Imamo krivu preseka: \(x^2+y^2-2y=0 \iff x^2+(y-1)^2=1\), tj. krug \(C(0,1)\), \(r=1\).

Takođe, preseci sa ravni \(z=0\) daju:

\[ z=2-\sqrt{x^2+y^2}, \quad z=0 \;\Rightarrow\; 2-\sqrt{x^2+y^2}=0 \]

\[ \Rightarrow\; \sqrt{x^2+y^2}=2 \;\Rightarrow\; x^2+y^2=4. \]

Tražena površina je

\[ P=\iint\limits_D \sqrt{\,1+(z'_x)^2+(z'_y)^2\,}\,dx\,dy, \]

gde je \(z=2-\sqrt{x^2+y^2}\).

Računamo parcijalne izvode:

\[ z_x = \frac{\partial}{\partial x}\!\left(2-\sqrt{x^2+y^2}\right) = -\,\frac{x}{\sqrt{x^2+y^2}},\qquad z_y = -\,\frac{y}{\sqrt{x^2+y^2}}. \]

Uvodimo:

\[ x=\rho\cos\varphi,\qquad y=\rho\sin\varphi,\qquad J=\rho. \]

Iz \(x^2+y^2=2y\) sledi \(\rho^2=2\rho\sin\varphi \;\Rightarrow\; 0\le \rho\le 2\sin\varphi,\) a pošto je \(2\sin\varphi\ge0\) dobijamo opseg uglova \(0\le \varphi\le \pi\).

\[ \iint\limits_D \sqrt{\,1+\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}\,}\,dx\,dy = \iint\limits_D \sqrt{\,\frac{2x^2+2y^2}{x^2+y^2}\,}\,dx\,dy = \]

\[ \begin{aligned} P &= \sqrt{2} \iint\limits_D \rho \, d\rho \, d\varphi \\[6pt] &= \sqrt{2} \int_{0}^{\pi} d\varphi \int_{0}^{2\sin\varphi} \rho \, d\rho \\[6pt] &= \sqrt{2} \int_{0}^{\pi} \left[ \frac{\rho^2}{2} \right]_{0}^{2\sin\varphi} d\varphi \\[6pt] &= \sqrt{2} \int_{0}^{\pi} \frac{(2\sin\varphi)^2}{2} \, d\varphi \\[6pt] &= \sqrt{2} \int_{0}^{\pi} 2\sin^2\varphi \, d\varphi \\[6pt] \end{aligned} \]

Rešenje: \[\boxed{\,P=\pi\sqrt{2}\,}.\]

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Presek sa \(z=0\) daje:

\[ 1-\sqrt{x^2+y^2}=0 \;\Rightarrow\; \sqrt{x^2+y^2}=1 \;\Rightarrow\; x^2+y^2=1. \]

Dakle, projekcija domena na \(xy\)–ravan je disk \(D:\ x^2+y^2\le 1\).

Tražena površina je

\[ P=\iint\limits_D \sqrt{\,1+(z_x)^2+(z_y)^2\,}\,dx\,dy, \qquad z=1-\sqrt{x^2+y^2}. \]

Računamo parcijalne izvode:

\[ z_x=-\,\frac{x}{\sqrt{x^2+y^2}},\qquad z_y=-\,\frac{y}{\sqrt{x^2+y^2}}. \]

\[ 1+(z_x)^2+(z_y)^2 =1+\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2} =1+\frac{x^2+y^2}{x^2+y^2} =2, \]

pa je

\[ P=\iint\limits_D \sqrt{2}\,dx\,dy. \]

Uvodimo polarne koordinate:

\[ x=\rho\cos\varphi,\qquad y=\rho\sin\varphi,\qquad J=\rho. \]

Za disk \(x^2+y^2\le 1\) dobijamo granice

\[ 0\le \rho\le 1,\qquad 0\le \varphi\le 2\pi. \]

\[ \begin{aligned} P &=\sqrt{2}\iint\limits_D dx\,dy =\sqrt{2}\int_{0}^{2\pi}\!\!\int_{0}^{1} \rho\, d\rho\, d\varphi \\[6pt] &=\sqrt{2}\int_{0}^{2\pi}\left[\frac{\rho^2}{2}\right]_{0}^{1} d\varphi =\sqrt{2}\int_{0}^{2\pi} \frac{1}{2}\, d\varphi \\[6pt] &=\sqrt{2}\cdot \frac{1}{2}\cdot 2\pi =\pi\sqrt{2}. \end{aligned} \]

Rešenje: \[\boxed{\,P=\pi\sqrt{2}\,}.\]

Rešenje

Granice oblasti

Preseci:

\[ \begin{aligned} &\text{sa } y=2:\quad xy=2 \Rightarrow x=1 \ \Rightarrow (1,2), \qquad x-2y=0 \Rightarrow x=4 \ \Rightarrow (4,2);\\ &\text{sa } x=2y \text{ i } xy=2:\quad 2y\cdot y=2 \Rightarrow y^2=1 \Rightarrow y=1,\ x=2 \ \Rightarrow (2,1). \end{aligned} \]

Dakle, za \(1\le y\le 2\) važi da \(x\) ide od hiperbole \(x=\dfrac{2}{y}\) do prave \(x=2y\).

Postavljanje integrala

\[ \begin{aligned} I &=\int_{y=1}^{2}\int_{x=\frac{2}{y}}^{\,2y} x^2\,y\,dx\,dy = \int_{1}^{2} y \left[\frac{x^3}{3}\right]_{\,x=\frac{2}{y}}^{\,x=2y}\,dy \\[6pt] &=\int_{1}^{2} y \cdot \frac{(2y)^3-\left(\frac{2}{y}\right)^3}{3}\,dy = \int_{1}^{2} \frac{y}{3}\bigl(8y^3- \frac{8}{y^3}\bigr)\,dy \\[6pt] &=\int_{1}^{2} \frac{8}{3}\Bigl(y^4 - \frac{1}{y^2}\Bigr)\,dy = \frac{8}{3}\left[\frac{y^5}{5}+ \frac{1}{y}\right]_{1}^{2} \\[6pt] &=\frac{8}{3}\left(\frac{32}{5}+\frac{1}{2} - \frac{1}{5}-1\right) =\frac{8}{3}\left(\frac{31}{5}-\frac{1}{2}\right) =\frac{8}{3}\cdot\frac{57}{10} =\frac{76}{5}. \end{aligned} \]

\[\boxed{I=\dfrac{76}{5}}\]

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Analiza oblasti

\[ x^{2}+y^{2}\ge x \;\;\Rightarrow\;\; x^{2}-x+y^{2}\ge 0 \;\;\Rightarrow\;\; x^{2}-x+\tfrac{1}{4}+y^{2}\ge \tfrac{1}{4} \;\;\Rightarrow\;\; (x-\tfrac{1}{2})^{2}+y^{2}\ge \tfrac{1}{4}. \]

Dakle, krug jednakosti ima centar \(A(\tfrac{1}{2},0)\), poluprečnik \(r=\tfrac{1}{2}\).

\[ x^{2}+y^{2}\le 2x \;\;\Rightarrow\;\; x^{2}-2x+y^{2}\le 0 \;\;\Rightarrow\;\; x^{2}-2x+1+y^{2}\le 1 \;\;\Rightarrow\;\; (x-1)^{2}+y^{2}\le 1. \]

Ovaj krug ima centar \(B(1,0)\), poluprečnik \(r=1\).

Uvođenje polarne smene

\[ x=\rho\cos\varphi,\qquad y=\rho\sin\varphi. \]

Iz uslova za oblast:

\[ x\le \rho^{2}\le 2x \;\;\Longrightarrow\;\; \rho\cos\varphi \le \rho^{2} \le 2\rho\cos\varphi. \]

Pošto je \(\rho\ge 0\), deljenjem sa \(\rho\) dobijamo:

\[ \cos\varphi \le \rho \le 2\cos\varphi. \]

Da bi imalo smisla, mora \(\cos\varphi\ge 0\), tj.

\[ -\frac{\pi}{2}\le \varphi \le \frac{\pi}{2}. \]

Zato je

\[ \Delta=\bigl\{(\rho,\varphi)\ \big|\ \cos\varphi \le \rho \le 2\cos\varphi,\; -\tfrac{\pi}{2}\le \varphi \le \tfrac{\pi}{2}\bigr\}. \]

Jakobijan

\[ J=\left|\frac{\partial(x,y)}{\partial(\rho,\varphi)}\right| =\begin{vmatrix} \frac{\partial x}{\partial \rho} & \frac{\partial x}{\partial \varphi}\\[2pt] \frac{\partial y}{\partial \rho} & \frac{\partial y}{\partial \varphi} \end{vmatrix} =\begin{vmatrix} \cos\varphi & -\rho\sin\varphi\\[2pt] \sin\varphi & \rho\cos\varphi \end{vmatrix} =\rho(\cos^{2}\varphi+\sin^{2}\varphi)=\rho. \]

\[ \sqrt{x^{2}+y^{2}}=\sqrt{(\rho\cos\varphi)^{2}+(\rho\sin\varphi)^{2}} =\sqrt{\rho^{2}\cos^{2}\varphi+\rho^{2}\sin^{2}\varphi} =\sqrt{\rho^{2}}=\rho. \]

Zato

\[ \sqrt{x^{2}+y^{2}}\;dx\,dy =\sqrt{(\rho\cos\varphi)^{2}+(\rho\sin\varphi)^{2}}\;\cdot J\,d\rho\,d\varphi =\rho\cdot \rho\,d\rho\,d\varphi=\rho^{2}\,d\rho\,d\varphi. \]

Računanje integrala

\[ \begin{aligned} \iint\limits_{D}\sqrt{x^{2}+y^{2}}\;dx\,dy &=\iint\limits_{\Delta}\sqrt{(\rho\cos\varphi)^{2}+(\rho\sin\varphi)^{2}}\;J\,d\rho\,d\varphi\\[4pt] &=\iint\limits_{\Delta}\sqrt{\rho^{2}}\,\rho\,d\rho\,d\varphi =\iint\limits_{\Delta}\rho^{2}\,d\rho\,d\varphi\\[6pt] &=\int_{-\pi/2}^{\pi/2}\int_{\rho=\cos\varphi}^{\,2\cos\varphi}\rho^{2}\,d\rho\,d\varphi\\[6pt] &=\int_{-\pi/2}^{\pi/2}\left[\frac{\rho^{3}}{3}\right]_{\rho=\cos\varphi}^{\,\rho=2\cos\varphi}\,d\varphi\\[6pt] &=\int_{-\pi/2}^{\pi/2}\frac{(2\cos\varphi)^{3}-(\cos\varphi)^{3}}{3}\,d\varphi\\[6pt] &=\frac{1}{3}\int_{-\pi/2}^{\pi/2}\bigl(8\cos^{3}\varphi-\cos^{3}\varphi\bigr)\,d\varphi\\[6pt] &=\frac{7}{3}\int_{-\pi/2}^{\pi/2}\cos^{3}\varphi\,d\varphi. \end{aligned} \]

Sada računamo trigonometrijski integral:

\[ \begin{aligned} \int_{-\pi/2}^{\pi/2}\cos^{3}\varphi\,d\varphi &=\int_{-\pi/2}^{\pi/2}\cos\varphi\,(1-\sin^{2}\varphi)\,d\varphi\\[4pt] &=\Bigl[\;t=\sin\varphi,\ dt=\cos\varphi\,d\varphi,\ t\in[-1,1]\;\Bigr]\\[2pt] &=\int_{-1}^{1}(1-t^{2})\,dt =\left[t-\frac{t^{3}}{3}\right]_{-1}^{1}\\[4pt] &=\Bigl(1-\frac{1}{3}\Bigr)-\Bigl(-1-\bigl(-\tfrac{1}{3}\bigr)\Bigr) =\frac{2}{3}-\Bigl(-\frac{2}{3}\Bigr)=\frac{4/3}. \end{aligned} \]

Zato je

\[ \iint\limits_{D}\sqrt{x^{2}+y^{2}}\;dx\,dy =\frac{7}{3}\cdot\frac{4}{3} =\boxed{\frac{28}{9}}. \]

Imamo krivu preseka:

\[ x^2+(y-2)^2=4 \;\;\Longleftrightarrow\;\; x^2+y^2-4y=0, \]

tj. krug sa centrom u \((0,2)\) i poluprečnikom \(r=2\).

Takođe, presek sa ravni \(z=0\) daje

\[ z=2-\sqrt{x^2+y^2}=0 \;\Rightarrow\; \sqrt{x^2+y^2}=2 \;\Rightarrow\; x^2+y^2=4. \]

(Ovo samo beležimo; račun ide preko date kružnice \(x^2+(y-2)^2=4\).)

Tražena površina je

\[ P=\iint\limits_D \sqrt{\,1+(z_x)^2+(z_y)^2\,}\,dx\,dy, \qquad z=2-\sqrt{x^2+y^2}. \]

Računamo parcijalne izvode:

\[ z_x=\frac{\partial}{\partial x}\!\left(2-\sqrt{x^2+y^2}\right) =-\,\frac{x}{\sqrt{x^2+y^2}},\qquad z_y=-\,\frac{y}{\sqrt{x^2+y^2}}. \]

\[ 1+(z_x)^2+(z_y)^2 =1+\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2} =2, \] pa je \[ P=\iint\limits_D \sqrt{2}\,dx\,dy. \]

Uvodimo polarne koordinate:

\[ x=\rho\cos\varphi,\qquad y=\rho\sin\varphi,\qquad J=\rho. \]

Iz \(x^2+(y-2)^2=4\) sledi

\[ \rho^2-4\rho\sin\varphi=0 \;\;\Rightarrow\;\; 0\le \rho\le 4\sin\varphi, \]

a pošto je \(4\sin\varphi\ge0\) dobijamo opseg uglova

\[ 0\le \varphi\le \pi. \]

\[ \begin{aligned} P &= \sqrt{2}\iint\limits_D dx\,dy = \sqrt{2}\int_{0}^{\pi}\int_{0}^{4\sin\varphi} \rho\, d\rho\, d\varphi \\[6pt] &= \sqrt{2}\int_{0}^{\pi}\left[\frac{\rho^2}{2}\right]_{0}^{\,4\sin\varphi} d\varphi = \sqrt{2}\int_{0}^{\pi} \frac{(4\sin\varphi)^2}{2}\, d\varphi \\[6pt] &= \sqrt{2}\int_{0}^{\pi} 8\sin^2\varphi \, d\varphi = \sqrt{2}\cdot 8\cdot \frac{\pi}{2} = 4\pi\sqrt{2}. \end{aligned} \]

Rešenje:

\[ \boxed{\,P=4\pi\sqrt{2}\,}. \]