Aptarimas:Matematika/Gauso formulė

Gauso formule atrodo kaip pilstymas is tuscio i kiaura.

Pavyzdžiai

Vaizdas:Pavirsinisintris427.jpg

427.

• Apskaičiuoti tūrį V kūno (piramidės), kurį riboja plokštuma $z=6-2x-3y$ ir sienos $x=0$, $y=0$, $z=0$. (pav. 427). Trikampio ABC Viršunės yra A(3; 0; 0), B(0; 2; 0) ir C(0; 0; 6). Piramidės AOBC viršunės yra A(3; 0; 0), O(0; 0; 0), B(0; 2; 0) ir C(0; 0; 6). Iš karto matosi, kad piramidės tūris yra $V={\frac {1}{3}}\cdot S_{\Delta ABC}\cdot h={\frac {1}{3}}\cdot {\frac {1}{2}}(OA\cdot OB)\cdot OC={\frac {1}{6}}\cdot (3\cdot 2)\cdot 6=6.$

Randame:

$x={\frac {6-3y-z}{2}}=3-{\frac {3y+z}{2}},$

$y={\frac {6-2x-z}{3}}=2-{\frac {2x+z}{3}}.$

Dar Randame:

$\iint _{S}xdydz=\int _{0}^{2}dy\int _{0}^{6}xdz=\int _{0}^{2}dy\int _{0}^{6}(3-{\frac {3y+z}{2}})dz=\int _{0}^{2}dy\int _{0}^{6}(3-{\frac {3y}{2}}-{\frac {z}{2}})dz=$

$=\int _{0}^{2}dy(3z-{\frac {3yz}{2}}-{\frac {z^{2}}{4}})|_{0}^{6}=\int _{0}^{2}(3\cdot 6-{\frac {3y\cdot 6}{2}}-{\frac {6^{2}}{4}})dy=\int _{0}^{2}(18-{\frac {18y}{2}}-{\frac {36}{4}})dy=\int _{0}^{2}(18-9y-9)dy=$

$=\int _{0}^{2}(9-9y)dy=(9y-{\frac {9y^{2}}{2}})|_{0}^{2}=9\cdot 2-{\frac {9\cdot 2^{2}}{2}}=18-18=0.$

$\iint _{S}ydzdx=\int _{0}^{3}dx\int _{0}^{6}ydz=\int _{0}^{3}dx\int _{0}^{6}(2-{\frac {2x+z}{3}})dz=\int _{0}^{3}dx\int _{0}^{6}(2-{\frac {2x}{3}}-{\frac {z}{3}})dz=$

$=\int _{0}^{3}dx(2z-{\frac {2xz}{3}}-{\frac {z^{2}}{6}})|_{0}^{6}=\int _{0}^{3}(2\cdot 6-{\frac {2x\cdot 6}{3}}-{\frac {6^{2}}{6}})dx=\int _{0}^{3}(12-4x-6)dx=\int _{0}^{3}(6-4x)dx=(6x-2x^{2})|_{0}^{3}=6\cdot 3-2\cdot 3^{2}=18-18=0.$

$V={\frac {1}{3}}\iint _{S}xdydz+ydzdx+zdxdy=0+0+0=0.$

Trikampio plotą galima surasti ir klasikiniu budu. Ieškomas plotas yra trikampis ABC, kurio taškai yra A(3; 0; 0), B(0; 2; 0) ir C(0; 0; 6). Pavadiname atkarpas AB=a, BC=b, CA=c; OA=d=3, OB=e=2, OC=f=6. Koordinačių pradžios taškas yra O(0; 0; 0). Randame trikampio ABC kraštinių ilgius:

$a={\sqrt {d^{2}+e^{2}}}={\sqrt {3^{2}+2^{2}}}={\sqrt {9+4}}={\sqrt {13}}=3,605551275;$

$b={\sqrt {e^{2}+f^{2}}}={\sqrt {2^{2}+6^{2}}}={\sqrt {4+36}}={\sqrt {40}}=2{\sqrt {10}}=6,32455532;$

$c={\sqrt {d^{2}+f^{2}}}={\sqrt {3^{2}+6^{2}}}={\sqrt {9+36}}={\sqrt {45}}=3{\sqrt {5}}=6,708203933.$

Toliau randame trikampio ABC pusperimetrį $p={\frac {a+b+c}{2}}={\frac {{\sqrt {13}}+{\sqrt {40}}+{\sqrt {45}}}{2}}=8,319155264$ ir plotą:

$S={\sqrt {p(p-a)(p-b)(p-c)}}={\sqrt {8,319155264(8,319155264-{\sqrt {13}})(8,319155264-{\sqrt {40}})(8,319155264-{\sqrt {45}})}}=$

$={\sqrt {8.319155264\cdot 4.713603989\cdot 1.994599944\cdot 1.610951332}}={\sqrt {126}}=11.22497216.$

Vadinasi šiame pavyzdyje ieškomas buvo ne trikampio plotas.

Integravime pagal paviršiaus plotą taikomos išvestinės todėl patikrinimas neteisingas

• Pavyzdis. Apskaičiuoti integralą $\iint _{S}x^{3}dydz+y^{3}dzdx+z^{3}dxdy$ pagal išorinę pusę sferos $x^{2}+y^{2}+z^{2}=R^{2}.$

Taikydami formulę Gauso - Ostragradksio, gauname:

$\iint _{S}x^{3}dydz+y^{3}dzdx+z^{3}dxdy=\iiint _{V}(3x^{2}+3y^{2}+3z^{2})dxdydz=3\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta \int _{0}^{R}\rho ^{4}d\rho =$

$=3\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta {\frac {\rho ^{5}}{5}}|_{0}^{R}=3\cdot {\frac {R^{5}}{5}}\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta ={\frac {3R^{5}}{5}}\int _{0}^{2\pi }d\phi \;\sin \theta |_{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}={\frac {3R^{5}}{5}}\int _{0}^{2\pi }(\sin {\frac {\pi }{2}}-\sin {\frac {-\pi }{2}})d\phi =$

$={\frac {3R^{5}}{5}}\int _{0}^{2\pi }(1-(-1))d\phi ={\frac {6R^{5}}{5}}\phi |_{0}^{2\pi }={\frac {6R^{5}}{5}}\cdot 2\pi ={\frac {12\pi R^{5}}{5}}.$

Patikrinsime apskaičiuodami $\iint _{S}x^{3}\mathbf {d} y\mathbf {d} z,\;\iint _{S}y^{3}\mathbf {d} z\mathbf {d} x$ ir $\iint _{S}x^{3}\mathbf {d} x\mathbf {d} y$ sumą.

$x^{2}=R^{2}-y^{2}-z^{2},$

$x={\sqrt {R^{2}-y^{2}-z^{2}}}.$

$V_{x}=\iint _{S}x^{3}\mathbf {d} y\mathbf {d} z=\iint _{S}\left({\sqrt {R^{2}-y^{2}-z^{2}}}\right)^{3}\mathbf {d} y\mathbf {d} z=\iint _{S}\left({\sqrt {R^{2}-\rho ^{2}}}\right)^{3}\mathbf {d} \rho \mathbf {d} \phi =$

$=\int _{0}^{2\pi }\left(\int _{0}^{R}{\sqrt {(R^{2}-\rho ^{2})^{3}}}\mathbf {d} \rho \right)\mathbf {d} \phi .$

Pasinaudodami internetiniu integratoriumi, gauname, kad

$\int _{0}^{R}{\sqrt {(R^{2}-\rho ^{2})^{3}}}\mathbf {d} \rho ={\frac {1}{8}}\left(\rho (5R^{2}-2\rho ^{2}){\sqrt {R^{2}-\rho ^{2}}}+3R^{4}\arctan {\frac {\rho }{\sqrt {R^{2}-\rho ^{2}}}}\right)|_{0}^{R}.$

Toliau pritaikydami ribas, kai $\rho$ artėja į $R$, gauname:

$\int _{0}^{R}{\sqrt {(R^{2}-\rho ^{2})^{3}}}\mathbf {d} \rho ={\frac {1}{8}}\left(R(5R^{2}-2R^{2}){\sqrt {R^{2}-R^{2}}}+3R^{4}\arctan {\frac {R}{\sqrt {R^{2}-\rho ^{2}}}}\right)|_{0}^{R}=$

$={\frac {1}{8}}\left(3R^{3}\cdot {\sqrt {0}}+3R^{4}\arctan {\frac {R}{\lim _{\rho \to R}({\sqrt {R^{2}-\rho ^{2}}})}}\right)|_{0}^{R}={\frac {1}{8}}\left(0+3R^{4}\arctan {\frac {R}{0.00000000001}}\right)=$

$={\frac {1}{8}}\left(3R^{4}\arctan(\infty )\right).$

Kadangi, $\lim _{a\to \infty }=\arctan(a)=\pi$, tai $\arctan(\infty )=\pi .$ Todėl,

$\int _{0}^{R}{\sqrt {(R^{2}-\rho ^{2})^{3}}}\mathbf {d} \rho ={\frac {3}{8}}R^{4}\pi .$

Vadinasi,

$V_{x}=\iint _{S}x^{3}\mathbf {d} y\mathbf {d} z=\int _{0}^{2\pi }\left(\int _{0}^{R}{\sqrt {(R^{2}-\rho ^{2})^{3}}}\mathbf {d} \rho \right)\mathbf {d} \phi =\int _{0}^{2\pi }{\frac {3}{8}}R^{4}\pi \mathbf {d} \phi ={\frac {3}{8}}R^{4}\pi \phi _{0}^{2\pi }={\frac {3}{8}}R^{4}\pi \phi |_{0}^{2\pi }={\frac {3}{8}}R^{4}\pi \cdot 2\pi ={\frac {3}{4}}R^{4}\pi ^{2}.$

$V=3V_{x}={\frac {9}{4}}R^{4}\pi ^{2}.$

Integravime pagal paviršiaus plotą taikomos išvestinės todėl patikrinimas neteisingas 2

• Pavyzdis. Apskaičiuoti integralą $\iint _{S}x^{3}dydz+y^{3}dzdx+z^{3}dxdy$ pagal išorinę pusę sferos $x^{2}+y^{2}+z^{2}=R^{2}.$

Taikydami formulę Gauso - Ostragradksio, gauname:

$\iint _{S}x^{3}dydz+y^{3}dzdx+z^{3}dxdy=\iiint _{V}(3x^{2}+3y^{2}+3z^{2})dxdydz=3\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta \int _{0}^{R}\rho ^{4}d\rho =$

$=3\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta {\frac {\rho ^{5}}{5}}|_{0}^{R}=3\cdot {\frac {R^{5}}{5}}\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta ={\frac {3R^{5}}{5}}\int _{0}^{2\pi }d\phi \;\sin \theta |_{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}={\frac {3R^{5}}{5}}\int _{0}^{2\pi }(\sin {\frac {\pi }{2}}-\sin {\frac {-\pi }{2}})d\phi =$

$={\frac {3R^{5}}{5}}\int _{0}^{2\pi }(1-(-1))d\phi ={\frac {6R^{5}}{5}}\phi |_{0}^{2\pi }={\frac {6R^{5}}{5}}\cdot 2\pi ={\frac {12\pi R^{5}}{5}}.$

Patikrinsime apskaičiuodami $\iint _{S}x^{3}\mathbf {d} y\mathbf {d} z,\;\iint _{S}y^{3}\mathbf {d} z\mathbf {d} x$ ir $\iint _{S}x^{3}\mathbf {d} x\mathbf {d} y$ sumą.

$x^{2}=R^{2}-y^{2}-z^{2},$

$x={\sqrt {R^{2}-y^{2}-z^{2}}}.$

$V_{x}=\iint _{S}x^{3}\mathbf {d} y\mathbf {d} z=\iint _{S}\left({\sqrt {R^{2}-y^{2}-z^{2}}}\right)^{3}\mathbf {d} y\mathbf {d} z=\iint _{S}\left({\sqrt {R^{2}-\rho ^{2}}}\right)^{3}\rho \mathbf {d} \rho \mathbf {d} \phi =$

$=\int _{0}^{2\pi }\left(\int _{0}^{R}\rho {\sqrt {(R^{2}-\rho ^{2})^{3}}}\mathbf {d} \rho \right)\mathbf {d} \phi .$

Pasinaudodami internetiniu integratoriumi, gauname, kad

$\int _{0}^{R}\rho {\sqrt {(R^{2}-\rho ^{2})^{3}}}\mathbf {d} \rho =-{\frac {1}{5}}((R-\rho )(R+\rho ))^{5/2}|_{0}^{R}=$

$=-{\frac {1}{5}}((R-R)(R+R))^{5/2}-\left(-{\frac {1}{5}}((R-0)(R+0))^{5/2}\right)=$

$=0+{\frac {1}{5}}((R-0)(R+0))^{5/2}={\frac {1}{5}}(R^{2})^{5/2}={\frac {R^{5}}{5}};$

$V_{x}=\int _{0}^{2\pi }\left(\int _{0}^{R}\rho {\sqrt {(R^{2}-\rho ^{2})^{3}}}\mathbf {d} \rho \right)\mathbf {d} \phi =\int _{0}^{2\pi }{\frac {R^{5}}{5}}\mathbf {d} \phi ={\frac {2\pi R^{5}}{5}}.$

Kadangi reikia dviejų rutulio pusrutulių (teigiama ir neigiama Ox kryptimi), tai

$V_{X}=2V_{x}=2\cdot {\frac {2\pi R^{5}}{5}}={\frac {4\pi R^{5}}{5}}.$

Kadangi, pagal sąlyga bus $V_{X}=V_{Y}=V_{Z},$ tai

$V=3V_{X}=3\cdot {\frac {4\pi R^{5}}{5}}={\frac {12\pi R^{5}}{5}}.$

Pasirodo įmanoma

• Pavyzdis. Apskaičiuoti integralą $\iint _{S}x^{3}dydz+y^{3}dzdx+z^{3}dxdy$ pagal išorinę pusę sferos $x^{2}+y^{2}+z^{2}=R^{2}.$

Taikydami formulę Gauso - Ostragradksio, gauname:

$\iint _{S}x^{3}dydz+y^{3}dzdx+z^{3}dxdy=\iiint _{V}(3x^{2}+3y^{2}+3z^{2})dxdydz=3\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta \int _{0}^{R}\rho ^{4}d\rho =$

$=3\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta {\frac {\rho ^{5}}{5}}|_{0}^{R}=3\cdot {\frac {R^{5}}{5}}\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta ={\frac {3R^{5}}{5}}\int _{0}^{2\pi }d\phi \;\sin \theta |_{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}={\frac {3R^{5}}{5}}\int _{0}^{2\pi }(\sin {\frac {\pi }{2}}-\sin {\frac {-\pi }{2}})d\phi =$

$={\frac {3R^{5}}{5}}\int _{0}^{2\pi }(1-(-1))d\phi ={\frac {6R^{5}}{5}}\phi |_{0}^{2\pi }={\frac {6R^{5}}{5}}\cdot 2\pi ={\frac {12\pi R^{5}}{5}}.$

Patikrinsime apskaičiuodami $\iint _{S}x^{3}\mathbf {d} y\mathbf {d} z,\;\iint _{S}y^{3}\mathbf {d} z\mathbf {d} x$ ir $\iint _{S}x^{3}\mathbf {d} x\mathbf {d} y$ sumą.

$x^{2}=R^{2}-y^{2}-z^{2},$

$x={\sqrt {R^{2}-y^{2}-z^{2}}};$

${\frac {\partial x}{\partial y}}={\frac {\partial ({\sqrt {R^{2}-y^{2}-z^{2}}})}{\partial y}}={\frac {-2y}{2{\sqrt {R^{2}-y^{2}-z^{2}}}}}={\frac {-y}{\sqrt {R^{2}-y^{2}-z^{2}}}};$

${\frac {\partial x}{\partial z}}={\frac {\partial ({\sqrt {R^{2}-y^{2}-z^{2}}})}{\partial z}}={\frac {-2z}{2{\sqrt {R^{2}-y^{2}-z^{2}}}}}={\frac {-z}{\sqrt {R^{2}-y^{2}-z^{2}}}}.$

$V_{x}=\iint _{S}x^{3}{\sqrt {1+\left({\frac {\partial x}{\partial y}}\right)^{2}+\left({\frac {\partial x}{\partial z}}\right)^{2}}}\mathbf {d} y\mathbf {d} z=$

$=\iint _{S}x^{3}{\sqrt {1+\left({\frac {-y}{\sqrt {R^{2}-y^{2}-z^{2}}}}\right)^{2}+\left({\frac {-z}{\sqrt {R^{2}-y^{2}-z^{2}}}}\right)^{2}}}\mathbf {d} y\mathbf {d} z=$

$=\iint _{S}x^{3}{\sqrt {1+{\frac {y^{2}}{R^{2}-y^{2}-z^{2}}}+{\frac {z^{2}}{R^{2}-y^{2}-z^{2}}}}}\mathbf {d} y\mathbf {d} z=$

$=\iint _{S}x^{3}{\sqrt {\frac {R^{2}-y^{2}-z^{2}+y^{2}+z^{2}}{R^{2}-y^{2}-z^{2}}}}\mathbf {d} y\mathbf {d} z=$

$=\iint _{S}x^{3}{\sqrt {\frac {R^{2}}{R^{2}-y^{2}-z^{2}}}}\mathbf {d} y\mathbf {d} z=\iint _{S}{\sqrt {R^{2}-y^{2}-z^{2}}}{\sqrt {\frac {R^{2}}{R^{2}-y^{2}-z^{2}}}}\mathbf {d} y\mathbf {d} z=\iint _{S}R\mathbf {d} y\mathbf {d} z=$

$=\iint _{S}R\rho \mathbf {d} \rho \mathbf {d} \phi =\int _{0}^{2\pi }\left(\int _{0}^{R}R\rho \mathbf {d} \rho \right)\mathbf {d} \phi =$

$=\int _{0}^{2\pi }\left(R\cdot {\frac {R^{2}}{2}}\right)\mathbf {d} \phi ={\frac {R^{3}}{2}}\int _{0}^{2\pi }\mathbf {d} \phi ={\frac {R^{3}}{2}}\cdot 2\pi =R^{3}\pi .$

$V=3V_{x}=3R^{3}\pi .$

Gautas tūris nesutampa su Gauso formulės logika.

Pastaba. Rutulio paviršiaus ploto neįmanoma apskaičiuoti nei cilindinėse, nei sferinėse koordinatėse. Bandydami, gauname:

$R^{2}=x^{2}+y^{2}+z^{2},$

$z={\sqrt {R^{2}-x^{2}-y^{2}}};$

$z_{x}'={\frac {-2x}{2{\sqrt {R^{2}-x^{2}-y^{2}}}}}={\frac {-x}{\sqrt {R^{2}-x^{2}-y^{2}}}};$

$z_{y}'={\frac {-2y}{2{\sqrt {R^{2}-x^{2}-y^{2}}}}}={\frac {-y}{\sqrt {R^{2}-x^{2}-y^{2}}}};$

${\sqrt {1+(z_{x}')^{2}+(z_{y}')^{2}}}={\sqrt {1+\left({\frac {-x}{\sqrt {R^{2}-x^{2}-y^{2}}}}\right)^{2}+\left({\frac {-y}{\sqrt {R^{2}-x^{2}-y^{2}}}}\right)^{2}}}=$

$={\sqrt {1+{\frac {x^{2}}{R^{2}-x^{2}-y^{2}}}+{\frac {y^{2}}{R^{2}-x^{2}-y^{2}}}}}={\sqrt {\frac {R^{2}-x^{2}-y^{2}+x^{2}+y^{2}}{R^{2}-x^{2}-y^{2}}}}={\sqrt {\frac {R^{2}}{R^{2}-x^{2}-y^{2}}}};$

$x^{2}+y^{2}=\rho ^{2}$ cilindinėse ir polinėse koordinatėse;

Iš internetinio integratoriaus $\int _{0}^{R}\rho {\sqrt {\frac {R^{2}}{R^{2}-\rho ^{2}}}}\;d\rho =(\rho ^{2}-R^{2}){\sqrt {\frac {R^{2}}{R^{2}-\rho ^{2}}}}|_{0}^{R}=-R{\sqrt {R^{2}-\rho ^{2}}}|_{0}^{R};$

$S_{pav.}=\iint _{s}{\sqrt {1+(z_{x}')^{2}+(z_{y}')^{2}}}\;dx\;dy=\iint _{s}{\sqrt {\frac {R^{2}}{R^{2}-x^{2}-y^{2}}}}\;dx\;dy=\int _{0}^{2\pi }d\phi \int _{0}^{R}\rho {\sqrt {\frac {R^{2}}{R^{2}-\rho ^{2}}}}\;d\rho =$

$=\int _{0}^{2\pi }-R{\sqrt {R^{2}-\rho ^{2}}}|_{0}^{R}d\phi =-R\int _{0}^{2\pi }({\sqrt {R^{2}-R^{2}}}-{\sqrt {R^{2}-0^{2}}})d\phi =-R\int _{0}^{2\pi }(0-{\sqrt {R^{2}}})d\phi =R^{2}\int _{0}^{2\pi }d\phi =2\pi R^{2}.$

Taigi, visas rutulio paviršiaus plotas susideda iš dviejų pusrutulių, todėl

$S_{visas}=2\cdot S_{pav.}=2\cdot 2\pi R^{2}=4\pi R^{2}.$

Kitaip patikrinsime apskaičiuodami $\iint _{S}x^{3}\mathbf {d} y\mathbf {d} z,\;\iint _{S}y^{3}\mathbf {d} z\mathbf {d} x$ ir $\iint _{S}x^{3}\mathbf {d} x\mathbf {d} y$ sumą.

$x^{2}=R^{2}-y^{2}-z^{2},$

$x={\sqrt {R^{2}-y^{2}-z^{2}}}.$

$V_{x}=\iint _{S}x^{3}\mathbf {d} y\mathbf {d} z=\iint _{S}\left({\sqrt {R^{2}-y^{2}-z^{2}}}\right)^{3}\mathbf {d} y\mathbf {d} z=\iint _{S}\left({\sqrt {R^{2}-\rho ^{2}}}\right)^{3}\rho \mathbf {d} \rho \mathbf {d} \phi =$

$=\int _{0}^{2\pi }\left(\int _{0}^{R}\rho {\sqrt {(R^{2}-\rho ^{2})^{3}}}\mathbf {d} \rho \right)\mathbf {d} \phi .$

Pasinaudodami internetiniu integratoriumi, gauname, kad

$\int _{0}^{R}\rho {\sqrt {(R^{2}-\rho ^{2})^{3}}}\mathbf {d} \rho =-{\frac {1}{5}}((R-\rho )(R+\rho ))^{5/2}|_{0}^{R}=$

$=-{\frac {1}{5}}((R-R)(R+R))^{5/2}-\left(-{\frac {1}{5}}((R-0)(R+0))^{5/2}\right)=$

$=0+{\frac {1}{5}}((R-0)(R+0))^{5/2}={\frac {1}{5}}(R^{2})^{5/2}={\frac {R^{5}}{5}};$

$V_{x}=\int _{0}^{2\pi }\left(\int _{0}^{R}\rho {\sqrt {(R^{2}-\rho ^{2})^{3}}}\mathbf {d} \rho \right)\mathbf {d} \phi =\int _{0}^{2\pi }{\frac {R^{5}}{5}}\mathbf {d} \phi ={\frac {2\pi R^{5}}{5}}.$

Kadangi reikia dviejų rutulio pusrutulių (teigiama ir neigiama Ox kryptimi), tai

$V_{X}=2V_{x}=2\cdot {\frac {2\pi R^{5}}{5}}={\frac {4\pi R^{5}}{5}}.$

Kadangi, pagal sąlyga bus $V_{X}=V_{Y}=V_{Z},$ tai

$V=3V_{X}=3\cdot {\frac {4\pi R^{5}}{5}}={\frac {12\pi R^{5}}{5}}.$

Gal blogas pavyzdis

• Pavyzdis. Apskaičiuoti integralą $\iint _{S}x^{2}\;dydz+0\;dzdx+0\;dxdy$ pagal išorinę pusę sferos $x^{2}+y^{2}+z^{2}=R^{2}.$

Taikydami formulę Gauso, gauname:

$\iint _{S}x^{2}dydz=\iiint _{V}(2x+0+0)dxdydz=2\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta \int _{0}^{R}\rho ^{3}d\rho =$

$=2\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta {\frac {\rho ^{4}}{4}}|_{0}^{R}=2\cdot {\frac {R^{4}}{4}}\int _{0}^{2\pi }d\phi \int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}\cos \theta d\theta ={\frac {R^{4}}{2}}\int _{0}^{2\pi }d\phi \;\sin \theta |_{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}={\frac {R^{4}}{2}}\int _{0}^{2\pi }(\sin {\frac {\pi }{2}}-\sin {\frac {-\pi }{2}})d\phi =$

$={\frac {R^{4}}{2}}\int _{0}^{2\pi }(1-(-1))d\phi =R^{4}\phi |_{0}^{2\pi }=R^{4}\cdot 2\pi =2\pi R^{4}.$

Dar viską reikia padalinti iš 3, kad gauti teisingai:

$\iint _{S}x^{2}\;dydz+0\;dzdx+0\;dxdy={\frac {2}{3}}\pi R^{4}.$

Kitaip patikrinsime apskaičiuodami $\iint _{S}x^{2}\mathbf {d} y\mathbf {d} z.$

$x^{2}=R^{2}-y^{2}-z^{2},$

$x={\sqrt {R^{2}-y^{2}-z^{2}}}.$

$V_{x}=\iint _{S}x^{2}\mathbf {d} y\mathbf {d} z=\iint _{S}\left({\sqrt {R^{2}-y^{2}-z^{2}}}\right)^{2}\mathbf {d} y\mathbf {d} z=\iint _{S}\left({\sqrt {R^{2}-\rho ^{2}}}\right)^{2}\rho \mathbf {d} \rho \mathbf {d} \phi =$

$=\int _{0}^{2\pi }\left(\int _{0}^{R}\rho (R^{2}-\rho ^{2})\mathbf {d} \rho \right)\mathbf {d} \phi .$

Pasinaudodami internetiniu integratoriumi, gauname, kad

$\int _{0}^{R}\rho (R^{2}-\rho ^{2})\mathbf {d} \rho =\int _{0}^{R}(\rho R^{2}-\rho ^{3})\mathbf {d} \rho =({\frac {\rho ^{2}R^{2}}{2}}-{\frac {\rho ^{4}}{4}})|_{0}^{R}=$

$={\frac {R^{2}\cdot R^{2}}{2}}-{\frac {R^{4}}{4}}={\frac {R^{4}}{4}}.$

$V_{x}=\int _{0}^{2\pi }\left(\int _{0}^{R}\rho {\sqrt {(R^{2}-\rho ^{2})^{2}}}\mathbf {d} \rho \right)\mathbf {d} \phi =\int _{0}^{2\pi }{\frac {R^{4}}{4}}\mathbf {d} \phi ={\frac {2\pi R^{4}}{4}}={\frac {\pi R^{4}}{2}}.$

Kadangi reikia dviejų rutulio pusrutulių (teigiama ir neigiama Ox kryptimi), tai

$V_{X}=2V_{x}=2\cdot {\frac {\pi R^{4}}{2}}=\pi R^{4}.$

Atsakymai ${\frac {2}{3}}\pi R^{4}$ ir $\pi R^{4}$ nesutampa, todėl tokie skaičiavimai neturi prasmės (Gauso formulė gali būti arba neteisinga arba turi labai mažai pritaikymo pavyzdžių).