Školska 2017/18, junski ispitni rok, II kolokvijum grupa 1

Data oblast: \[ x^2 + y^2 = 4y, \quad z = 0, \quad z = 4 – x^2 – y^2 \] Cilindar je: \[ x^2 + y^2 = 4y \quad \Leftrightarrow \quad x^2 + (y-2)^2 = 4 \] Krug ima radijus 2 sa centrom u tački (0,2). Za tačku na tom krugu možemo uzeti parametar \(\theta\) (ugao meren od pozitivne x-ose): \[ \{x = 2 \cos \theta, \quad y = 2 + 2 \sin \theta, \quad 0 \le \theta \le 2\pi \} \] Za takvu tačku izračunamo: \[ x^2 + y^2 = 4 \cos^2 \theta + (2 + 2 \sin \theta)^2 = 4 \cos^2 \theta + (2 + 2 \sin \theta)^2 = 8(1+\sin \theta) \] Paraboloid: \[ z_p = 4 – x^2 – y^2 = 4 – 8(1+\sin \theta) = -4 – 8 \sin \theta \] Donja ravan je \(z=0\) \[ -4 – 8 \sin \theta \ge 0 \quad \Leftrightarrow \quad \sin \theta \le -\frac{1}{2} \] Granice za \(\theta\): \[ \theta = \frac{7\pi}{6}, \quad \frac{11\pi}{6} \] Površina je tada: \[ \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} 2(-4 – 8 \sin \theta) \, d\theta = \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} (-8 -16 \sin \theta) \, d\theta = \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} -8 \, d\theta – \int_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} 16 \sin \theta \, d\theta \] \[ = [-8 \theta + 16 \cos \theta]_{\frac{7\pi}{6}}^{\frac{11\pi}{6}} = \left(-\frac{44\pi}{3} + 8 \sqrt{3}\right) – \left(-\frac{28\pi}{3} – 8 \sqrt{3}\right) \] \[ = -\frac{16 \pi}{3} + 16 \sqrt{3} \]

\[ P = 16\sqrt{3} – \frac{16\pi}{3} \]

Iz teksta zadatka zaključujemo da je oblast:\[ D=\{(x,y)\in\mathbb{R}^2:\;x\le y,\;x\ge -y\}. \]Zbog \(x\le y\) i \(x\ge -y\) sledi da je nužno \( -y \le y \), tj. \( y \le 0 \). Uvedimo smenu \( t=-y\ge 0 \). Tada su granice \( 0\le t\le 1 \) (prema zadatom opsegu), pa \[ y\in[-1,0],\qquad x\in[-y,\;y]. \] Integral postaje: \[ I=\int_{-1}^{0}\left(\int_{-y}^{y}\frac{x^2}{x^2+y^2}\,dx\right)\!dy. \] Za unutrašnji integral: \[ \frac{x^2}{x^2+y^2}=1-\frac{y^2}{x^2+y^2}, \] pa \[ \int \frac{x^2}{x^2+y^2}\,dx =\int 1\,dx – y^2\!\int\frac{dx}{x^2+y^2} = x – |y|\,\mathrm{arctg}\!\frac{x}{|y|}. \] Ovde je \( |y|=-y \) jer \(y\le 0\). Za granice \(a=-y\), \(b=y\) (uz \(t=-y\)): \[ \int_{-y}^{y}\frac{x^2}{x^2+y^2}\,dx = (y-(-y)) – |y|\left(\mathrm{arctg}\frac{y}{|y|}-\mathrm{arctg}\frac{-y}{|y|}\right). \] Pošto je \(t=-y\ge 0\), \[ \mathrm{arctg}(-1)=-\frac{\pi}{4},\qquad \mathrm{arctg}(1/t) = \mathrm{arctg}\frac{1}{t}. \] Zato se izraz svodi na integral u promenljivoj \(t\): \[ I=\int_{0}^{1}\left(\frac{\pi}{4}t – t\,\mathrm{arctg}\frac{1}{t}\right)\,dt. \] Razložimo: \[ \frac{\pi}{4}\int_{0}^{1} t\,dt = \frac{\pi}{8}. \] Uvedimo \(J=\int_{0}^{1} t\,\mathrm{arctg}(1/t)\,dt.\) Smena \(t=u^2\): \[ J= \int_{0}^{1} 2u^3\,\mathrm{arctg}\frac{1}{u}\,du = \frac{\pi}{4} – 2\int_{0}^{1} u^3\,\mathrm{arctg} u\,du = \frac{\pi}{4} – 2K. \] Za \(K=\int_{0}^{1} u^3\,\mathrm{arctg} u\,du\): \[ K=\frac{\pi}{16}-\frac{1}{4}\!\int_{0}^{1}\frac{u^4}{1+u^2}\,du =\frac{\pi}{16}-\frac{\pi}{16}+\frac{1}{6} =\frac{1}{6}. \] Zato \(J=\frac{\pi}{4}-\frac{1}{3}\).

\[ I = \frac{1}{2} – \frac{\pi}{8} \]

Masa \(M\) tela gustine \(\rho(x,y,z)=2z\) je trojni integral:\[ M = \iiint_V \rho(x,y,z)\, dV \]Neka su cilindrične koordinate: \[ x = r \cos\theta, \quad y = r \sin\theta, \quad z = z \] Tada je: \[ x^2+y^2 = r^2 \] \[ x^2+y^2-z^2 = r^2 – z^2 \le 16 \implies r^2 \le 16+z^2, \quad r \in [0, \sqrt{16+z^2}] \] Jakobijan za cilindrične koordinate je: \[ dV = r\, dr\, d\theta\, dz \] Granice integracije: \[ z \in [1,3], \quad r \in [0, \sqrt{16+z^2}], \quad \theta \in [0, \pi] \] Integral postaje: \[ M = \int_1^3 \int_0^\pi \int_0^{\sqrt{16+z^2}} 2z \, r \, dr \, d\theta \, dz \] Unutrašnji integral po \(r\): \[ \int_0^{\sqrt{16+z^2}} 2z \, r \, dr = 2z \cdot \frac{r^2}{2} \Big|_0^{\sqrt{16+z^2}} = z(16+z^2) \] Srednji integral po \(\theta\): \[ \int_0^\pi z(16+z^2) \, d\theta = z(16+z^2) \cdot \pi \] Preostaje integral po \(z\): \[ \int_1^3 z(16+z^2) \pi \, dz = \pi \int_1^3 (16 z + z^3) \, dz \] Završni integral: \[ \pi \int_1^3 (16 z + z^3) \, dz = \pi \left[ 8 z^2 + \frac{z^4}{4} \right]_1^3 \] Za \(z=3\): \[ 8\cdot 9 + \frac{81}{4} = 72 + 20,25 = 92,25 \] Za \(z=1\): \[ 8\cdot 1 + \frac{1}{4} = 8 + 0,25 = 8,25 \] Razlika je: \[ 92,25 – 8,25 = 84\pi \]

\[ M = 84\pi \]