Zadatak 1. Naći površinu kardioide \( r = 2a(1+\cos\varphi) \)
Rešenje:
U polarnim koordinatama \(x=r\cos\varphi,\ y=r\sin\varphi\) Jakobijan je \(J=r\), pa je \(dx\,dy=r\,dr\,d\varphi\). Oblast je data kao \(0\le \varphi\le 2\pi\) i \(0\le r\le 2a(1+\cos\varphi)\).
\[ P=\iint\limits_{D} dx\,dy = \int_{0}^{2\pi}\!\!\int_{0}^{\,2a(1+\cos\varphi)} r\,dr\,d\varphi. \] \[ P=\int_{0}^{2\pi}\!\frac{r^{2}}{2}\Big|_{0}^{\,2a(1+\cos\varphi)} d\varphi = \frac{1}{2}\int_{0}^{2\pi} \bigl(2a(1+\cos\varphi)\bigr)^{2}\,d\varphi = 2a^{2}\int_{0}^{2\pi} \bigl(1+2\cos\varphi+\cos^{2}\varphi\bigr)\,d\varphi. \]Koristimo \(\cos^{2}\varphi=\frac{1+\cos 2\varphi}{2}\). Integral kosinusa jednak je nuli na \([0,2\pi]\), pa dobijamo:
\[ P = 2a^{2}\left(\int_{0}^{2\pi}1\,d\varphi + \int_{0}^{2\pi}\cos^{2}\varphi\,d\varphi\right) = 2a^{2}\left(2\pi + \pi\right) = 6\pi a^{2}. \]Zadatak 2. Naći zapreminu ispod površi \( z=5\left(1-\frac{r^{2}}{9}\right) \) nad oblasti \( 1\le r\le 3 \), \( 0\le\varphi\le \pi/2 \)
Rešenje:
U polarnim koordinatama \(x=r\cos\varphi,\ y=r\sin\varphi\) je Jakobijan \(J=r\). Data oblast je \(1\le r\le 3\), \(0\le\varphi\le \pi/2\). Tada
\[ V=\iint_{D} z\,dx\,dy =\int_{0}^{\pi/2}\!\!\int_{1}^{3} 5\left(1-\frac{r^{2}}{9}\right)\,r\,dr\,d\varphi. \]Računamo integral:
\[ \int_{1}^{3}\!\Bigl(r-\frac{r^{3}}{9}\Bigr)\,dr =\Bigl[\frac{r^{2}}{2}-\frac{r^{4}}{36}\Bigr]_{1}^{3} =\Bigl(\frac{9}{2}-\frac{81}{36}\Bigr)-\Bigl(\frac{1}{2}-\frac{1}{36}\Bigr) =\frac{16}{9}. \]Zato
\[ V=5\cdot\frac{16}{9}\cdot\int_{0}^{\pi/2}\!d\varphi =\frac{80}{9}\cdot\frac{\pi}{2} =\frac{40}{9}\,\pi. \]Zadatak 3. Naći površinu između \(x^2+y^2=2x\), prave \(y=0\) i prave \(y=x\)
Rešenje:
Uvedimo eliptične koordinate \(x=2r\cos\varphi,\ y=3r\sin\varphi\), pa je Jakobian \(J=\left|\frac{\partial(x,y)}{\partial(r,\varphi)}\right|=6r\). Oblast je \(0\le r\le 1,\ 0\le \varphi\le 2\pi\), a \[ 1-\frac{x^2}{4}-\frac{y^2}{9}=1-r^2. \] Zato je \[ V=\iint_D 5\left(1-\frac{x^2}{4}-\frac{y^2}{9}\right)\,dx\,dy =\int_{0}^{2\pi}\int_{0}^{1} 5(1-r^2)\,(6r)\,dr\,d\varphi. \] Računamo: \[ \int_{0}^{1} (1-r^2)\,r\,dr=\int_{0}^{1} (r-r^3)\,dr =\left(\frac{r^2}{2}-\frac{r^4}{4}\right)\Big|_{0}^{1}=\frac{1}{4}. \] Pa sledi \[ V = 6\cdot 5 \cdot \int_{0}^{2\pi}\frac{1}{4}\,d\varphi = 30 \cdot \frac{1}{4}\cdot 2\pi = 15\pi. \]Na početku je sve isključeno. Uključi elemente po potrebi.
Zadatak 4. \(\displaystyle \iint\limits_D (3x^2+2y^2)\,dx\,dy\), gde je \(D:\ 1\le r\le 3,\ -\frac{\pi}{6}\le \varphi\le \frac{\pi}{3}\)
Rešenje:
U polarnim koordinatama je \(x=r\cos\varphi,\ y=r\sin\varphi\), pa \[ 3x^2+2y^2 = 3r^2\cos^2\varphi + 2r^2\sin^2\varphi = r^2\big(3\cos^2\varphi + 2\sin^2\varphi\big), \] a \(dx\,dy = r\,dr\,d\varphi\). Dakle \[ I=\int_{-\pi/6}^{\pi/3}\int_{1}^{3} r^2\big(3\cos^2\varphi + 2\sin^2\varphi\big)\,r\,dr\,d\varphi =\Big(\int_{1}^{3} r^{3}\,dr\Big)\!\cdot\! \Big(\int_{-\pi/6}^{\pi/3}\big(3\cos^2\varphi + 2\sin^2\varphi\big)\,d\varphi\Big). \] \[ \int_{1}^{3} r^{3}\,dr=\left(\frac{r^4}{4}\right)\Big|_{1}^{3}=20. \] Za ugaoni deo koristimo \(3\cos^2\varphi + 2\sin^2\varphi =\frac{5}{2}+\frac{1}{2}\cos 2\varphi\): \[ \int_{-\pi/6}^{\pi/3}\!\left(\frac{5}{2}+\frac{1}{2}\cos 2\varphi\right)\,d\varphi =\frac{5}{2}\cdot\frac{\pi}{2}+\frac{1}{4}\left(\sin 2\varphi\right)\Big|_{-\pi/6}^{\pi/3} =\frac{5\pi}{4}+\frac{\sqrt{3}}{4}. \] Konačno, \[ I=20\left(\frac{5\pi}{4}+\frac{\sqrt{3}}{4}\right) = 25\pi+5\sqrt{3}. \]Granice: \(1\le x^2+y^2\le 9\), \(y\ge -x/\sqrt{3}\), \(y\le \sqrt{3}x\).
Zadatak 5. Površina oblasti između \(r=1\) i \(r=4\cos\varphi\) za \(0\le \varphi\le \frac{\pi}{3}\)
Rešenje:
Pošto je na intervalu \(0\le\varphi\le\pi/3\) važeće \(4\cos\varphi\ge 1\), površina je \[ P=\int_{0}^{\pi/3}\int_{1}^{\,4\cos\varphi} r\,dr\,d\varphi =\int_{0}^{\pi/3}\left(\frac{r^2}{2}\right)\Big|_{1}^{\,4\cos\varphi}\,d\varphi =\frac{1}{2}\int_{0}^{\pi/3}\big(16\cos^2\varphi-1\big)\,d\varphi. \] \[ \int_{0}^{\pi/3}\cos^2\varphi\,d\varphi =\left(\frac{\varphi}{2}+\frac{\sin 2\varphi}{4}\right)\Big|_{0}^{\pi/3} =\frac{\pi}{6}+\frac{\sqrt{3}}{8}. \] Zato \[ P=\frac{1}{2}\left(16\left(\frac{\pi}{6}+\frac{\sqrt{3}}{8}\right)-\frac{\pi}{3}\right) =\frac{1}{2}\left(\frac{8\pi}{3}+2\sqrt{3}-\frac{\pi}{3}\right) =\frac{7\pi}{6}+\sqrt{3}. \]Granice: \(x^2+y^2\ge 1\), \(x^2+y^2\le 4x\), \(y\ge0\), \(y\le \sqrt{3}x\).