Divergensiya teoremasidan quyidagi yo'nalishda foydalanamiz.

\[\iint\limits_{S}{{\vec F\centerdot d\vec S}} = \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F\,dV}}\]

Bu erda \(E\) 1-qadamdagi eskizlarda ko'rsatilgan qattiq narsadir.

\(E\) sharning bir qismi bo'lgani uchun biz integratsiya uchun sferik koordinatalardan foydalanishni xohlaymiz. Mana bu mintaqa uchun foydalanishimiz kerak bo'lgan sharsimon chegaralar.

\[\begin{array}{c}\pi \le \theta \le 2\pi \\ \displaystyle \frac{\pi }{2} \le \varphi \le \pi \\ 0 \le \rho \le 4\end{array}\]

Muammo bayonotida mintaqaga nisbatan limitlardan biri \(y \le 0\) edi. Bu shuni anglatadiki, agar biz buni yuqoridan ko'rib chiqsak, biz \(x\) o'qi ostidagi 4 radiusli doira qismini ko'ramiz va shuning uchun bizga ushbu mintaqani qoplash uchun yuqoridagi \(\theta \) diapazoni kerak bo'ladi.

Muammo bayonotida bizga \(z \le 0\) ham aytildi va shuning uchun biz faqat \(xy\) tekisligi ostidagi sharning qismini xohlaymiz. Shunday qilib, biz faqat \(xy\) tekisligidan pastroq ekanligimizga ishonch hosil qilish uchun berilgan \(\varphi \) diapazoniga muhtojmiz.

Bizga vektor maydonining divergentsiyasi ham kerak bo'ladi, shuning uchun bu erda.

\[{\mathop{\rm div}\nolimits} \vec F = \frac{\partial }{{\partial x}}\left( {y{x^2}} \right) + \frac{\partial }{{\partial y}}\left( {x{y^2} - 3{z^4}} \right) + \frac{\partial }{{\partial z}}\left( {{x^3} + {y^2}} \right) = 4xy\]

Keling, disvergentsiya teoremasini integralga qo'llaymiz va uni sferik koordinatalarga aylantiramiz.

\[\begin{align*}\iint\limits_{S}{{\vec F\centerdot d\vec S}} & = \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F\,dV}}\\ & = \iiint\limits_{E}{{4xy\,dV}}\\ & = \int_{\pi }^{{2\pi }}{{\int_{{\frac{1}{2}\pi }}^{\pi }{{\int_{0}^{4}{{4\left( {\rho \sin \varphi \cos \theta } \right)\left( {\rho \sin \varphi \sin \theta } \right)\left( {{\rho ^2}\sin \varphi } \right)\,d\rho }}\,d\varphi }}\,d\theta }}\\ & = \int_{\pi }^{{2\pi }}{{\int_{{\frac{1}{2}\pi }}^{\pi }{{\int_{0}^{4}{{4{\rho ^4}{{\sin }^3}\varphi \cos \theta \sin \theta \,d\rho }}\,d\varphi }}\,d\theta }}\end{align*}\]

\({\rho ^2}\sin \varphi \)-ni sferik koordinatalarga o'tkazishda \({\rho ^2}\sin \varphi \)-ni olishni unutmang.

Keyin integralni baholashda qilishimiz kerak.

\[\begin{align*}\iint\limits_{S}{{\vec F\centerdot d\vec S}} & = \int_{\pi }^{{2\pi }}{{\int_{{\frac{1}{2}\pi }}^{\pi }{{\int_{0}^{4}{{4{\rho ^4}{{\sin }^3}\varphi \cos \theta \sin \theta \,d\rho }}\,d\varphi }}\,d\theta }}\\ & = \int_{\pi }^{{2\pi }}{{\int_{{\frac{1}{2}\pi }}^{\pi }{{\left. {\left( {\frac{4}{5}{\rho ^5}{{\sin }^3}\varphi \cos \theta \sin \theta } \right)} \right|_0^4\,d\varphi }}\,d\theta }}\\ & = \int_{\pi }^{{2\pi }}{{\int_{{\frac{1}{2}\pi }}^{\pi }{{\frac{{4096}}{5}\sin \varphi \left( {1 - {{\cos }^2}\varphi } \right)\cos \theta \sin \theta \,d\varphi }}\,d\theta }}\\ & = \int_{\pi }^{{2\pi }}{{\left. {\left( { - \frac{{4096}}{5}\left( {\cos \varphi - \frac{1}{3}{{\cos }^3}\varphi } \right)\cos \theta \sin \theta } \right)} \right|_{\frac{1}{2}\pi }^\pi \,d\theta }}\\ & = \int_{\pi }^{{2\pi }}{{\frac{{8192}}{{15}}\cos \theta \sin \theta \,d\theta }}\\ & = \int_{\pi }^{{2\pi }}{{\frac{{4096}}{{15}}\sin \left( {2\theta } \right)\,d\theta }}\\ & = \left. {-\frac{{2048}}{{15}}\cos \left( {2\theta } \right)} \right|_\pi ^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{0}\end{align*}\]

Ushbu turdagi integrallar bilan ishlash uchun biz bu erda qilganimiz kabi trig formulalaridan foydalanishingiz mumkinligiga ishonch hosil qiling!