Površinski integral I vrste – druga strana

\[ x^{2}+y^{2}=2x\ \Longrightarrow\ (x-1)^{2}+y^{2}=1, \]

pa je projekcija

\[ D=\{(x,y)\in\mathbb{R}^{2}:(x-1)^{2}+y^{2}\leq1\}. \]

Kao u prethodnom primeru, integral se svodi na:

\[ \iint\limits_{S}\frac{x+1}{\sqrt{1+4z}}\,dS=\iint\limits_{D}(x+1)dx\,dy. \]

Prelaskom na polarne koordinate

\[ x=r\cos\theta,\quad y=r\sin\theta,\quad dx\,dy=r\,dr\,d\theta, \]

krug postaje

\[ r^{2}=2r\cos\theta\quad\Rightarrow\quad r=2\cos\theta, \]

pa su granice oblasti \(D\)

\[ 0\le r\le2\cos\theta,\;-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}. \]

Tada integral postaje

\[ \iint\limits_{D}(x+1)dx\,dy=\int_{-\pi/2}^{\pi/2}\int_{0}^{2\cos\theta}(r\cos\theta+1)r\,dr\,d\theta. \]

Unutrašnji integral je

\[ \int_{0}^{2\cos\theta}(r^{2}\cos\theta+r)\,dr=\left[\frac{\cos\theta}{3}r^{3}+\frac{1}{2}r^{2}\right]\bigg|_{0}^{2\cos\theta}=\frac{8}{3}\cos^{4}\theta+2\cos^{2}\theta, \]

pa je

\[ \int_{-\pi/2}^{\pi/2}\left(\frac{8}{3}\cos^{4}\theta+2\cos^{2}\theta\right)d\theta=\frac{8}{3}\int_{-\pi/2}^{\pi/2}\cos^{4}\theta\,d\theta+2\int_{-\pi/2}^{\pi/2}\cos^{2}\theta\,d\theta. \]

Koristeći identitete

\[ \cos^{2}\theta=\frac{1+\cos2\theta}{2},\quad\cos^{4}\theta=\frac{3+4\cos2\theta+\cos4\theta}{8}, \]

dobijamo

\[ \int_{-\pi/2}^{\pi/2}\cos^{2}\theta\,d\theta=\frac{\pi}{2},\quad\int_{-\pi/2}^{\pi/2}\cos^{4}\theta\,d\theta=\frac{3}{8}\pi. \]

Sada je

\[ \iint\limits_{S}\frac{x+1}{\sqrt{1+4z}}\,dS=\frac{8}{3}\cdot\frac{3}{8}\pi+2\cdot\frac{\pi}{2}=\pi+\pi=2\pi. \]

Kako je \(z=x^{2}+y^{2}\) dobijamo \(z’_{x}=2x\) i \(z’_{y}=2y\) pa je

\[ dS=\sqrt{1+(z’_{x})^{2}+(z’_{y})^{2}}\,dx\,dy=\sqrt{1+4x^{2}+4y^{2}}\,dx\,dy. \]

Dakle

\[ \iint\limits_{S}\sqrt{1+4z}\,dS =\iint\limits_{D}\bigl(1+4(x^{2}+y^{2})\bigr)\,dxdy. \]

Oblast \(D\) je definisana uslovima:

\[ y\le x^{2}+y^{2}\le2y. \]

Prelazimo na polarne koordinate

\[ x=r\cos\theta,\quad y=r\sin\theta,\quad dx\,dy=r\,dr\,d\theta, \]

i dobijamo

\[ r\sin\theta\le r^{2}\le2r\sin\theta \quad\Rightarrow\quad r\in[\sin\theta,2\sin\theta],\quad0\le\theta\le\pi. \]

Sada

\[ \iint\limits_{D}\bigl(1+4(x^{2}+y^{2})\bigr)\,dxdy =\int_{0}^{\pi}\int_{\sin\theta}^{2\sin\theta}\bigl(1+4r^{2}\bigr)\,r\,dr\,d\theta. \]

Kako je

\[ \int_{\sin\theta}^{2\sin\theta}\bigl(1+4r^{2}\bigr)\,r\,dr =\left[\frac{r^{2}}{2}+r^{4}\right]_{\sin\theta}^{2\sin\theta} =\frac{3}{2}\sin^{2}\theta+15\sin^{4}\theta, \]

analogno prethodnom zadatku dobijamo

\[ \int_{0}^{\pi}\left(\frac{3}{2}\sin^{2}\theta+15\sin^{4}\theta\right)d\theta =\frac{3}{2}\cdot\frac{\pi}{2}+15\cdot\frac{3\pi}{8} =\frac{3\pi}{4}+\frac{45\pi}{8} =\frac{51}{8}\pi, \]

pa je

\[ \iint\limits_{S}\sqrt{1+4z}\,dS=\frac{51}{8}\pi. \]

Površ \(S\) možemo zapisati kao

\[ z=2-\frac{1}{3}x-\frac{2}{3}y, \]

pa je

\[ dS=\sqrt{1+(z’_{x})^{2}+(z’_{y})^{2}}\,dx\,dy=\frac{\sqrt{14}}{3}\,dx\,dy. \]

Integrand sada postaje:

\[ 6x+4y+3z=6x+4y+3\Bigl(2-\frac{1}{3}x-\frac{2}{3}y\Bigr) =5x+2y+6. \]

Projekcija na \(xy\) ravan je trougao

\[ D=\{(x,y)\mid x\ge0,\; y\ge0,\; x+2y\le6\}, \]

pa integral postaje

\[ \iint\limits_{S}(6x+4y+3z)\,dS =\frac{\sqrt{14}}{3}\iint\limits_{D}(5x+2y+6)\,dx\,dy. \]

Sada je

\[ \iint\limits_{D}(5x+2y+6)\,dx\,dy =\int_{0}^{3}\left(\int_{0}^{6-2y}(5x+2y+6)\,dx\right)dy. \]

Unutrašnji integral je

\[ \int_{0}^{6-2y}(5x+2y+6)\,dx =\left[\frac{5}{2}x^{2}+(2y+6)x\right]_{0}^{6-2y} =\frac{5}{2}(6-2y)^{2}+(2y+6)(6-2y) \]

\[ =\frac{5}{2}(36-24y+4y^{2})+(12y+36-4y^{2}-12y) =126-60y+6y^{2}. \]

Zatim

\[ \int_{0}^{3}\bigl(126-60y+6y^{2}\bigr)\,dy =\left[126y-30y^{2}+2y^{3}\right]_{0}^{3} =378-270+54 =162. \]

Dakle

\[ \iint\limits_{S}(6x+4y+3z)\,dS =\frac{\sqrt{14}}{3}\cdot162 =54\sqrt{14}. \]

Izračunaćemo integral nad gornjom polovinom \(S^{+}\) sfere pa na kraju to pomnožiti sa \(2\) kako bismo dobili nad celom sferom.

Dakle, posmatramo

\[ z=\sqrt{a^{2}-x^{2}-y^{2}}. \]

Kako je

\[ dS=\sqrt{1+z_{x}'{}^{2}+z_{y}'{}^{2}}\,dx\,dy, \]

računamo parcijalne izvode

\[ z_{x}'=\frac{-x}{\sqrt{a^{2}-x^{2}-y^{2}}},\qquad z_{y}'=\frac{-y}{\sqrt{a^{2}-x^{2}-y^{2}}}. \]

Sada je

\[ 1+z_{x}'{}^{2}+z_{y}'{}^{2}=1+\frac{x^{2}}{a^{2}-x^{2}-y^{2}}+\frac{y^{2}}{a^{2}-x^{2}-y^{2}}=\frac{a^{2}}{a^{2}-x^{2}-y^{2}}, \]

\[ \Rightarrow dS=\frac{a}{\sqrt{a^{2}-x^{2}-y^{2}}}\,dx\,dy. \]

Integral postaje

\[ \iint\limits_{S^{+}}(x^{2}+y^{2})\,dS=\iint\limits_{D}(x^{2}+y^{2})\frac{a}{\sqrt{a^{2}-x^{2}-y^{2}}}\,dx\,dy=a\iint\limits_{D}\frac{x^{2}+y^{2}}{\sqrt{a^{2}-x^{2}-y^{2}}}\,dx\,dy, \]

gde je oblast \(D\) projekcija sfere na \(xy\) ravan:

\[ D=\{(x,y)\mid x^{2}+y^{2}\le a^{2}\}. \]

Prelazimo na polarne koordinate

\[ x=r\cos\theta,\quad y=r\sin\theta,\quad dx\,dy=r\,dr\,d\theta. \]

Sada integral postaje

\[ a\int_{0}^{2\pi}\int_{0}^{a}\frac{r^{2}}{\sqrt{a^{2}-r^{2}}}\,r\,dr\,d\theta=a\int_{0}^{2\pi}\int_{0}^{a}\frac{r^{3}}{\sqrt{a^{2}-r^{2}}}\,dr\,d\theta. \]

Integral po \(r\) rešavamo smenom

\[ u=a^{2}-r^{2}\Rightarrow du=-2r\,dr\Rightarrow r\,dr=-\frac{du}{2} \]

i dobijamo

\[ \int_{0}^{a}\frac{r^{3}}{\sqrt{a^{2}-r^{2}}}\,dr=\int_{u=a^{2}}^{0}\frac{(a^{2}-u)(-du/2)}{\sqrt{u}}=\frac{1}{2}\int_{0}^{a^{2}}\frac{a^{2}-u}{\sqrt{u}}\,du \]

\[ =\frac{1}{2}\int_{0}^{a^{2}}\left(\frac{a^{2}}{\sqrt{u}}-\sqrt{u}\right)du=\frac{1}{2}\left[2a^{2}\sqrt{u}-\frac{2}{3}u^{3/2}\right]\bigg|_{0}^{a^{2}}=\frac{2}{3}a^{3}. \]

Kako je

\[ \int_{0}^{2\pi}d\theta=2\pi \]

dobijamo

\[ \iint\limits_{S^{+}}(x^{2}+y^{2})\,dS=a\cdot2\pi\cdot\frac{2}{3}a^{3}=\frac{4\pi a^{4}}{3}. \]

Dakle

\[ \iint\limits_{S}(x^{2}+y^{2})\,dS=2\cdot\frac{4\pi a^{4}}{3}=\frac{8\pi a^{4}}{3}. \]