Školska 2017/18, septembarski ispitni rok, II kolokvijum grupa B

Presek: \(x^2+y^2+z=6,\ x^2+y^2+z^2=6z \Rightarrow z\in\{1,6\}\); relevantno je \(z=1\) pa \(x^2+y^2=5\).

Uslovi \(x\le0,\ y\le0,\ y\le x\) biraju luk u III kvadrantu sa \(\theta\in[\frac{5\pi}{4},\frac{3\pi}{2}]\); krajevi su \(P_1:(x_1,y_1)=(-\frac{\sqrt{10}}{2},-\frac{\sqrt{10}}{2})\) i \(P_2:(x_2,y_2)=(0,-\sqrt5)\).

Na luku je \(z\equiv1\) i \(dz=0\), zato \(\displaystyle \int_{\ell}(y\,dx+z\,dy+x\,dz)=\int_{\ell}(y\,dx+dy)\).

Njutn–Lajbnic: pošto je \(d(xy)=x\,dy+y\,dx\), imamo \(y\,dx=d(xy)-x\,dy\). Otuda \(\displaystyle \int_{\ell}(y\,dx+dy)=[xy]_{P_1}^{P_2}+\int_{\ell}(1-x)\,dy\) — prvi deo zavisi samo od krajnjih tačaka.

Vrednosti u krajevima: \([xy]_{P_1}^{P_2}=x_2y_2-x_1y_1=0-{\frac{10}{4}}=-\frac{5}{2}\), a \(y_2-y_1=-\sqrt5+\frac{\sqrt{10}}{2}\).

Račun \(\int_{\ell}x\,dy\): sa \(x^2+y^2=5\) i negativnom granom \(x=-\sqrt{5-y^2}\) dobijamo \(\displaystyle \int_{\ell}x\,dy=\int_{y_1}^{y_2}(-\sqrt{5-y^2})\,dy=-\Big[\frac{y}{2}\sqrt{5-y^2}+\frac{5}{2}\arcsin\!\frac{y}{\sqrt5}\Big]_{y_1}^{y_2}=-\frac{5}{4}+\frac{5\pi}{8}\).

Sastavljanjem: \(\displaystyle \int_{\ell}(y\,dx+dy)=[xy]+(y_2-y_1)-\int_{\ell}x\,dy=-\frac{5}{2}+(-\sqrt5+\frac{\sqrt{10}}{2})-\big(-\frac{5}{4}+\frac{5\pi}{8}\big)\).

Rezultat: \(\displaystyle \int_{\ell}(y\,dx+z\,dy+x\,dz)=-\frac{5\pi}{8}-\frac{5}{4}-\sqrt5+\frac{\sqrt{10}}{2}\).

Traži se \( \iint_{D} x\,dx\,dy \), gde je
\( D = \{(x,y)\mid -4\le xy\le 4,\ y\ge -4,\ x\ge 6y+10\} \).

Posmatrajmo samo pojas \( 0 < y \le \frac{1}{3} \). Tada iz \( -4 \le xy \le 4 \) sledi \( -\frac{4}{y} \le x \le \frac{4}{y} \), a uslov \( x \ge 6y + 10 \) daje efektivne granice \( x \in [6y + 10,\ \frac{4}{y}] \), jer \( 6y + 10 \le \frac{4}{y} \Leftrightarrow 6y^2 + 10y - 4 \le 0 \), što važi za \( 0 < y \le \frac{1}{3} \).

Njutn–Lajbnic korak (po \( x \)): \( \int_{6y+10}^{\,4/y} x\,dx = \frac{1}{2} \left[ \left( \frac{4}{y} \right)^{2} – (6y + 10)^{2} \right] \).

Za \( 0 < y \le \frac{1}{3} \) važi \( (6y + 10)^2 \le 12^2 = 144 \), pa \( \int_{6y+10}^{4/y} x\,dx \ge \frac{8}{y^{2}} - 72 \).

Prema tome, za svako \( \varepsilon \in (0,\frac{1}{3}] \) dobijamo donju ogradu celog dvostrukog integrala: \( \iint_{D} x\,dx\,dy \ge \int_{0}^{\varepsilon} \left[ \int_{6y+10}^{\,4/y} x\,dx \right] dy \ge \int_{0}^{\varepsilon} \left( \frac{8}{y^{2}} – 72 \right) dy = \left[ -\frac{8}{y} – 72y \right]_0^{\varepsilon} = +\infty \).

Zaključak: \( \iint_{D} x\,dx\,dy \) diverguje (ide u \( +\infty \)).

Oblast \(G\) je sferni pojas između sfera poluprečnika 2 i 3, uz uslove \(0\le x\le y\) i \(z\ge 0\). U sfernim koordinatama: \(x=r\sin\varphi\cos\theta,\ y=r\sin\varphi\sin\theta,\ z=r\cos\varphi\); gustina je \(\rho=\sqrt{x^2+y^2}=r\sin\varphi\), a Jakobijan je \(r^2\sin\varphi\), pa \(\rho\,dV=r^3\sin^2\varphi\,dr\,d\varphi\,d\theta\).

Granice promenljivih (obrazloženje): iz \(4\le r^2\le 9\) sledi \(2\le r\le 3\); uslov \(z\ge0\Rightarrow r\cos\varphi\ge0\) daje \(\varphi\in[0,\frac{\pi}{2}]\); iz \(0\le x\le y\) dobijamo \(\sin\varphi\cos\theta\le \sin\varphi\sin\theta\) (za \(\sin\varphi\ge0\)), tj. \(\cos\theta\le \sin\theta\Rightarrow \tan\theta\ge1\) uz \(\theta\in[0,\frac{\pi}{2}]\), pa \(\theta\in[\frac{\pi}{4},\,\frac{\pi}{2}]\). Dakle: \(2\le r\le 3,\ 0\le\varphi\le\frac{\pi}{2},\ \frac{\pi}{4}\le\theta\le\frac{\pi}{2}\).

Masa: \(M=\displaystyle\int_{2}^{3}\int_{0}^{\pi/2}\int_{\pi/4}^{\pi/2} r^{3}\sin^{2}\varphi\,d\theta\,d\varphi\,dr\).

Jedno po jedno: \(\displaystyle \int_{\pi/4}^{\pi/2}\!d\theta=\frac{\pi}{4}\). Zatim \(\displaystyle \int_{0}^{\pi/2}\!\sin^{2}\varphi\,d\varphi=\int_{0}^{\pi/2}\!\frac{1-\cos2\varphi}{2}\,d\varphi=\Big[\frac{\varphi}{2}-\frac{\sin2\varphi}{4}\Big]_{0}^{\pi/2}=\frac{\pi}{4}\). Konačno \(\displaystyle \int_{2}^{3}\!r^{3}dr=\frac{3^{4}-2^{4}}{4}=\frac{65}{4}\).

Iz toga sledi: \(M=\frac{65}{4}\cdot\frac{\pi}{4}\cdot\frac{\pi}{4}=\frac{65\pi^{2}}{64}\).

Rezultat: \(M=\frac{65\pi^{2}}{64}\).

Površina \(S:z=x^2+y^2\) odsečena ravni \(z=2y\). Projekcija \(D\) na \(xy\)-ravan: \(x^2+y^2\le 2y\iff x^2+(y-1)^2\le 1\) (disk radijusa 1, centar \((0,1)\)).

Za \(z=g(x,y)=x^2+y^2\) važi \(dS=\sqrt{1+g_x^2+g_y^2}\,dA=\sqrt{1+4x^2+4y^2}\,dA\)
(jer su \(g_x=2x,\ g_y=2y\)).

Prevođenjem na \(D\): \(\displaystyle \iint_{S}\frac{x+1}{\sqrt{1+4z}}\,dS=\iint_{D}\frac{x+1}{\sqrt{1+4(x^2+y^2)}}\sqrt{1+4x^2+4y^2}\,dA=\iint_{D}(x+1)\,dA\)
(faktori se skraćuju).

Rastavimo: \(\displaystyle \iint_{D}(x+1)\,dA=\iint_{D}x\,dA+\iint_{D}1\,dA\).

Simetrija: \(\iint_{D}x\,dA=0\)
(za svaku tačku \((x,y)\in D\) i \((-x,y)\in D\); sabiranjem doprinosi \(x\) i \(-x\) se ponište).
Površina: \(\text{P}(D)=\pi\cdot 1^2=\pi\)
(disk poluprečnika 1).

Rezultat: \(\displaystyle \iint_{S}\frac{x+1}{\sqrt{1+4z}}\,dS=\pi\).

Razdvajamo: \(\frac{y^{2}}{\sqrt{1-x^{2}}}
=\underbrace{\frac{1}{\sqrt{1-x^{2}}}}_{\text{zavisno od }x}
\cdot\underbrace{y^{2}}_{\text{zavisno od }y}\).
Pošto je domen kvadrat \(D=[0,1]\times[0,1]\) i izraz je produkt funkcije od \(x\) i funkcije od \(y\),
možemo račun podeliti na dva odvojena integrala:
\[
I=\Big(\int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}}\Big)\Big(\int_{0}^{1}y^{2}\,dy\Big).
\]

Kratak komentar oko krajnje tačke \(x=1\):
iako \(\frac{1}{\sqrt{1-x^{2}}}\) raste kada se \(x\) približava 1, rast je dovoljno blag (kao \(1/\sqrt{1-x}\))
pa je površina ispod krive konačna.

Prvi integral računamo standardnom zamenom \(x = \sin t\) (\(dx = \cos t\,dt\)):
\(\int_{0}^{1} \frac{dx}{\sqrt{1 – x^{2}}} = \int_{0}^{\pi/2} dt = \frac{\pi}{2}.\)

Drugi integral:
\(\displaystyle \int_{0}^{1}y^{2}\,dy=\Big[\frac{y^{3}}{3}\Big]_{0}^{1}=\frac{1}{3}\).

Kombinovanjem dobijamo:
\(I=\frac{\pi}{2}\cdot\frac{1}{3}=\frac{\pi}{6}\).

Rezultat: \(I=\frac{\pi}{6}\).

Proizvod: \(\big(\iint_{D}(x^2+y^2)\,dx\,dy\big)\big(\int_{2}^{3}dz\big)\), gde je \(D:x^2+y^2\le 4\).

Deo po \(z\): \(\int_{2}^{3}dz=3-2=1\), pa ostaje \(\iint_{D}(x^2+y^2)\,dx\,dy\).

Prelaz na polarne koordinate. Iz \(x^2+y^2\le 4\) sledi \(r^2\le 4\Rightarrow 0\le r\le 2\), a ceo disk pokriva \(\theta\in[0,2\pi]\). Važe \(x=r\cos\theta,\ y=r\sin\theta,\ x^2+y^2=r^2,\ dx\,dy=r\,dr\,d\theta\).

Računamo: \(\iint_{D}(x^2+y^2)\,dx\,dy=\int_{0}^{2\pi}\int_{0}^{2} r^{2}\cdot r\,dr\,d\theta=\int_{0}^{2\pi}\int_{0}^{2} r^{3}\,dr\,d\theta.\)

Računamo po \(r\): \(\int_{0}^{2}r^{3}\,dr=\big[\frac{r^{4}}{4}\big]_{0}^{2}=\frac{16}{4}=4.\)

Računamo po \(\theta\): \(\int_{0}^{2\pi}4\,d\theta=4\cdot 2\pi=8\pi.\) Dakle, \(\iint_{D}(x^2+y^2)\,dx\,dy=8\pi\), a kako je \(\int_{2}^{3}dz=1\), ceo proizvod je \(8\pi\).

Rezultat: \(8\pi\).

Kriva: \(y^2=2x\). Od \(O(0,0)\) do \(A(4,\sqrt{8})\) uzimamo \(y\ge0\).
(Provera: \((\sqrt8)^2=8=2\cdot4\Rightarrow A\) leži na krivi.)
Parametrizacija: \(y=t,\ x=\frac{t^2}{2}\), pa kako je \(y=t\ge0\) sledi \(t\in[0,2\sqrt2]\).
Izvodi: \(\frac{dx}{dt}=t,\ \frac{dy}{dt}=1\).

Koristimo formulu za ds: \(ds=\sqrt{\big(\frac{dx}{dt}\big)^2+\big(\frac{dy}{dt}\big)^2}\,dt
=\sqrt{t^2+1}\,dt\)
(jer \((dx/dt,dy/dt)=(t,1)\)).

Integral: \(\displaystyle \int_{\ell} y\,ds=\int_{0}^{2\sqrt2} t\sqrt{1+t^2}\,dt\).

Substitucija: \(u=1+t^2,\ du=2t\,dt\Rightarrow t\,dt=\frac12\,du\);
granice: \(t=0\Rightarrow u=1\), \(t=2\sqrt2\Rightarrow u=1+8=9\).

Zato
\(
\int_{0}^{2\sqrt2} t\sqrt{1+t^2}\,dt
=\frac12\int_{1}^{9} u^{1/2}\,du
=\frac12\cdot\frac{2}{3}\,u^{3/2}\Big|_{1}^{9}
=\frac13\big(9^{3/2}-1^{3/2}\big).
\)

Ovde je \(9^{3/2}=(\sqrt9)^3=3^3=27\), \(1^{3/2}=1\).

Rezultat: \(\displaystyle \int_{\ell} y\,ds=\frac{1}{3}(27-1)=\frac{26}{3}.\)