Keling, endi integralni o'rnatamiz. Bunday holda, biz shar ustida birlashamiz va shuning uchun biz sirt uchun parametrlarni o'rnatishimiz kerak.

Biz ushbu bobning birinchi qismida sharni qanday parametrlashtirishni ko'rib chiqdik, shuning uchun parametrlash uchun batafsil ma'lumot bermaymiz. Parametrlash - bu,

\[\vec r\left( {\theta ,\varphi } \right) = \left\langle {3\sin \varphi \cos \theta ,3\sin \varphi \sin \theta ,3\cos \varphi } \right\rangle \hspace{0.25in}\,\,\,\,\frac{1}{2}\pi \le \theta \le \pi ,\,\,\,\,\,0 \le \varphi \le \frac{1}{2}\pi \]

Sfera yuqori yarmining faqat bir qismini olishimizga ishonch hosil qilish uchun bizga \(\varphi \)-da cheklov kerak edi (i.e.\(z \ge 0\)). Xuddi shunday, \(\theta \)-dagi cheklov faqat 2-da bo'lgan qismni olish uchun kerak edi.^nd \(xy\) tekisligi kvadranti (i.e. \(x \le 0\) va \(y \ge 0\)).

Keyinchalik, biz o'zaro mahsulotni hisoblashimiz kerak.

\[{\vec r_\theta } = \left\langle { - 3\sin \varphi \sin \theta ,3\sin \varphi \cos \theta ,0} \right\rangle \hspace{0.25in}\hspace{0.25in}{\vec r_\varphi } = \left\langle {3\cos \varphi \cos \theta ,3\cos \varphi \sin \theta , - 3\sin \varphi } \right\rangle \] \[\begin{align*}{{\vec r}_\theta } \times {{\vec r}_\varphi } & = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{ - 3\sin \varphi \sin \theta }&{3\sin \varphi \cos \theta }&0\\{3\cos \varphi \cos \theta }&{3\cos \varphi \sin \theta }&{ - 3\sin \varphi }\end{array}} \right|\\ & = - 9{\sin ^2}\varphi \cos \theta \,\vec i - 9\sin \varphi \cos \varphi {\sin ^2}\theta \vec k - 9\sin \varphi \cos \varphi {\cos ^2}\theta \vec k - 9{\sin ^2}\varphi \sin \theta \vec j\\ & = - 9{\sin ^2}\varphi \cos \theta \,\vec i - 9{\sin ^2}\varphi \sin \theta \vec j - 9\sin \varphi \cos \varphi \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)\vec k\\ & = - 9{\sin ^2}\varphi \cos \theta \,\vec i - 9{\sin ^2}\varphi \sin \theta \vec j - 9\sin \varphi \cos \varphi \vec k\end{align*}\]

U holda sirt uchun birlik normal vektor,

\[\vec n = \frac{{{{\vec r}_\theta } \times {{\vec r}_\varphi }}}{{\left\| {{{\vec r}_\theta } \times {{\vec r}_\varphi }} \right\|}} = \frac{{\left\langle { - 9{{\sin }^2}\varphi \cos \theta , - 9{{\sin }^2}\varphi \sin \theta , - 9\sin \varphi \cos \varphi } \right\rangle }}{{\left\| {{{\vec r}_\theta } \times {{\vec r}_\varphi }} \right\|}}\]

Biz o'zaro mahsulotning kattaligini hisoblamadik, chunki biz integral bilan ishlashni boshlaganimizda u bekor qilinishini bilamiz.

Endi biz ushbu vektor to'g'ri yo'nalishga ega yoki yo'qligini aniqlashimiz kerak. Biz bilamizki, normal vektor kelib chiqishiga ishora qilishi kerak. Keling, bu sharning yuqori yarmidagi normal vektor uchun nimani anglatishini o'ylab ko'raylik va biz \(xy\) tekisligida qaysi kvadrantda ekanligimiz muhim emas.

Agar biz sharning yuqori yarmida bo'lsak va normal vektorlar boshlang'ichni ko'rsatishi kerak bo'lsa, ularning barchasi pastga yo'naltirilishi kerakligini bilamiz. Ular \(xy\) tekisligidan qaysi kvadrantda ekanligimizga qarab ijobiy yoki salbiy \(x\) (yoki \(y\)) yo'nalishini ko'rsatishi mumkin, ammo ularning barchasi pastga qarab turishi kerak. Boshqacha qilib aytganda, \(z\) komponenti salbiy bo'lishi kerak.

Shunday qilib, yuqoridagi normal vektorning \(z\) komponenti \( - 9\sin \varphi \cos \varphi \) va biz ushbu muammoda ishlayotgan sohaning bir qismi uchun \(0 \le \varphi \le \frac{1}{2}\pi \) bilan cheklanganligini bilamiz. \(\varphi \) ning ushbu diapazonida biz ham sinus, ham kosinus ijobiy ekanligini bilamiz va shuning uchun \(z\) komponenti har doim salbiy bo'lishi kerak. Bu shuni anglatadiki, yuqoridagi normal vektor ushbu muammo uchun to'g'ri yo'nalishga ega.

E'tibor bering, agar biz sharning pastki yarmida bo'lsak (bu muammo uchun tegishli emas, lekin baribir o'ylash foydali bo'lsa) va normal vektor boshlang'ichga ishora qilgan bo'lar edi va shuning uchun ularning barchasi yuqoriga ishora qilishi kerak edi.

Shuni ham yodda tutingki, agar normal vektorlarning barchasi boshlang'ichdan uzoqda bo'lsa, biz kerakli normal vektorni olish uchun yuqoridagi normal vektorni minus birga ko'paytirishimiz kerak.

Keyinchalik, biz quyidagi nuqta mahsulotini hisoblashimiz kerak.

\[\begin{align*}\vec F\left( {\vec r\left( {\theta ,\varphi } \right)} \right)\centerdot \vec n & = \left\langle {1,3\cos \varphi ,18\sin \varphi \cos \theta } \right\rangle \centerdot \frac{{\left\langle { - 9{{\sin }^2}\varphi \cos \theta , - 9{{\sin }^2}\varphi \sin \theta , - 9\sin \varphi \cos \varphi } \right\rangle }}{{\left\| {{{\vec r}_\theta } \times {{\vec r}_\varphi }} \right\|}}\\ & = \frac{1}{{\left\| {{{\vec r}_\theta } \times {{\vec r}_\varphi }} \right\|}}\left( { - 9{{\sin }^2}\varphi \cos \theta - 27{{\sin }^2}\varphi \cos \varphi \sin \theta - 162{{\sin }^2}\varphi \cos \varphi \cos \theta } \right)\\ & = \frac{1}{{\left\| {{{\vec r}_\theta } \times {{\vec r}_\varphi }} \right\|}}\left( { - \frac{9}{2}\left( {1 - \cos \left( {2\varphi } \right)} \right)\cos \theta - {{\sin }^2}\varphi \cos \varphi \left( {27\sin \theta + 162\cos \theta } \right)} \right)\end{align*}\]

E'tibor bering, biz yuqoridagi oxirgi bosqichda integratsiya jarayoni uchun biroz soddalashtirdik.

U holda integral,

\[\begin{align*}\iint\limits_{S}{{\vec F\centerdot \,d\vec S}} & = \iint\limits_{S}{{\frac{1}{{\left\| {{{\vec r}_\theta } \times {{\vec r}_\varphi }} \right\|}}\left( { - \frac{9}{2}\left( {1 - \cos \left( {2\varphi } \right)} \right)\cos \theta - {{\sin }^2}\varphi \cos \varphi \left( {27\sin \theta + 162\cos \theta } \right)} \right)\,dS}}\\ & = \iint\limits_{D}{{\frac{1}{{\left\| {{{\vec r}_\theta } \times {{\vec r}_\varphi }} \right\|}}\left( { - \frac{9}{2}\left( {1 - \cos \left( {2\varphi } \right)} \right)\cos \theta - {{\sin }^2}\varphi \cos \varphi \left( {27\sin \theta + 162\cos \theta } \right)} \right)\left\| {{{\vec r}_\theta } \times {{\vec r}_\varphi }} \right\|\,dA}}\\ & = \iint\limits_{D}{{ - \frac{9}{2}\left( {1 - \cos \left( {2\varphi } \right)} \right)\cos \theta - {{\sin }^2}\varphi \cos \varphi \left( {27\sin \theta + 162\cos \theta } \right)\,dA}}\end{align*}\]

Yuqorida ta'kidlab o'tilganidek, biz o'zaro mahsulotning kattaligini hisoblashimiz shart emas edi, chunki biz sirt integralini "normal" ikki karrali integralga aylantirganimizda, u bekor qilinadi.

Shuni ham yodda tutingki, \(D\) biz 2-bosqich boshida topilgan \(\theta \) va \(\varphi \) limitlari bilan berilgan.

Endi biz qilishimiz kerak bo'lgan yagona narsa - ikki karrali integralni baholash va bu hozir juda qiyin bo'lmasligi kerak.

U holda integral,

\[\begin{align*}\iint\limits_{S}{{\vec F\centerdot \,d\vec S}} & = \iint\limits_{D}{{ - \frac{9}{2}\left( {1 - \cos \left( {2\varphi } \right)} \right)\cos \theta - {{\sin }^2}\varphi \cos \varphi \left( {27\sin \theta + 162\cos \theta } \right)\,dA}}\\ & = \int_{{\frac{1}{2}\pi }}^{\pi }{{\int_{0}^{{\frac{1}{2}\pi }}{{ - \frac{9}{2}\left( {1 - \cos \left( {2\varphi } \right)} \right)\cos \theta - {{\sin }^2}\varphi \cos \varphi \left( {27\sin \theta + 162\cos \theta } \right)\,d\varphi }}\,d\theta }}\\ & = \int_{{\frac{1}{2}\pi }}^{\pi }{{\left. { - \frac{9}{2}\left( {\varphi - \frac{1}{2}\sin \left( {2\varphi } \right)} \right)\cos \theta - \frac{1}{3}{{\sin }^3}\varphi \left( {27\sin \theta + 162\cos \theta } \right)} \right|_0^{\frac{1}{2}\pi }\,d\theta }}\\ & = \int_{{\frac{1}{2}\pi }}^{\pi }{{ - \frac{9}{4}\pi \cos \theta - 9\sin \theta - 54\cos \theta \,d\theta }}\\ & = \left. {\left( { - \frac{9}{4}\pi \sin \theta + 9\cos \theta - 54\sin \theta } \right)} \right|_{\frac{1}{2}\pi }^\pi = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{9}{4}\pi + 45}}\end{align*}\]