Kolokvijumi — simulacije sa rešenjima

Oblik $y = x\,f(y') + \varphi(y')$ sa $f(p) = 1+p \neq p$ → Lagranžova. Smena $y' = p$:

$$y = x(1+p) + p^2 \;\Big/{}'_x \qquad (\text{levo } \tfrac{dy}{dx} = p)$$

$$p = 1 + p + (x + 2p)\,\frac{dp}{dx} \;\Longrightarrow\; -1 = (x+2p)\frac{dp}{dx}$$

Pređi na $x(p)$ (podeli sa $\frac{dp}{dx}$; $\frac{dx}{dp} = x'$):

$$x' = -(x + 2p) \;\Longrightarrow\; x' + x = -2p \quad\text{— linearna po } x(p)$$

Smena $x = uv$: iz $v' + v = 0$: $v = e^{-p}$; iz $u'e^{-p} = -2p$: $u' = -2pe^{p}$, pa parcijalnom integracijom (↗ p.i.):

$$u = -2\int pe^p dp = -2e^p(p-1) + c$$

$$x = uv = ce^{-p} - 2(p-1) = ce^{-p} - 2p + 2$$

$x = ce^{-p} - 2p + 2, \qquad y = x(1+p) + p^2$ (opšte rešenje u parametarskom obliku)

Poeni se dobijaju za: prepoznavanje tipa (1), diferenciranje + prelaz na $x(p)$ (2), rešenu linearnu (1), zapisano parametarsko rešenje (1).

Homogena: KJ $r^2 - 4r + 8 = 0$: $r_{1,2} = \frac{4 \pm \sqrt{16-32}}{2} = 2 \pm 2i$ (↗ kompleksni koreni):

$$y_h = e^{2x}(C_1\cos 2x + C_2\sin 2x)$$

Superpozicija — desna strana je zbir dva tipa: $y = y_h + y_{p_1} + y_{p_2}$.

1° $Q_1 = e^{2x}$: $a = 2$; koreni KJ su $2\pm2i \neq 2$ → nije rezonanca → $y_{p_1} = Ae^{2x}$:

$$y_{p_1}'' - 4y_{p_1}' + 8y_{p_1} = (4A - 8A + 8A)e^{2x} = 4Ae^{2x} = e^{2x} \;\Longrightarrow\; A = \frac14$$

2° $Q_2 = \sin 2x$: $\alpha + i\beta = 0 + 2i$; nije koren KJ ($2\pm2i$) → $y_{p_2} = B\cos 2x + D\sin 2x$:

$$y_{p_2}'' = -4B\cos2x - 4D\sin2x, \qquad y_{p_2}' = -2B\sin2x + 2D\cos2x$$

Uvrsti i grupiši uz $\cos2x$ i $\sin2x$:

$$(4B - 8D)\cos2x + (8B + 4D)\sin2x = \sin2x \;\Longrightarrow\; \begin{cases}4B - 8D = 0\\ 8B + 4D = 1\end{cases} \;\Longrightarrow\; B = \frac{1}{10},\; D = \frac{1}{20}$$

$y = e^{2x}(C_1\cos 2x + C_2\sin 2x) + \frac14e^{2x} + \frac{1}{10}\cos 2x + \frac{1}{20}\sin 2x$

Zamka: iako $Q_2$ ima samo $\sin$, u $y_{p_2}$ idu i $\cos$ i $\sin$ (i zaista $B \neq 0$!).

Faktorijel + stepeni → Dalamber:

$$\frac{a_{n+1}}{a_n} = \frac{3^{n+1}(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{3^nn!} = 3\Big(\frac{n}{n+1}\Big)^n = \frac{3}{\big(1+\frac1n\big)^n} \longrightarrow \frac3e > 1$$

Red divergira ($\ell = \frac3e \approx 1{,}1 > 1$) — ključ: prepoznati broj $e$ (↗ poznati limesi).

Prvo sredi $a_n$ — racionalizacija brojioca:

$$\sqrt{n+1}-\sqrt n = \frac{(\sqrt{n+1}-\sqrt n)(\sqrt{n+1}+\sqrt n)}{\sqrt{n+1}+\sqrt n} = \frac{1}{\sqrt{n+1}+\sqrt n} \;\Longrightarrow\; a_n = \frac{1}{(\sqrt{n+1}+\sqrt n)\sqrt[3]{n}}$$

Apsolutna konvergencija? Glavni deo: $\sqrt{n+1}+\sqrt n \sim 2\sqrt n$, pa $a_n \sim \frac{1}{2n^{1/2}\cdot n^{1/3}} = \frac{1}{2n^{5/6}}$. Drugi poredbeni sa $\frac{1}{n^{5/6}}$:

$$\lim\frac{a_n}{1/n^{5/6}} = \lim\frac{n^{5/6}}{(\sqrt{n+1}+\sqrt n)n^{1/3}} = \lim\frac{n^{1/2}}{\sqrt{n+1}+\sqrt n} = \frac12 \in (0,\infty)$$

$\sum\frac{1}{n^{5/6}}$ divergira ($\alpha = \frac56 \le 1$) ⇒ nije apsolutno konvergentan.

Lajbnic: $a_n = \frac{1}{(\sqrt{n+1}+\sqrt n)\sqrt[3]n}$ — imenilac raste ⇒ $(a_n)\downarrow$; i $a_n \to 0$. ✓

Red uslovno konvergira

Poluprečnik (Dalamber po koeficijentima):

$$R = \lim\Big|\frac{a_n}{a_{n+1}}\Big| = \lim\frac{(2n+3)^2}{(2n+1)^2} = 1$$

Red konvergira za $|x - 1| < 1$, tj. $0 < x < 2$. Krajevi:

• $x = 2$: $\;\sum\frac{1}{(2n+1)^2}$ — poredbeni sa $\frac{1}{n^2}$ ($\ell = \frac14$) → konvergira.
• $x = 0$: $\;\sum\frac{(-1)^n}{(2n+1)^2}$ — apsolutno konvergira (isti poredbeni) → konvergira.

$D = [0, 2]$ (oba kraja unutra — zato je zadatak „lak” za 6 poena: ne zaboravi krajeve!)

Kvadratna po $y$ → Rikatijeva. Korak 1 — odredi $a$: uvrsti $y_1 = a\sin x$, $y_1' = a\cos x$:

$$a\cos x = \frac{2\cos^2x - \sin^2x + a^2\sin^2x}{2\cos x} \;\Big/\cdot 2\cos x$$

$$2a\cos^2x = 2\cos^2x + (a^2 - 1)\sin^2x$$

Da bi važilo za svako $x$: uz $\cos^2x$: $2a = 2 \Rightarrow a = 1$; uz $\sin^2x$: $a^2 = 1$ ✓. Dakle $y_1 = \sin x$.

Korak 2 — smena $y = \sin x + \frac1z$, $y' = \cos x - \frac{z'}{z^2}$. Posle uvrštavanja i skraćivanja (koristi $\sin^2x + \cos^2x = 1$, ↗ identiteti):

$$z' + z\tan x = -\frac{1}{2\cos x} \quad\text{— linearna po } z$$

Korak 3 — $z = uv$: iz $v' + v\tan x = 0$: $\frac{dv}{v} = -\tan x\,dx \Rightarrow \ln|v| = \ln|\cos x| \Rightarrow v = \cos x$; iz $u'\cos x = -\frac{1}{2\cos x}$: $u' = -\frac{1}{2\cos^2x} \Rightarrow u = -\frac{\tan x}{2} + C$:

$$z = \cos x\Big(C - \frac{\tan x}{2}\Big) = C\cos x - \frac{\sin x}{2}$$

$y = \sin x + \dfrac{1}{C\cos x - \frac{\sin x}{2}} = \sin x + \dfrac{2}{C_1\cos x - \sin x}$

Provera koraka 3: $z' + z\tan x = -C\sin x - \frac{\cos x}{2} + C\sin x - \frac{\sin^2x}{2\cos x} = -\frac{\cos^2x + \sin^2x}{2\cos x} = -\frac{1}{2\cos x}$ ✓

Homogena: KJ $r^3 - 6r^2 + 12r - 8 = (r-2)^3 = 0$ (kub binoma — zadatak 6 sa 3. vežbi!): $r = 2$ trostruki:

$$y_h = (C_1 + C_2x + C_3x^2)e^{2x}$$

Superpozicija: $Q = 1 + 2e^{2x}$ — dva sabirka.

1° $Q_1 = 1 = e^{0x}$: $a = 0$ nije koren KJ → $y_{p_1} = A$: uvrsti ($y' = y'' = y''' = 0$): $-8A = 1 \Rightarrow A = -\frac18$.

2° $Q_2 = 2e^{2x}$: $a = 2$ JESTE koren KJ višestrukosti $p = 3$ → faktor $x^3$:

$$y_{p_2} = Bx^3e^{2x}$$

Izvodi: $y_{p_2}' = B(3x^2 + 2x^3)e^{2x}$; $\;y_{p_2}'' = B(6x + 12x^2 + 4x^3)e^{2x}$; $\;y_{p_2}''' = B(6 + 36x + 36x^2 + 8x^3)e^{2x}$.

Uvrsti — svi članovi sa $x$, $x^2$, $x^3$ se potiru (provera: uz $x$: $36 - 36 = 0$; uz $x^2$: $36 - 72 + 36 = 0$; uz $x^3$: $8 - 24 + 24 - 8 = 0$):

$$6Be^{2x} = 2e^{2x} \;\Longrightarrow\; B = \frac13$$

$y = (C_1 + C_2x + C_3x^2)e^{2x} - \frac18 + \frac{x^3}{3}e^{2x}$

Ovo je NAJTEŽA rezonanca koja se pojavila: $a$ = trostruki koren → $x^3$. Ako zaboraviš faktor — sve se poništi i nema rešenja.

Stepen $n^2$ → koreni kriterijum:

$$\sqrt[n]{a_n} = 3\Big(\frac{n}{n+1}\Big)^n = \frac{3}{\big(1+\frac1n\big)^n} \longrightarrow \frac3e > 1$$

Red divergira (isti limes $\frac3e$ kao na redovnom roku — samo Koši umesto Dalambera!)

Apsolutna? $\ln(n+1) < n+1 \Rightarrow \frac{1}{\ln(n+1)} > \frac{1}{n+1}$, a $\sum\frac{1}{n+1}$ divergira → po 1. poredbenom $\sum\frac{1}{\ln(n+1)}$ divergira ⇒ nije apsolutna.

Lajbnic: $\ln$ raste ⇒ $a_n = \frac{1}{\ln(n+1)}\downarrow$; $\;a_n \to 0$ ✓.

Red uslovno konvergira

Poluprečnik:

$$R = \lim\Big|\frac{a_n}{a_{n+1}}\Big| = \lim\frac{(n+1)(n+2)(n+3)}{n(n+1)(n+2)} = \lim\frac{n+3}{n} = 1$$

Konvergira za $|x+2| < 1$, tj. $-3 < x < -1$. Krajevi:

• $x = -1$: $\;\sum\frac{1}{n(n+1)(n+2)}$ — poredbeni sa $\frac{1}{n^3}$ → konvergira.
• $x = -3$: $\;\sum\frac{(-1)^n}{n(n+1)(n+2)}$ — apsolutno konvergira → konvergira.

$D = [-3, -1]$

$$F = x - 2y + 2z + \lambda(x^2+y^2+z^2-1)$$

$$F_x = 1 + 2\lambda x = 0, \quad F_y = -2 + 2\lambda y = 0, \quad F_z = 2 + 2\lambda z = 0 \;\Longrightarrow\; x = -\frac{1}{2\lambda},\; y = \frac1\lambda,\; z = -\frac1\lambda$$

Uvrsti u uslov:

$$\frac{1}{4\lambda^2} + \frac{1}{\lambda^2} + \frac{1}{\lambda^2} = 1 \;\Longrightarrow\; \frac{9}{4\lambda^2} = 1 \;\Longrightarrow\; \lambda = \pm\frac32$$

• $\lambda = \frac32$: $M_1\big({-\frac13}, \frac23, {-\frac23}\big)$; $\;d^2F = 2\lambda\big((dx)^2+(dy)^2+(dz)^2\big) > 0$ → uslovni minimum, $f(M_1) = -\frac13 - \frac43 - \frac43 = -3$.
• $\lambda = -\frac32$: $M_2\big(\frac13, {-\frac23}, \frac23\big)$; $\;d^2F < 0$ → uslovni maksimum, $f(M_2) = 3$.

$f_{\min} = -3$ u $M_1$; $\;f_{\max} = 3$ u $M_2$

Brza kontrola: linearna funkcija na jediničnoj sferi ima ekstreme $\pm\|(1,-2,2)\| = \pm\sqrt{1+4+4} = \pm3$ ✓

Oblast je „kosa traka” između dve paralelne prave, odsečena vertikalama — projekcija na $x$-osu daje jedan komad:

$$D = \{0 \le x \le 2,\; x \le y \le x+1\}$$

$$\iint_D 2y\,dxdy = \int_0^2 dx\int_x^{x+1}2y\,dy = \int_0^2 y^2\Big|_x^{x+1}dx = \int_0^2\big((x+1)^2 - x^2\big)dx = \int_0^2(2x+1)\,dx$$

$$= \big(x^2 + x\big)\Big|_0^2 = 4 + 2 = \text{<span class="rezultat">$6$</span>}$$

Poenta zadatka: pravilno opisati oblast (skiciraj!); sam račun je kratak.

Integrand na krivoj: $x^2 + y^2 = 4\cos^2t + 4\sin^2t = 4$; $\;z^2 = 9t^2$ → integrand $= 4 + 9t^2$.

Element luka: $x' = -2\sin t$, $y' = 2\cos t$, $z' = 3$:

$$ds = \sqrt{4\sin^2t + 4\cos^2t + 9}\,dt = \sqrt{13}\,dt$$

$$I = \sqrt{13}\int_0^{2\pi}(4 + 9t^2)\,dt = \sqrt{13}\Big(4t + 3t^3\Big)\Big|_0^{2\pi} = \sqrt{13}\,\big(8\pi + 24\pi^3\big)$$

$I = 8\sqrt{13}\,\pi\big(1 + 3\pi^2\big)$

Tipično za heliks: $ds$ je konstanta — sva „težina” zadatka je u urednom integraljenju po $t$.

Element površi za konus: $p = \frac{x}{\sqrt{x^2+y^2}}$, $q = \frac{y}{\sqrt{x^2+y^2}}$, pa $\sqrt{1+p^2+q^2} = \sqrt2$ (uvek za ovaj konus!):

$$dS = \sqrt2\,dxdy$$

Projekcija: $a \le z = \sqrt{x^2+y^2} \le b$ ⇒ prsten $a^2 \le x^2+y^2 \le b^2$.

Integrand: $z^2 = x^2+y^2$, pa $x^2z^2 = x^2(x^2+y^2)$. Polarne koordinate: $x^2 = \rho^2\cos^2\varphi$:

$$I = \sqrt2\iint_{\text{prsten}}x^2(x^2+y^2)\,dxdy = \sqrt2\int_0^{2\pi}\cos^2\varphi\,d\varphi\int_a^b\rho^4\cdot\rho\,d\rho$$

$$= \sqrt2\cdot\pi\cdot\frac{\rho^6}{6}\Big|_a^b \qquad\Big(\int_0^{2\pi}\cos^2\varphi\,d\varphi = \pi,\; \text{<a class="cs" href="cheatsheet.html#trig-identiteti">↗ identiteti</a>}\Big)$$

$I = \dfrac{\sqrt2\,\pi\,(b^6 - a^6)}{6}$

$$F = x^2 + xy + y^2 + \lambda(x^2+y^2-1): \qquad F_x = (2+2\lambda)x + y = 0, \quad F_y = x + (2+2\lambda)y = 0$$

Homogen sistem ima netrivijalno rešenje kad je determinanta nula:

$$(2+2\lambda)^2 - 1 = 0 \;\Longrightarrow\; 2+2\lambda = \pm1 \;\Longrightarrow\; \lambda = -\frac12 \;\text{ili}\; \lambda = -\frac32$$

$\lambda = -\frac12$: iz $x + y = 0$: $y = -x$; uslov $2x^2 = 1$: tačke $\big({\pm\frac{\sqrt2}{2}}, {\mp\frac{\sqrt2}{2}}\big)$. Vrednost: $f = \underbrace{x^2+y^2}_{=1} + xy = 1 - \frac12 = \frac12$.

$\lambda = -\frac32$: iz $-x + y = 0$: $y = x$; tačke $\big({\pm\frac{\sqrt2}{2}}, {\pm\frac{\sqrt2}{2}}\big)$. Vrednost: $f = 1 + \frac12 = \frac32$.

Klasifikacija ($d^2F = (2+2\lambda)\big((dx)^2+(dy)^2\big) + 2\,dx\,dy$, uz diferenciranu vezu $x\,dx + y\,dy = 0$):

• $\lambda = -\frac12$: $d^2F = (dx+dy)^2$; veza u $\big(\frac{\sqrt2}{2}, -\frac{\sqrt2}{2}\big)$ daje $dx = dy$, pa $d^2F = 4(dx)^2 > 0$ → uslovni minimum, $f = \frac12$.
• $\lambda = -\frac32$: $d^2F = -(dx-dy)^2$; veza u $\big(\frac{\sqrt2}{2}, \frac{\sqrt2}{2}\big)$ daje $dy = -dx$, pa $d^2F = -4(dx)^2 < 0$ → uslovni maksimum, $f = \frac32$.

$f_{\min} = \frac12$ u $\big({\pm\frac{\sqrt2}{2}}, {\mp\frac{\sqrt2}{2}}\big)$; $\;f_{\max} = \frac32$ u $\big({\pm\frac{\sqrt2}{2}}, {\pm\frac{\sqrt2}{2}}\big)$

Zamka: $d^2F$ je ovde semi-definitan sam po sebi — mora se uvrstiti diferencirana veza (korak koji nosi poene!).

Presek: $2 + x^2 + y^2 = 6 \Rightarrow x^2 + y^2 = 4$. Zbog $y \ge 0$ projekcija je polukrug poluprečnika 2.

$$V = \iint_{D}\big(6 - (2 + x^2 + y^2)\big)\,dxdy = \iint_D\big(4 - (x^2+y^2)\big)\,dxdy$$

Polarne koordinate — polukrug znači $\varphi \in [0, \pi]$ (ne $2\pi$!):

$$V = \int_0^{\pi}d\varphi\int_0^2(4-\rho^2)\,\rho\,d\rho = \pi\Big(2\rho^2 - \frac{\rho^4}{4}\Big)\Big|_0^2 = \pi(8 - 4)$$

$V = 4\pi$

Zamka: uslov $y \ge 0$ prepolovljava telo — ceo krug bi dao $8\pi$.

Tri stranice (orijentacija nebitna za 1. vrstu):

• $\overline{OA}$: $y = 0$, $ds = dx$: $\;\int_0^1x\,dx = \frac12$
• $\overline{AB}$: $y = 1-x$, $y' = -1$, $ds = \sqrt2\,dx$; integrand $x + (1-x) = 1$: $\;\sqrt2\int_0^1dx = \sqrt2$
• $\overline{BO}$: $x = 0$, $ds = dy$: $\;\int_0^1y\,dy = \frac12$

$I = \frac12 + \sqrt2 + \frac12 = 1 + \sqrt2$

Površ: $z = \frac{6 - x - 2y}{3} = 2 - \frac x3 - \frac{2y}{3}$: $\;p = -\frac13$, $q = -\frac23$:

$$dS = \sqrt{1 + \frac19 + \frac49}\,dxdy = \sqrt{\frac{14}{9}}\,dxdy = \frac{\sqrt{14}}{3}\,dxdy$$

Integrand na ravni — uprosti pre integraljenja (trik!): $3z = 6 - x - 2y$, pa

$$6x + 4y + 3z = 6x + 4y + 6 - x - 2y = 5x + 2y + 6$$

Projekcija ($z \ge 0 \Rightarrow x + 2y \le 6$; $x, y \ge 0$): trougao sa temenima $(0,0)$, $(6,0)$, $(0,3)$: $\;D = \{0 \le x \le 6,\; 0 \le y \le \frac{6-x}{2}\}$.

$$\iint_D(5x + 2y + 6)\,dxdy = \int_0^6\Big(5xy + y^2 + 6y\Big)\Big|_0^{\frac{6-x}{2}}dx = \int_0^6\Big(\frac{5x(6-x)}{2} + \frac{(6-x)^2}{4} + 3(6-x)\Big)dx$$

Tri dela: $\;\frac52\int_0^6(6x - x^2)dx = \frac52\cdot36 = 90$; $\;\frac14\int_0^6(6-x)^2dx = \frac14\cdot\frac{216}{3} = 18$; $\;3\int_0^6(6-x)dx = 3\cdot18 = 54$. Zbir: $162$.

$$I = \frac{\sqrt{14}}{3}\cdot162$$

$I = 54\sqrt{14}$

Trik: izrazi $y$ i grupiši uz $x$:

$$y = y'^2 + (x-2)y' + 1 = xy' + \big(y'^2 - 2y' + 1\big) = xy' + (y'-1)^2$$

— Klerova sa $\varphi(p) = (p-1)^2$!

Opšte ($y' \to C$): $y = Cx + (C-1)^2$

Singularno: $x = -\varphi'(p) = -2(p-1) \Rightarrow p = 1 - \frac x2$; $\;y = xp + (p-1)^2 = x\big(1-\frac x2\big) + \frac{x^2}{4}$:

$y = x - \dfrac{x^2}{4}$ (provera uvrštavanjem: ✓)

Homogena: KJ $(r-1)^2 = 0 \Rightarrow r = 1$ dvostruki: $y_h = (C_1 + C_2x)e^x$.

1° $Q_1 = 4e^x$: $a = 1$ je koren višestrukosti 2 → faktor $x^2$: $y_{p_1} = Ax^2e^x$.

Uvrsti ($y_{p_1}'' = A(2 + 4x + x^2)e^x$, $y_{p_1}' = A(2x+x^2)e^x$):

$$A e^x\big[(2+4x+x^2) - 2(2x+x^2) + x^2\big] = 2Ae^x = 4e^x \;\Longrightarrow\; A = 2$$

2° $Q_2 = 2\sin x$: $i$ nije koren → $y_{p_2} = B\cos x + D\sin x$; uvrštavanjem: uz $\cos$: $-2D = 0$… pažljivo:

$$(-B\cos - D\sin) - 2(-B\sin + D\cos) + (B\cos + D\sin) = -2D\cos x + 2B\sin x = 2\sin x$$

$D = 0$, $B = 1$: $y_{p_2} = \cos x$.

$y = (C_1 + C_2x)e^x + 2x^2e^x + \cos x$

Integralni kriterijum, smena $t = \ln x$: $\;\int_2^{\infty}\frac{dx}{x\ln^2x} = \int_{\ln2}^{\infty}\frac{dt}{t^2} = \frac{1}{\ln 2} < \infty$.

Red konvergira

Apsolutni red $\sim \frac2n$ (2. poredbeni sa $\frac1n$, $\ell = 2$) → divergira. Lajbnic: $a_{n+1} - a_n = -\frac{2}{n(n+2)} < 0$ i $a_n\to0$ ✓.

Uslovno konvergira

Zbog $x^{2n}$ radi direktno Dalamberom na ceo član:

$$\lim\Big|\frac{x^{2n+2}/(2n+2)!}{x^{2n}/(2n)!}\Big| = \lim\frac{x^2}{(2n+1)(2n+2)} = 0 < 1 \quad\text{za svako } x$$

$D = (-\infty, +\infty) = \mathbb{R}$ (ovo je red za $\cos x - 1$ — konvergira svuda!)

$F = 6 - 4x - 3y + \lambda(x^2+y^2-1)$: iz $F_x = -4 + 2\lambda x = 0$, $F_y = -3 + 2\lambda y = 0$: $x = \frac{2}{\lambda}$, $y = \frac{3}{2\lambda}$.

Uslov: $\frac{4}{\lambda^2} + \frac{9}{4\lambda^2} = 1 \Rightarrow \frac{25}{4\lambda^2} = 1 \Rightarrow \lambda = \pm\frac52$.

• $\lambda = \frac52$: $M_1\big(\frac45, \frac35\big)$, $d^2F = 2\lambda(\ldots) > 0$ → min, $f = 6 - \frac{16}{5} - \frac95 = 1$.
• $\lambda = -\frac52$: $M_2\big({-\frac45}, {-\frac35}\big)$ → max, $f = 11$.

$f_{\min} = 1$, $f_{\max} = 11$ (kontrola: $6 \mp \|(-4,-3)\| = 6\mp5$ ✓)

Presek: $y^2 = 1 \Rightarrow y = \pm1$. Projekcija na $y$-osu (parabola sleva? ne — parabola je leva granica $x = \frac{y^2}{2}$, prava desna):

$$D = \Big\{{-1} \le y \le 1,\; \frac{y^2}{2} \le x \le \frac12\Big\}$$

$$\iint = \int_{-1}^1 y^2\,dy\int_{y^2/2}^{1/2}x\,dx = \int_{-1}^1y^2\cdot\frac{x^2}{2}\Big|_{y^2/2}^{1/2}dy = \frac18\int_{-1}^1\big(y^2 - y^6\big)dy = \frac18\cdot2\Big(\frac13 - \frac17\Big) = \frac14\cdot\frac{4}{21}$$

$\dfrac{1}{21}$ (parnost: $2\int_0^1$!)

Parametrizacija: $x = t$, $y = -1-t$, $t\in[-1, 0]$; $y' = -1 \Rightarrow ds = \sqrt2\,dt$.

$$I = \sqrt2\int_{-1}^0\big(t - 2(1+t)^2\big)dt = \sqrt2\Big[\frac{t^2}{2}\Big|_{-1}^0\cdot\ldots\Big] = \sqrt2\Big({-\frac12} - \frac23\Big)$$

($\int_{-1}^0t\,dt = -\frac12$; $\int_{-1}^0 2(1+t)^2dt = \frac23(1+t)^3\big|_{-1}^0\cdot\ldots = \frac23$)

$I = -\dfrac{7\sqrt2}{6}$ (može biti negativan — integrand menja znak, $ds > 0$!)

Donji konus — ali $dS$ je isti: $p = \frac{-x}{\sqrt{x^2+y^2}}$, $q = \frac{-y}{\sqrt{x^2+y^2}}$, $\sqrt{1+p^2+q^2} = \sqrt2$, i $z^2 = x^2+y^2$.