Keling, endi integralni o'rnatamiz. Bunday holda, biz silindrni birlashtiramiz va shuning uchun biz sirt uchun parametrlarni o'rnatishimiz kerak.

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

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

Biz \(\theta \) ning to‘liq diapazonidan foydalanamiz, chunki biz uning butun atrofida aylanishiga ruxsat beramiz. x‑o'qi. \(x\) chegaralari silindrni "bog'laydigan" ikkita tekislikdan kelib chiqadi va biz parametrlash yordamida yuqori chegarani aylantirishimiz kerak.

Keyinchalik, biz o'zaro mahsulotni hisoblashimiz kerak.

\[{\vec r_x} = \left\langle {1,0,0} \right\rangle \hspace{0.25in}\hspace{0.25in}{\vec r_\theta } = \left\langle {0,2\cos \theta , - 2\sin \theta } \right\rangle \] \[{\vec r_x} \times {\vec r_\theta } = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\1&0&0\\0&{2\cos \theta }&{ - 2\sin \theta }\end{array}} \right| = 2\sin \theta \vec j + 2\cos \theta \vec k\]

U holda sirt uchun birlik normal vektor,

\[\vec n = \frac{{{{\vec r}_x} \times {{\vec r}_\theta }}}{{\left\| {{{\vec r}_x} \times {{\vec r}_\theta }} \right\|}} = \frac{{\left\langle {0,2\sin \theta ,2\cos \theta } \right\rangle }}{{\left\| {{{\vec r}_x} \times {{\vec r}_\theta }} \right\|}}\]

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. Avval silindrni silindrning oldidan va to'g'ridan-to'g'ri \(x\) o'qi bo'ylab ko'rib chiqaylik. Bu biz ko'rgan narsadir.

Ushbu eskizda \(x\) o'qi to'g'ridan-to'g'ri eskizdan kelib chiqadi. Parametrlashtirishga \(\theta \) ning bir nechta qiymatini kiritsak, biz yuqorida sanab o'tilgan nuqtalarda bo'lishimizni ko'rishimiz mumkin.

Endi, \(0 \le \theta \le \frac{1}{2}\pi \) diapazonida biz bilamizki, sinus va kosinus ham ijobiy, shuning uchun normal vektorda \(y\) va \(z\) komponentlari ham ijobiy bo'ladi. Bu shuni anglatadiki, 1-bandda^st normal vektor ustidagi kvadrant koordinatadan uzoqroqda bo'lishi kerak. Aynan shu narsani ko'rishimiz kerak, chunki yo'nalish \(x\) o'qidan uzoqqa ishora qilib berilgan va \(x\) o'qi to'g'ridan-to'g'ri eskizdan kelib chiqishini eslatib o'tamiz.

Keyinchalik, agar biz \(\frac{1}{2}\pi \le \theta \le \pi \) ga qarasak (demak, biz 4-da^th Yuqoridagi grafikning kvadranti….) biz bilamizki, bu diapazonda sinus hali ham ijobiy, lekin kosinus hozir manfiy. Yuqoridagi birlik vektorimizdan ma'lumki, \(y\) komponenti ijobiy (shuning uchun musbat \(y\) yo'nalishiga ishora qiladi) va \(z\) komponenti manfiy (shuning uchun manfiy \(z\) yo'nalishiga ishora qiladi). Birgalikda bu yana 4-dagi asl manbadan uzoqlashishimiz kerakligini anglatadi^th kvadrant, bu yana biz xohlagan yo'nalishdir.

Biz qolgan ikkita kvadrantga qarab shu tarzda davom etishimiz mumkin, biroq biz juftlikni bajarib, to'g'ri yo'nalishni olganimizdan so'ng, qolganlari uchun to'g'ri yo'nalishni olishda davom etishimizni bilamiz.

Keyinchalik, biz quyidagi nuqta mahsulotini hisoblashimiz kerak.

\[\begin{align*}\vec F\left( {\vec r\left( {x,\theta } \right)} \right)\centerdot \vec n & = \left\langle {{x^2},2\left( {2\cos \theta } \right), - 3\left( {2\sin \theta } \right)} \right\rangle \centerdot \frac{{\left\langle {0,2\sin \theta ,2\cos \theta } \right\rangle }}{{\left\| {{{\vec r}_x} \times {{\vec r}_\theta }} \right\|}}\\ & = \frac{1}{{\left\| {{{\vec r}_x} \times {{\vec r}_\theta }} \right\|}}\left( { - 4\sin \theta \cos \theta } \right)\end{align*}\]

Esda tutingki, biz vektor maydoniga sirt uchun parametrlarni kiritishimiz kerak edi!

U holda integral,

\[\begin{align*}\iint\limits_{S}{{\vec F\centerdot \,d\vec S}} & = \iint\limits_{S}{{\frac{1}{{\left\| {{{\vec r}_x} \times {{\vec r}_\theta }} \right\|}}\left( { - 4\sin \theta \cos \theta } \right)\,dS}}\\ & = \iint\limits_{D}{{\frac{1}{{\left\| {{{\vec r}_x} \times {{\vec r}_\theta }} \right\|}}\left( { - 4\sin \theta \cos \theta } \right)\left\| {{{\vec r}_x} \times {{\vec r}_\theta }} \right\|\,dA}}\\ & = \iint\limits_{D}{{ - 4\sin \theta \cos \theta \,dA}}\end{align*}\]

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, \(D\) 2-bosqich boshida topilgan \(x\) va \(\theta \) 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,

\[\begin{align*}\iint\limits_{S}{{\vec F\centerdot \,d\vec S}} & = \iint\limits_{D}{{ - 4\sin \theta \cos \theta \,dA}}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{3 - 2\cos \theta }}{{ - 4\sin \theta \cos \theta \,dx}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\left. { - 4x\sin \theta \cos \theta } \right|_0^{3 - 2\cos \theta }\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{ - 4\left( {3 - 2\cos \theta } \right)\sin \theta \cos \theta \,d\theta }}\\ & = \int_{0}^{{2\pi }}{{ - 12\sin \theta \cos \theta + 8\sin \theta {{\cos }^2}\theta \,d\theta }}\\ & = \int_{0}^{{2\pi }}{{ - 6\sin \left( {2\theta } \right) + 8\sin \theta {{\cos }^2}\theta \,d\theta }}\\ & = \left. {\left( { 3\cos \left( {2\theta } \right) - \frac{8}{3}{{\cos }^3}\theta } \right)} \right|_0^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{0}\end{align*}\]