April 22, 2018

Dvostruki integral 1

Izračunajte dvostruki integral na području


Najprije lijepo nacrtamo sliku područja. Slika područja Očito trebamo prijeći na eliptične koordinate. \[ \begin{cases} x = \rho cos \varphi\\ y = \rho sin \varphi \end{cases} \quad 0 < \rho < 1, \quad 0 < \varphi < \pi \] Sad kad imamo eliptične koordinate i znamo granice možemo rješavati.

\[\int_{0}^{1} \,d\rho \int_{0}^{\pi} \frac{1-\rho^2 cos^2 \varphi-\rho^2 sin^2 \varphi}{1+\rho^2 cos^2 \varphi+\rho^2 sin^2 \varphi} \cdot \rho \,d\varphi\] \[\int_{0}^{1} \rho \,d\rho \int_{0}^{\pi} \frac{1-\rho^2}{1+\rho^2} \,d\varphi\] \[\int_{0}^{1} \rho \cdot \frac{1-\rho^2}{1+\rho^2} \,d\rho \cdot \int_{0}^{\pi} \,d\varphi\] Uvodimo supstituciju , pa to deriviramo i dobijemo

I granice se mijenjaju pa imamo i

\[\frac{1}{2}\int_{0}^{1} \frac{1-t}{1+t} \,dt \cdot \Big(\varphi\Big\vert_{0}^{\pi}\] \[\frac{1}{2}\int_{0}^{1} \frac{1+t-2t}{1+t} \,dt \cdot \pi\] \[\frac{\pi}{2}\Big(\int_{0}^{1} \,dt - 2\int_{0}^{1}\frac{t}{1+t} \,dt \Big)\] \[\frac{\pi}{2}\Big(1 - 2\int_{0}^{1}\frac{1+t-1}{1+t} \,dt \Big)\] \[\frac{\pi}{2}\bigg(1 - 2\Big(\int_{0}^{1} \,dt - \int_{0}^{1}\frac{\,dt}{1+t} \Big)\bigg)\] \[\frac{\pi}{2}\bigg(1 - 2 + 2\Big(ln(1+t)\Big\vert_{0}^{1}\bigg)\] \[\frac{\pi}{2}\Big(-1 + 2ln2\Big)\] \[\frac{\pi}{2}\Big(ln4-1\Big)\]