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 $V=\frac{1}{3}\cdot S_{\mathrm{\Delta }ABC}\cdot h=\frac{1}{3}\cdot \frac{1}{2}(OA\cdot OB)\cdot OC=\frac{1}{6}\cdot (3\cdot 2)\cdot 6=6.$

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:

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

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

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:

${\iint }_S{x}^{3}dydz+y^{3}dzdx+z^{3}dxdy={\iiint }_V(3x^2+3y^2+3z^2)dxdydz=3{\int }_0^{2\pi }d\varphi {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos \theta d\theta {\int }_0^R{\rho }^{4}d\rho =$

$=3{\int }_0^{2\pi }d\varphi {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos \theta d\theta \frac{{\rho }^5}{5}{|}_0^R=3\cdot \frac{R^5}{5}{\int }_0^{2\pi }d\varphi {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos \theta d\theta =\frac{3R^5}{5}{\int }_0^{2\pi }d\varphi \sin \theta {|}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}=\frac{3R^5}{5}{\int }_0^{2\pi }(\sin \frac{\pi }{2}-\sin \frac{-\pi }{2})d\varphi =$

$=\frac{3R^5}{5}{\int }_0^{2\pi }(1-(-1))d\varphi =\frac{6R^5}{5}\varphi {|}_0^{2\pi }=\frac{6R^5}{5}\cdot 2\pi =\frac{12\pi R^5}{5}.$

Patikrinsime apskaičiuodami ${\iint }_S{x}^3𝐝y𝐝z,{\iint }_S{y}^3𝐝z𝐝x$ ir ${\iint }_S{x}^3𝐝x𝐝y$ sumą.

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

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

$V_x={\iint }_S{x}^3𝐝y𝐝z={\iint }_S{\left(\sqrt{R^2-y^2-z^2}\right)}^3𝐝y𝐝z={\iint }_S{\left(\sqrt{R^2-{\rho }^2}\right)}^3𝐝\rho 𝐝\varphi =$

$={\int }_0^{2\pi }\left({\int }_0^R\sqrt{(R^2-{\rho }^2{)}^3}𝐝\rho \right)𝐝\varphi .$

Pasinaudodami internetiniu integratoriumi, gauname, kad

${\int }_0^R\sqrt{(R^2-{\rho }^2{)}^3}𝐝\rho =\frac{1}{8}(\rho (5R^2-2{\rho }^2)\sqrt{R^2-{\rho }^2}+3R^4\arctan \frac{\rho }{\sqrt{R^2-{\rho }^2}}){|}_0^R.$

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

${\int }_0^R\sqrt{(R^2-{\rho }^2{)}^3}𝐝\rho =\frac{1}{8}(R(5R^2-2R^2)\sqrt{R^2-R^2}+3R^4\arctan \frac{R}{\sqrt{R^2-{\rho }^2}}){|}_0^R=$

$=\frac{1}{8}(3R^3\cdot \sqrt{0}+3R^4\arctan \frac{R}{\underset{\rho \to R}{lim}(\sqrt{R^2-{\rho }^2})}){|}_0^R=\frac{1}{8}(0+3R^4\arctan \frac{R}{0.00000000001})=$

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

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

${\int }_0^R\sqrt{(R^2-{\rho }^2{)}^3}𝐝\rho =\frac{3}{8}{R}^4\pi .$

Vadinasi,

$V_x={\iint }_S{x}^3𝐝y𝐝z={\int }_0^{2\pi }\left({\int }_0^R\sqrt{(R^2-{\rho }^2{)}^3}𝐝\rho \right)𝐝\varphi ={\int }_0^{2\pi }\frac{3}{8}{R}^4\pi 𝐝\varphi =\frac{3}{8}{R}^4\pi {\varphi }_0^{2\pi }=\frac{3}{8}{R}^4\pi \varphi {|}_0^{2\pi }=\frac{3}{8}{R}^4\pi \cdot 2\pi =\frac{3}{4}{R}^4{\pi }^2.$

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

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:

${\iint }_S{x}^{3}dydz+y^{3}dzdx+z^{3}dxdy={\iiint }_V(3x^2+3y^2+3z^2)dxdydz=3{\int }_0^{2\pi }d\varphi {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos \theta d\theta {\int }_0^R{\rho }^{4}d\rho =$

$=3{\int }_0^{2\pi }d\varphi {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos \theta d\theta \frac{{\rho }^5}{5}{|}_0^R=3\cdot \frac{R^5}{5}{\int }_0^{2\pi }d\varphi {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos \theta d\theta =\frac{3R^5}{5}{\int }_0^{2\pi }d\varphi \sin \theta {|}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}=\frac{3R^5}{5}{\int }_0^{2\pi }(\sin \frac{\pi }{2}-\sin \frac{-\pi }{2})d\varphi =$

$=\frac{3R^5}{5}{\int }_0^{2\pi }(1-(-1))d\varphi =\frac{6R^5}{5}\varphi {|}_0^{2\pi }=\frac{6R^5}{5}\cdot 2\pi =\frac{12\pi R^5}{5}.$

Patikrinsime apskaičiuodami ${\iint }_S{x}^3𝐝y𝐝z,{\iint }_S{y}^3𝐝z𝐝x$ ir ${\iint }_S{x}^3𝐝x𝐝y$ sumą.

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

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

$V_x={\iint }_S{x}^3𝐝y𝐝z={\iint }_S{\left(\sqrt{R^2-y^2-z^2}\right)}^3𝐝y𝐝z={\iint }_S{\left(\sqrt{R^2-{\rho }^2}\right)}^3\rho 𝐝\rho 𝐝\varphi =$

$={\int }_0^{2\pi }\left({\int }_0^R\rho \sqrt{(R^2-{\rho }^2{)}^3}𝐝\rho \right)𝐝\varphi .$

Pasinaudodami internetiniu integratoriumi, gauname, kad

${\int }_0^R\rho \sqrt{(R^2-{\rho }^2{)}^3}𝐝\rho =-\frac{1}{5}((R-\rho )(R+\rho ){)}^{5/2}{|}_0^R=$

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

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

$V_x={\int }_0^{2\pi }\left({\int }_0^R\rho \sqrt{(R^2-{\rho }^2{)}^3}𝐝\rho \right)𝐝\varphi ={\int }_0^{2\pi }\frac{R^5}{5}𝐝\varphi =\frac{2\pi R^5}{5}.$

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:

${\iint }_S{x}^{3}dydz+y^{3}dzdx+z^{3}dxdy={\iiint }_V(3x^2+3y^2+3z^2)dxdydz=3{\int }_0^{2\pi }d\varphi {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos \theta d\theta {\int }_0^R{\rho }^{4}d\rho =$

$=3{\int }_0^{2\pi }d\varphi {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos \theta d\theta \frac{{\rho }^5}{5}{|}_0^R=3\cdot \frac{R^5}{5}{\int }_0^{2\pi }d\varphi {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos \theta d\theta =\frac{3R^5}{5}{\int }_0^{2\pi }d\varphi \sin \theta {|}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}=\frac{3R^5}{5}{\int }_0^{2\pi }(\sin \frac{\pi }{2}-\sin \frac{-\pi }{2})d\varphi =$

$=\frac{3R^5}{5}{\int }_0^{2\pi }(1-(-1))d\varphi =\frac{6R^5}{5}\varphi {|}_0^{2\pi }=\frac{6R^5}{5}\cdot 2\pi =\frac{12\pi R^5}{5}.$

Patikrinsime apskaičiuodami ${\iint }_S{x}^3𝐝y𝐝z,{\iint }_S{y}^3𝐝z𝐝x$ ir ${\iint }_S{x}^3𝐝x𝐝y$ sumą.

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

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

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

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

$V_x={\iint }_S{x}^3\sqrt{1+{\left(\frac{\partial x}{\partial y}\right)}^2+{\left(\frac{\partial x}{\partial z}\right)}^2}𝐝y𝐝z=$

$={\iint }_S{x}^3\sqrt{1+{\left(\frac{-y}{\sqrt{R^2-y^2-z^2}}\right)}^2+{\left(\frac{-z}{\sqrt{R^2-y^2-z^2}}\right)}^2}𝐝y𝐝z=$

$={\iint }_S{x}^3\sqrt{1+\frac{y^2}{R^2-y^2-z^2}+\frac{z^2}{R^2-y^2-z^2}}𝐝y𝐝z=$

$={\iint }_S{x}^3\sqrt{\frac{R^2-y^2-z^2+y^2+z^2}{R^2-y^2-z^2}}𝐝y𝐝z=$

$={\iint }_S{x}^3\sqrt{\frac{R^2}{R^2-y^2-z^2}}𝐝y𝐝z={\iint }_S\sqrt{R^2-y^2-z^2}\sqrt{\frac{R^2}{R^2-y^2-z^2}}𝐝y𝐝z={\iint }_{S}R𝐝y𝐝z=$

$={\iint }_{S}R\rho 𝐝\rho 𝐝\varphi ={\int }_0^{2\pi }\left({\int }_0^{R}R\rho 𝐝\rho \right)𝐝\varphi =$

$={\int }_0^{2\pi }(R\cdot \frac{R^2}{2})𝐝\varphi =\frac{R^3}{2}{\int }_0^{2\pi }𝐝\varphi =\frac{R^3}{2}\cdot 2\pi =R^3\pi .$

$V=3V_x=3R^3\pi .$

Gautas tūris nesutampa su Gauso formulės logika.

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

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

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

${z_x}^{\prime }=\frac{-2x}{2\sqrt{R^2-x^2-y^2}}=\frac{-x}{\sqrt{R^2-x^2-y^2}};$

${z_y}^{\prime }=\frac{-2y}{2\sqrt{R^2-x^2-y^2}}=\frac{-y}{\sqrt{R^2-x^2-y^2}};$

$\sqrt{1+({z_x}^{\prime }{)}^2+({z_y}^{\prime }{)}^2}=\sqrt{1+{\left(\frac{-x}{\sqrt{R^2-x^2-y^2}}\right)}^2+{\left(\frac{-y}{\sqrt{R^2-x^2-y^2}}\right)}^2}=$

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

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

Iš internetinio integratoriaus ${\int }_0^R\rho \sqrt{\frac{R^2}{R^2-{\rho }^2}}d\rho =({\rho }^2-R^2)\sqrt{\frac{R^2}{R^2-{\rho }^2}}{|}_0^R=-R\sqrt{R^2-{\rho }^2}{|}_0^R;$

$S_{pav.}={\iint }_s\sqrt{1+({z_x}^{\prime }{)}^2+({z_y}^{\prime }{)}^2}dxdy={\iint }_s\sqrt{\frac{R^2}{R^2-x^2-y^2}}dxdy={\int }_0^{2\pi }d\varphi {\int }_0^R\rho \sqrt{\frac{R^2}{R^2-{\rho }^2}}d\rho =$

$={\int }_0^{2\pi }-R\sqrt{R^2-{\rho }^2}{|}_0^{R}d\varphi =-R{\int }_0^{2\pi }(\sqrt{R^2-R^2}-\sqrt{R^2-0^2})d\varphi =-R{\int }_0^{2\pi }(0-\sqrt{R^2})d\varphi =R^2{\int }_0^{2\pi }d\varphi =2\pi R^2.$

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 ${\iint }_S{x}^3𝐝y𝐝z,{\iint }_S{y}^3𝐝z𝐝x$ ir ${\iint }_S{x}^3𝐝x𝐝y$ sumą.

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

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

$V_x={\iint }_S{x}^3𝐝y𝐝z={\iint }_S{\left(\sqrt{R^2-y^2-z^2}\right)}^3𝐝y𝐝z={\iint }_S{\left(\sqrt{R^2-{\rho }^2}\right)}^3\rho 𝐝\rho 𝐝\varphi =$

$={\int }_0^{2\pi }\left({\int }_0^R\rho \sqrt{(R^2-{\rho }^2{)}^3}𝐝\rho \right)𝐝\varphi .$

Pasinaudodami internetiniu integratoriumi, gauname, kad

${\int }_0^R\rho \sqrt{(R^2-{\rho }^2{)}^3}𝐝\rho =-\frac{1}{5}((R-\rho )(R+\rho ){)}^{5/2}{|}_0^R=$

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

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

$V_x={\int }_0^{2\pi }\left({\int }_0^R\rho \sqrt{(R^2-{\rho }^2{)}^3}𝐝\rho \right)𝐝\varphi ={\int }_0^{2\pi }\frac{R^5}{5}𝐝\varphi =\frac{2\pi R^5}{5}.$

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 ${\iint }_S{x}^{2}dydz+0dzdx+0dxdy$ pagal išorinę pusę sferos $x^2+y^2+z^2=R^2.$

Taikydami formulę Gauso, gauname:

${\iint }_S{x}^{2}dydz={\iiint }_V(2x+0+0)dxdydz=2{\int }_0^{2\pi }d\varphi {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos \theta d\theta {\int }_0^R{\rho }^{3}d\rho =$

$=2{\int }_0^{2\pi }d\varphi {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos \theta d\theta \frac{{\rho }^4}{4}{|}_0^R=2\cdot \frac{R^4}{4}{\int }_0^{2\pi }d\varphi {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos \theta d\theta =\frac{R^4}{2}{\int }_0^{2\pi }d\varphi \sin \theta {|}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}=\frac{R^4}{2}{\int }_0^{2\pi }(\sin \frac{\pi }{2}-\sin \frac{-\pi }{2})d\varphi =$

$=\frac{R^4}{2}{\int }_0^{2\pi }(1-(-1))d\varphi =R^4\varphi {|}_0^{2\pi }=R^4\cdot 2\pi =2\pi R^4.$

Dar viską reikia padalinti iš 3, kad gauti teisingai:

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

Kitaip patikrinsime apskaičiuodami ${\iint }_S{x}^2𝐝y𝐝z.$

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

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

$V_x={\iint }_S{x}^2𝐝y𝐝z={\iint }_S{\left(\sqrt{R^2-y^2-z^2}\right)}^2𝐝y𝐝z={\iint }_S{\left(\sqrt{R^2-{\rho }^2}\right)}^2\rho 𝐝\rho 𝐝\varphi =$

$={\int }_0^{2\pi }({\int }_0^R\rho (R^2-{\rho }^2)𝐝\rho )𝐝\varphi .$

Pasinaudodami internetiniu integratoriumi, gauname, kad

${\int }_0^R\rho (R^2-{\rho }^2)𝐝\rho ={\int }_0^R(\rho R^2-{\rho }^3)𝐝\rho =(\frac{{\rho }^2{R}^2}{2}-\frac{{\rho }^4}{4}){|}_0^R=$

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

$V_x={\int }_0^{2\pi }\left({\int }_0^R\rho \sqrt{(R^2-{\rho }^2{)}^2}𝐝\rho \right)𝐝\varphi ={\int }_0^{2\pi }\frac{R^4}{4}𝐝\varphi =\frac{2\pi R^4}{4}=\frac{\pi R^4}{2}.$

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).