3.(4 boda) Prelaskom na polarne koordinate izračunajte \( \int_{S}(2 x y-x) d S \) po donjoj polovini kružnog vijenca \( x^{2}+y^{2} \geq 1 ; x^{2}+y^{2} \leq 4 \).

Da bismo izračunali integral \( \int_{S}(2xy - x) \, dS \) preko donje polovice kružnog vijenca \( x^{2} + y^{2} \geq 1 \) i \( x^{2} + y^{2} \leq 4 \) pomoću polarnih koordinata, slijedimo sljedeće korake: ### 1. Pretvorba u polarne koordinate U polarnoj koordinati: \[ x = r \cos \theta, \quad y = r \sin \theta \] Element površine \( dS \) u polarnoj koordinati je: \[ dS = r \, dr \, d\theta \] Dakle, izraz integranda postaje: \[ 2xy - x = 2(r \cos \theta)(r \sin \theta) - r \cos \theta = 2r^{2} \cos \theta \sin \theta - r \cos \theta \] ### 2. Definiranje granica integracije Kružni vijenc je definiran s \( 1 \leq r \leq 2 \). Donja polovica vijenca odgovara uglovima \( \theta \) od \( \frac{\pi}{2} \) do \( \frac{3\pi}{2} \): \[ \theta \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \] ### 3. Postavljanje integrala Integral postaje: \[ \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \int_{1}^{2} \left(2r^{2} \cos \theta \sin \theta - r \cos \theta\right) r \, dr \, d\theta = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \int_{1}^{2} \left(2r^{3} \cos \theta \sin \theta - r^{2} \cos \theta\right) dr \, d\theta \] ### 4. Izračunavanje unutarnjeg integrala po \( r \) \[ \int_{1}^{2} \left(2r^{3} \cos \theta \sin \theta - r^{2} \cos \theta\right) dr = 2 \cos \theta \sin \theta \left[\frac{r^{4}}{4}\right]_1^2 - \cos \theta \left[\frac{r^{3}}{3}\right]_1^2 \] \[ = 2 \cos \theta \sin \theta \left(\frac{16}{4} - \frac{1}{4}\right) - \cos \theta \left(\frac{8}{3} - \frac{1}{3}\right) \] \[ = 2 \cos \theta \sin \theta \cdot \frac{15}{4} - \cos \theta \cdot \frac{7}{3} = \frac{15}{2} \cos \theta \sin \theta - \frac{7}{3} \cos \theta \] ### 5. Izračunavanje vanjskog integrala po \( \theta \) \[ \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \left(\frac{15}{2} \cos \theta \sin \theta - \frac{7}{3} \cos \theta\right) d\theta \] Razdvajamo integral: \[ \frac{15}{2} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \cos \theta \sin \theta \, d\theta - \frac{7}{3} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \cos \theta \, d\theta \] Prvo računamo: \[ \int \cos \theta \sin \theta \, d\theta = \frac{1}{2} \sin^{2} \theta \] Dakle: \[ \frac{15}{2} \left[\frac{1}{2} \sin^{2} \theta \right]_{\frac{\pi}{2}}^{\frac{3\pi}{2}} = \frac{15}{4} (\sin^{2}(\frac{3\pi}{2}) - \sin^{2}(\frac{\pi}{2})) = \frac{15}{4} (1 - 1) = 0 \] Zatim računamo: \[ \frac{7}{3} \left[ \sin \theta \right]_{\frac{\pi}{2}}^{\frac{3\pi}{2}} = \frac{7}{3} (\sin(\frac{3\pi}{2}) - \sin(\frac{\pi}{2})) = \frac{7}{3} (-1 - 1) = -\frac{14}{3} \] Ukupno: \[ 0 - \left(-\frac{14}{3}\right) = \frac{14}{3} \] ### Zaključak Vrijednost integrala je: \[ \boxed{\dfrac{14}{3}} \]