3.(4 boda) Prelaskom na polarne koordinate izračunajte po donjoj
polovini kružnog vijenca .

Ask by Williams Mcguire. in Croatia

Jan 21,2025

Question

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

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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:
$$
\boxed{\dfrac{14}{3}}
$$
Vrijednost integrala je $$ \frac{14}{3} $$.

3.(4 boda) Prelaskom na polarne koordinate izračunajte po donjoj
polovini kružnog vijenca .

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3.(4 boda) Prelaskom na polarne koordinate izračunajte po donjoj polovini kružnog vijenca .