Trojni integral – izračunavanje i smena promenljivih
Neka je \(f(x,y,z)\) neprekidna u oblasti
\[G=\{(x,y,z)\mid z_{1}(x,y)\le z\le z_{2}(x,y),\ (x,y)\in D\ \}\]
gde je
\[D=\{(x,y)\mid y_{1}(x)\le y\le y_{2}(x),\ a\leq x\leq b\ \}.\]
Tada je
\[\iiint\limits_{G}f(x,y,z)\,dx\,dy\,dz=\iint\limits_{D}dx\,dy\,\int_{z_{1}(x,y)}^{z_{2}(x,y)}f(x,y,z)\,dz=\int_{a}^{b}dx\,\int_{y_{1}(x)}^{y_{2}(x)}dy\,\int_{z_{1}(x,y)}^{z_{2}(x,y)}f(x,y,z)\,dz.\]
Primer 1. Izračunati integral \(\iiint\limits_{G}x\,z\;dx\,dy\,dz\) gde je telo
\(G\) ograničeno sa \(x=1,\,y=x^{2}\) i \(z=x+y\) u I oktantu.
Prikaži rešenje
Prikaži parabolički cilindar
Prikaži ravan \( x = 1 \)
Prikaži ravan \( z + x + y \)
Prikaži telo \( G \)
Rešenje:
Vidimo da je telo
\[G=\{(x,y,z)\;:\;0\le x\le1,\;0\le y\le x^{2},\;0\le z\le x+y\},\]
Dakle
\[\iiint\limits_{G}x\,z\;dx\,dy\,dz=\int_{0}^{1}\,dx\int_{0}^{x^{2}}\,dy\int_{0}^{x+y}x\,z\;dz.\]
Kako je
\[\int_{0}^{x+y}x\,z\,dz=x\frac{z^{2}}{2}\bigg|_{0}^{x+y}=\frac{x}{2}(x+y)^{2},\]
imamo
\[I=\int_{0}^{1}\,dx\int_{0}^{x^{2}}\frac{x}{2}(x+y)^{2}\;dy=\frac{1}{2}\int_{0}^{1}x\left(\int_{0}^{x^{2}}(x+y)^{2}dy\right)dx.\]
Dalje,
\[\int_{0}^{x^{2}}(x+y)^{2}dy=\int_{0}^{x^{2}}\bigl(x^{2}+2xy+y^{2}\bigr)dy=\bigg(x^{2}y+xy^{2}+\frac{y^{3}}{3}\bigg)\Big|_{0}^{x^{2}}=x^{4}+x^{5}+\frac{x^{6}}{3}.\]
Dakle
\[I=\frac{1}{2}\int_{0}^{1}x\left(x^{4}+x^{5}+\frac{x^{6}}{3}\right)dx=\frac{1}{2}\int_{0}^{1}\left(x^{5}+x^{6}+\frac{x^{7}}{3}\right)dx,\]
pa je
\[I=\frac{1}{2}\left(\frac{1}{6}+\frac{1}{7}+\frac{1}{24}\right)=\frac{1}{2}\cdot\frac{59}{168}=\frac{59}{336}.\]
Smena promenljivih u trojnom integralu
Neka je sistemom funkcija
\[x=x(u,v,w),\quad y=y(u,v,w),\quad z=z(u,v,w)\]
data bijekcija oblasti \(\Gamma\) na oblast \(G\) i neka su funkcije \(x(u,v,w), y(u,v,w), z(u,v,w)\) neprekidne zajedno sa svojim parcijalnim izvodima na oblasti \(\Gamma\). Jakobijan preslikavanja je:
\[J(u,v,w)=\begin{vmatrix}x'_{u} & x'_{v} & x'_{w}\\ y'_{u} & y'_{v} & y'_{w}\\ z'_{u} & z'_{v} & z'_{w}\end{vmatrix}.\]
Tada važi formula za smenu promenljivih:
\[\iiint\limits_{G}f(x,y,z)\,dx\,dy\,dz=\iiint\limits_{G'}f(x(u,v,w),y(u,v,w),z(u,v,w))\,\big|J(u,v,w)\big|\,du\,dv\,dw,\]
Polarno cilindrične koordinate:
Koriste se kod cilindra, paraboloida, hiperboloida...
\[
\begin{cases}
x=\rho\cos\varphi, & 0\leq\rho\lt +\infty\\
y=\rho\sin\varphi, & 0\leq\varphi\leq2\pi\\
z=z & J=\rho
\end{cases}
\]
Sferne koordinate:
Koriste se kada imamo sferu ili kombinaciju sfere i konusa...
\[
\begin{cases}
x=\rho\sin\theta\cos\varphi, & 0\leq\rho\lt +\infty\\
y=\rho\sin\theta\sin\varphi, & 0\leq\varphi\leq2\pi,\\
z=\rho\cos\theta & 0\leq\theta\leq\pi
\end{cases}\quad J=\rho^{2}\sin\theta.
\]
Eliptičke koordinate:
Koriste se kada imamo elipsoid ili kombinaciju elipsoida i konusa...
\[
\begin{cases}
x=a\rho\sin\theta\cos\varphi, & 0\leq\rho\lt +\infty\\
y=b\rho\sin\theta\sin\varphi, & 0\leq\varphi\leq2\pi,\\
z=c\rho\cos\theta & 0\leq\theta\leq\pi
\end{cases}\quad J=abc\rho^{2}\sin\theta.
\]
Primer 2. Izračunati \(\iiint\limits_{G} z\,dx\,dy\,dz\) gde je \(G\) deo cilindra \(x^{2}+y^{2}\le1,\ y\ge0\) ograničenog ravnima \(z=0\) i \(z=2\).
Prikaži rešenje
Prikaži cilindar
Prikaži ravan \( y = 0 \)
Prikaži ravan \( z = 0 \)
Prikaži ravan \( z = 2 \)
Prikaži telo \( G \)
Rešenje:
Prelazimo u cilindrične koordinate:
\[
\begin{cases}
x=\rho\cos\varphi, & 0\leq\rho\leq1\\
y=\rho\sin\varphi, & 0\leq\varphi\leq\pi\\
z=z & J=\rho
\end{cases}
\]
Tada integral postaje:
\[\iiint\limits_{G}z\,dx\,dy\,dz=\int_{0}^{\pi}\int_{0}^{1}\int_{0}^{2}z\,\rho\,dz\,d\rho\,d\varphi\]
Kako je
\[\int_{0}^{2}zdz=\bigg(\frac{z^{2}}{2}\bigg)\bigg|_{0}^{2}=2\]
dobijamo
\[
\iiint\limits_{G} z\,dx\,dy\,dz
= \int_{0}^{\pi}\int_{0}^{1} 2\rho \, d\rho \, d\varphi
= \int_{0}^{\pi} 1 \, d\varphi
= \pi.
\]
Primer 3. Izračunati \(\iiint\limits_{G}dx\,dy\,dz\) gde je \(G:\;\frac{x^{2}}{4}+\frac{y^{2}}{9}+\frac{z^{2}}{16}\le1.\)
Rešenje:
Uvodimo uopštene sferne koordinate
\[\begin{cases}
x=2\rho\sin\theta\cos\varphi, & 0\leq\rho\leq1\\
y=3\rho\sin\theta\sin\varphi, & 0\leq\varphi\leq2\pi,\\
z=4\rho\cos\theta & 0\leq\theta\leq\pi
\end{cases}\quad J=24\rho^{2}\sin\theta.\]
Dakle,
\[\iiint\limits_{G}dx\,dy\,dz=\int_{0}^{2\pi}\!\int_{0}^{\pi}\!\int_{0}^{1}24\rho^{2}\sin\theta\,d\rho\,d\theta\,d\varphi.\]
Kako su
\[\int_{0}^{1}\rho^{2}\,d\rho=\frac{1}{3},\qquad\int_{0}^{\pi}\sin\theta\,d\theta=2,\qquad\int_{0}^{2\pi}d\varphi=2\pi,\]
dobijamo
\[\iiint_{G} dx\,dy\,dz = 24 \cdot \frac{1}{3} \cdot 2 \cdot 2\pi = 32\pi\]
