Trojni integral – izračunavanje i smena promenljivih — strana 2
Zadatak 1. Izračunati \(\iiint\limits_{G}\frac{dx\,dy\,dz}{(x+y+z+1)^{3}}\)
ako je \(G\) ograničena sa
\(x+y+z=1,\ x=0,\ y=0,\ z=0\).
Prikaži rešenje
Prikaži ravan \( x + y + z = 1 \)
Prikaži ravan \( x = 0 \)
Prikaži ravan \( y = 0 \)
Prikaži ravan \( z = 0 \)
Prikaži telo \( G \)
Rešenje:
Na osnovu slike vidimo da je
\[\iiint\limits_{G}\frac{dx\,dy\,dz}{(x+y+z+1)^{3}}=\int_{0}^{1}dx\int_{0}^{1-x}dy\int_{0}^{1-x-y}\frac{dz}{(x+y+z+1)^{3}}\]
Računamo najpre integral po \(z\):
\[\int_{0}^{1-x-y}(x+y+z+1)^{-3}dz=\left[-\frac{1}{2}(x+y+z+1)^{-2}\right]_{0}^{1-x-y}=\frac{1}{2}\left(\frac{1}{(x+y+1)^{2}}-\frac{1}{4}\right)\]
Dalje
\[\int_{0}^{1-x}\frac{1}{2}\left(\frac{1}{(x+y+1)^{2}}-\frac{1}{4}\right)dy=\frac{1}{2}\int_{0}^{1-x}\frac{dy}{(x+y+1)^{2}}-\frac{1}{8}\int_{0}^{1-x}dy\]
Računamo pojedinačno ove integrale
\[\int_{0}^{1-x}\frac{dy}{(x+y+1)^{2}}=\left[-\frac{1}{x+y+1}\right]\Big|_{0}^{1-x}=\frac{1}{x+1}-\frac{1}{2}\]
\[\int_{0}^{1-x}dy=1-x\]
Dakle
\[\int_{0}^{1-x}\frac{1}{2}\left(\frac{1}{(x+y+1)^{2}}-\frac{1}{4}\right)dy=\frac{1}{2(x+1)}+\frac{x-3}{8}\]
Na kraju integralimo po \(x\) i dobijamo
\[\int_{0}^{1}\left(\frac{1}{2(x+1)}+\frac{x-3}{8}\right)dx=\frac{1}{2}\int_{0}^{1}\frac{dx}{x+1}+\frac{1}{8}\int_{0}^{1}(x-3)dx\]
Kako je
\[\frac{1}{2}\int_{0}^{1}\frac{dx}{x+1}=\frac{1}{2}\ln(x+1)\Big|_{0}^{1}=\frac{1}{2}\ln2\]
\[\frac{1}{8}\int_{0}^{1}(x-3)dx=\frac{1}{8}\left[\frac{x^{2}}{2}-3x\right]_{0}^{1}=\frac{1}{8}\left(\frac{1}{2}-3\right)=-\frac{5}{16}\]
dobijamo
\[\iiint\limits_{G}\frac{dx\,dy\,dz}{(x+y+z+1)^{3}}=\frac{1}{2}\ln2-\frac{5}{16}.\]
Zadatak 2. Izračunati \(\iiint\limits_{G}dx\,dy\,dz\) ako je \(G\)
ograničena sa
\(x=0,\ y=0,\ 6x+3y+2z=6,\ 2x+y+2z=2\).
Prikaži rešenje
Prikaži ravan \( x = 0 \)
Prikaži ravan \( y = 0 \)
Prikaži ravan \( 6x + 3y + 2z = 6 \)
Prikaži ravan \( 2x + y + 2z = 2 \)
Prikaži telo \( G \)
Prikaži oblast \( D \)
Rešenje:
Sa slike vidimo da je
\[G=\{(x,y,z)\mid 1-x-\frac{1}{2}y\le z\le3-3x-\frac{3}{2}y,\ (x,y)\in
D\ \},\]
gde je
\[D=\{(x,y)\mid0\le x\le1,\ 0\le y\le2-2x\}\]
Dakle, sada imamo
\[\iiint_{G}\,dx\,dy\,dz=\iint_{D}\,dx\,dy\int_{1-x-\frac{1}{2}y}^{3-3x-\frac{3}{2}y}dz=\int_{0}^{1}\!\int_{0}^{2-2x}(2-2x-y)\,dy\,dx\]
Računamo unutrašnji integral:
\[\int_{0}^{2-2x}(2-2x-y)\,dy=\bigg((2-2x)y-\frac{y^{2}}{2}\bigg)\Big|_{0}^{\,2-2x}=\frac{1}{2}(2-2x)^{2}=2(1-x)^{2}\]
Zatim spoljašnji:
\[\int_{0}^{1}2(1-x)^{2}\,dx=2\int_{0}^{1}(1-2x+x^{2})\,dx=2\bigg(x-x^{2}+\frac{x^{3}}{3}\bigg)\bigg|_{0}^{1}=2\cdot\frac{1}{3}=\frac{2}{3}\]
Dakle, dobijamo
\[\iiint\limits_{G}dx\,dy\,dz=\frac{2}{3}.\]
Zadatak 3. Izračunati \(\iiint\limits_{G}ydx\,dy\,dz\) gde je \(G\) telo određeno relacijama
\(x^{2}+y^{2}+z^{2}\leq2,\ z\geq x^{2}+y^{2},\ y\geq0\).
Prikaži rešenje
Prikaži sferu
Prikaži eliptički paraboloid
Prikaži ravan \( y = 0 \)
Prikaži telo \( G \)
Rešenje:
Uvodimo cilindrične koordinate:
\[x=\rho\cos\varphi,\qquad y=\rho\sin\varphi,\qquad z=z\]
U ovim koordinatama paraboloid je \(z=\rho^{2}\), a gornja polusfera je \(z=\sqrt{2-\rho^{2}}\).
Granica preseka dobijamo izjednačavanjem:
\[\rho^{2}=\sqrt{2-\rho^{2}}\quad\Longrightarrow\quad\rho^{4}+\rho^{2}-2=0\quad\Longrightarrow\quad\rho^{2}=-2\:\lor\:\rho^{2}=1\]
Pošto je \(y\geq0\) dobijamo \(0\leq\varphi\leq\pi\). Dakle, imamo granice
\[\varphi\in[0,\pi],\quad\rho\in[0,1],\quad z\in[\rho^{2},\sqrt{2-\rho^{2}}]\]
Sada integral postaje:
\[\iiint\limits_{G}ydx\,dy\,dz=\int_{0}^{\pi}\int_{0}^{1}\int_{\rho^{2}}^{\sqrt{2-\rho^{2}}}\rho^{2}\sin\varphi\,dz\,d\rho\,d\varphi\]
Dalje imamo:
\[\iiint_{G}y\,dx\,dy\,dz=\int_{0}^{\pi}\sin\varphi\,d\varphi\cdot\int_{0}^{1}\rho^{2}\big(\sqrt{2-\rho^{2}}-\rho^{2}\big)\,d\rho\]
Računamo pojedinačno svaki integral:
\[\int_{0}^{\pi}\sin\varphi\,d\varphi=2,\qquad\int_{0}^{1}\rho^{4}\,d\rho=\frac{1}{5}\]
Za \(\int_{0}^{1}\rho^{2}\sqrt{2-\rho^{2}}\,d\rho\) koristimo smenu \(\rho=\sqrt{2}\sin t\), \(t\in[0,\pi/4]\) i dobijamo
\[\int_{0}^{1}\rho^{2}\sqrt{2-\rho^{2}}\,d\rho=\int_{0}^{\pi/4}4\sin^{2}t\cos^{2}t\,dt=\int_{0}^{\pi/4}\sin^{2}(2t)\,dt=\int_{0}^{\pi/4}\frac{1-\cos4t}{2}\,dt=\frac{\pi}{8}\]
Dakle, dobijamo
\[\iiint_{G}y\,dx\,dy\,dz=2\left(\frac{\pi}{8}-\frac{1}{5}\right)=\frac{\pi}{4}-\frac{2}{5}.\]
Zadatak 4. Izračunati \(\iiint\limits_{G}\sqrt{x^{2}+y^{2}+z^{2}}\;dx\,dy\,dz,\) ako je \(G:x^{2}+y^{2}+z^{2}\le z.\)
Rešenje:
Uvodimo sferne koordinate
\[
\begin{cases}
x=\rho\sin\theta\cos\varphi,\\
y=\rho\sin\theta\sin\varphi,\\
z=\rho\cos\theta
\end{cases}\quad J=\rho^{2}\sin\theta.
\]
Neposredno iz nejednačine dobijamo \(\rho^{2}\le\rho\cos\theta\), tj. za \(\rho\geq0\) važi \(\rho\le\cos\theta\). Pošto je \(\rho\ge0\), mora biti i \(\cos\theta\ge0\),
pa je \(0\le\theta\le\frac{\pi}{2}.\) Dakle granice su
\[0\le\varphi\le2\pi,\qquad0\le\theta\le\frac{\pi}{2},\qquad0\le\rho\le\cos\theta.\]
Sada je
\[I=\int_{0}^{2\pi}\!\int_{0}^{\pi/2}\!\int_{0}^{\cos\theta}\rho\cdot\rho^{2}\sin\theta\;d\rho\,d\theta\,d\varphi=\int_{0}^{2\pi}d\varphi\int_{0}^{\pi/2}\sin\theta\left(\int_{0}^{\cos\theta}\rho^{3}d\rho\right)d\theta.\]
Kako je
\[\int_{0}^{2\pi}d\varphi=2\pi,\qquad\int_{0}^{\cos\theta}\rho^{3}d\rho=\frac{\cos^{4}\theta}{4}.\]
dobijamo
\[I=2\pi\int_{0}^{\pi/2}\frac{\cos^{4}\theta}{4}\sin\theta\,d\theta=\frac{\pi}{2}\int_{0}^{\pi/2}\cos^{4}\theta\sin\theta\,d\theta.\]
Ovaj integral rešavamo smenom \(t=\cos\theta\), \(dt=-\sin\theta\,d\theta\). Kada je \(\theta=0,\) dobijamo \(t=1,\) a za \(\theta=\pi/2\) imamo \(t=0\). Stoga
\[\int_{0}^{\pi/2}\cos^{4}\theta\sin\theta\,d\theta=\int_{1}^{0}t^{4}(-dt)=\int_{0}^{1}t^{4}dt=\frac{t^{5}}{5}\bigg|_{0}^{1}=\frac{1}{5}.\]
Dakle, dobijamo
\[I=\frac{\pi}{2}\cdot\frac{1}{5}=\frac{\pi}{10}.\]
Zadatak 5. Izračunati \(\iiint\limits_{G}dx\,dy\,dz\) gde je \(G\) telo određeno relacijama
\(x^{2}+y^{2}+z^{2}\leq16,\ z\sqrt{3}\geq\sqrt{x^{2}+y^{2}}\)
Prikaži rešenje
Prikaži sferu
Prikaži deo konusa
Prikaži telo \( G \)
Rešenje:
Uvodimo sferne koordinate
\[
\begin{cases}
x=\rho\sin\theta\cos\varphi,\\
y=\rho\sin\theta\sin\varphi,\\
z=\rho\cos\theta
\end{cases}\quad J=\rho^{2}\sin\theta.
\]
Sa slike vidimo da nam sfera \(x^{2}+y^{2}+z^{2}\leq16\) određuje granice za \(\rho\), tj. zamenom sfernih koordinata jednačina sfere postaje \(\rho^{2}\leq16\).
Kako je telo simetrično oko \(z\)-ose, vidimo da ugao \(\varphi\) pravi pun krug u \(xy\)-ravni pa je \(0\leq\varphi\leq2\pi\).
Na kraju, ugao \(\theta\) je određen konusom: uslov \(z\sqrt{3}\geq\sqrt{x^{2}+y^{2}}\) u sfernim koordinatama postaje \(\rho\cos\theta\,\sqrt{3}\geq\rho\sin\theta\).
Kako je \(\rho>0\) možemo podeliti sa \(\rho\) i dobijamo
\[\sqrt{3}\cos\theta\geq\sin\theta\Longrightarrow\tan\theta\leq\sqrt{3}\Longrightarrow0\leq\theta\leq\frac{\pi}{3}.\]
Dakle, telo \(G\) u sfernim koordinatama je opisano granicama:
\[0\leq\rho\leq4,\quad0\leq\theta\leq\frac{\pi}{3},\quad0\leq\varphi\leq2\pi.\]
Sada je
\[\iiint\limits_{G}dx\,dy\,dz=\int_{0}^{2\pi}\int_{0}^{\pi/3}\int_{0}^{4}\rho^{2}\sin\theta\,d\rho\,d\theta\,d\varphi.\]
Kako su granice konstantne a promenljive unutar integrala nezavisne, možemo razdvojiti na proizvod tri integrala
\[\int_{0}^{4}\rho^{2}\,d\rho=\frac{\rho^{3}}{3}\bigg|_{0}^{4}=\frac{64}{3},\]
\[\int_{0}^{\pi/3}\sin\theta\,d\theta=-\cos\theta\bigg|_{0}^{\pi/3}=1-\cos\frac{\pi}{3}=1-\frac{1}{2}=\frac{1}{2},\]
\[\int_{0}^{2\pi}d\varphi=2\pi.\]
Dakle, dobijamo
\[\iiint\limits_{G}dx\,dy\,dz=\frac{64}{3}\cdot\frac{1}{2}\cdot2\pi=\frac{64\pi}{3}.\]
