%7B3%7D++%5Cend%7Barray%7D&bg=ffffff&fg=333333&s=0 "\begin{array}{rcl} \displaystyle \int_k \frac{z^2}{x^2+y^2}\, ds &=& \displaystyle \int_0^{2\pi}\frac{(at)^2}{(a\cos t)^2 + (a\sin t)^2}\, a\sqrt{2}\, dt\\[20pt] &=& \displaystyle a\sqrt{2}\int_0^{2\pi} t^2\, dt = \frac{8\sqrt{2} \pi^3 a}{3} \end{array}")
