17.4-bo'lim : Vektor maydonlarining sirt integrallari
6. \( \displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}\) ni baholang, bu erda \(\vec F = yz\,\vec i + x\,\vec j + 3{y^2}\,\vec k\) va \(S\) - manfiy yo'nalishli \({x^2} + {y^2} = 4\), \(z = x - 3\) va \(z = x + 2\) bilan chegaralangan qattiq jismning yuzasi. E'tibor bering, ushbu qattiq jismning uchta yuzasi \(S\) tarkibiga kiritilgan.
Yechim
Keling, ushbu muammoni hal qilishda biz ishlayotgan sirtning qisqacha eskizi bilan boshlaylik.
Muammo bayonotida ta'kidlanganidek, bu erda uchta sirt mavjud. Tsilindrning "yuqori" qismini ko'rish biroz qiyin. Biz silindrning devorlarini biroz shaffof qilib qo'ydik va silindrning yuqori qismini sirtning yuqori qismi bo'ylab quyuqroq ellips sifatida ko'rish mumkin.
Silindrning yuqori va pastki qismi o'rtasidagi munosabatni tasavvur qilish uchun bu erda sirtning boshqa ko'rinishi mavjud.
Ushbu ko'rinishdan biz silindrni "qopqoq" qiladigan yuqori va pastki tekisliklar parallel ekanligini ko'rishimiz mumkin.
Sketchdagi uchta sirtni quyidagicha aniqlaymiz.
\({S_1}:{\mbox{ Cylinder given by }}{x^2} + {y^2} = 4\,\,\,\left( {i.e{\mbox{ the walls of the solid}}} \right)\)
\({S_2}:{\mbox{ Plane given by }}z = x + 2{\mbox{ }}\left( {i.e{\mbox{ the top cap of the cylinder}}} \right)\)
\({S_3}:{\mbox{ Plane given by }}z = x - 3{\mbox{ }}\left( {i.e{\mbox{ the bottom cap of the cylinder}}} \right)\)
Endi biz ushbu sirtlarning har biri uchun integralni ko'rib chiqishimiz va bajarishimiz kerak va biz ularni ushbu bo'limdagi birinchi bir nechta muammolarga qaraganda biroz tezroq o'tkazamiz.
2-qadam
\({S_1}\) bilan boshlaylik. Bu holda sirt silindrdir va shuning uchun biz uni parametrlashimiz kerak. Sirtning parametrizatsiyasi,
\[\vec r\left( {z,\theta } \right) = \left\langle {2\cos \theta ,2\sin \theta ,z} \right\rangle \]\(z\) va \(\theta \) limitlari:
\[0 \le \theta \le 2\pi ,\,\,\,\,\,2\cos \theta - 3 = x - 3 \le z \le x + 2 = 2\cos \theta + 2\]\(z\) chegaralari bilan biz \(x\) ni parametrlangan shaklga aylantirganimizga ishonch hosil qilishimiz kerak.
Bu holda integralni baholash uchun bizga \({\vec r_z} \times {\vec r_\theta }\) o'zaro mahsuloti kerak bo'ladi, shuning uchun bu ish.
\[{\vec r_z} = \left\langle {0,0,1} \right\rangle \hspace{0.25in}\hspace{0.25in}{\vec r_\theta } = \left\langle { - 2\sin \theta ,2\cos \theta ,0} \right\rangle \] \[{\vec r_z} \times {\vec r_\theta } = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\0&0&1\\{ - 2\sin \theta }&{2\cos \theta }&0\end{array}} \right| = - 2\cos \theta \vec i - 2\sin \theta \vec j\]U holda sirt uchun birlik normal vektor,
\[\vec n = \frac{{{{\vec r}_z} \times {{\vec r}_\theta }}}{{\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\|}} = \frac{{\left\langle { - 2\cos \theta , - 2\sin \theta ,0} \right\rangle }}{{\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\|}}\]Biz o'zaro mahsulotning kattaligini hisoblamadik, chunki biz integral bilan ishlashni boshlaganimizda u bekor qilinishini bilamiz.
Sirt salbiy yo'nalishga ega va shuning uchun sirt bilan o'ralgan hududga ishora qilishi kerak. Bu shuni anglatadiki, silindrdagi oddiy vektorlar \(z\) o'qiga ishora qilishi kerak va bu vektor shu yo'nalishni ko'rsatadi.
Ushbu vektor \(z\) o'qi tomon ishora qilishini ko'rish uchun \(0 \le \theta \le \frac{\pi }{2}\) ni ko'rib chiqing. Bu diapazonda sinus ham, kosinus ham musbat va shuning uchun normal vektorning \(x\) va \(y\) komponentlari manfiy bo'ladi va shuning uchun \(z\) o'qiga ishora qiladi.
Endi bizga quyidagi nuqta mahsuloti kerak bo'ladi va vektor maydoniga sirt parametrlarini kiritishni unutmang.
\[\begin{align*}\vec F\left( {\vec r\left( {z,\theta } \right)} \right)\centerdot \vec n & = \left\langle {2z\sin \theta ,2\cos \theta ,12{{\sin }^2}\theta } \right\rangle \centerdot \frac{{\left\langle { - 2\cos \theta , - 2\sin \theta ,0} \right\rangle }}{{\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\|}}\\ & = \frac{1}{{\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\|}}\left( { - 4z\sin \theta \cos \theta - 4\sin \theta \cos \theta } \right)\\ & = \frac{1}{{\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\|}}\left[ { - 4\sin \theta \cos \theta \left( {z + 1} \right)} \right]\end{align*}\]U holda integral,
\[\begin{align*}\iint\limits_{{{S_1}}}{{\vec F\centerdot \,d\vec S}} & = \iint\limits_{{{S_1}}}{{\frac{1}{{\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\|}}\left[ { - 2\sin \left( {2\theta } \right)\left( {z + 1} \right)} \right]\,dS}}\\ & = \iint\limits_{D}{{\frac{1}{{\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\|}}\left[ { - 2\sin \left( {2\theta } \right)\left( {z + 1} \right)} \right]\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\|\,dA}}\\ & = \iint\limits_{D}{{ - 2\sin \left( {2\theta } \right)\left( {z + 1} \right)\,dA}}\end{align*}\]Bu holda \(D\) biz yuqorida bergan \(z\) va \(\theta \) limitlaridan boshqa narsa emas va shuning uchun endi integralni tugatishimiz mumkin.
\[\begin{align*}\iint\limits_{{{S_1}}}{{\vec F\centerdot \,d\vec S}} & = \iint\limits_{D}{{ - 4\sin \theta \cos \theta \left( {z + 1} \right)\,dA}}\\ & = \int_{0}^{{2\pi }}{{\int_{{2\cos \theta - 3}}^{{2\cos \theta + 2}}{{ - 4\sin \theta \cos \theta \left( {z + 1} \right)\,dz}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\left. { - 4\sin \theta \cos \theta \left( {\frac{1}{2}{z^2} + z} \right)} \right|_{2\cos \theta - 3}^{2\cos \theta + 2}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{ - 10\sin \theta \cos \theta - 40\sin \theta {{\cos }^2}\theta \,d\theta }}\\ & = \int_{0}^{{2\pi }}{{ - 5\sin \left( {2\theta } \right) - 40\sin \theta {{\cos }^2}\theta \,d\theta }}\\ & = \left. {{{\left( {\frac{5}{2}\cos \left( {2\theta } \right) + \frac{{40}}{3}{{\cos }^3}\theta } \right)}^3}} \right|_0^{2\pi } = \underline 0 \end{align*}\]3-qadam
Keyinchalik biz \({S_2}\)-ga g'amxo'rlik qilamiz. Bu holda sirt uchun tenglama \(z - x - 2 = 0\) sifatida yozilishi mumkin va \(D\) \({x^2} + {y^2} \le 4\) diskidir.
\({S_2}\) uchun normal birlik vektor u holda,
\[\vec n = \frac{{\nabla f}}{{\left\| {\nabla f} \right\|}} = \frac{{\left\langle { - 1,0,1} \right\rangle }}{{\left\| {\nabla f} \right\|}}\]Mintaqa salbiy yo'nalishga ega va shuning uchun yopiq hududga ishora qilishi kerak va shuning uchun pastga yo'naltirilishi kerak (chunki bu silindrning yuqori "qopqog'i"). Yuqoridagi normal vektor yuqoriga ishora qiladi (u musbat \(z\) komponentiga ega) va shuning uchun biz ushbu sirt uchun kerakli normal vektorni olish uchun uni minus birga ko'paytirishimiz kerak bo'ladi.
To'g'ri normal vektor u holda,
\[\vec n = - \frac{{\nabla f}}{{\left\| {\nabla f} \right\|}} = \frac{{\left\langle {1,0, - 1} \right\rangle }}{{\left\| {\nabla f} \right\|}}\]Biz gradientning kattaligini hisoblamadik, chunki biz integral bilan ishlashni boshlaganimizda u bekor qilinishini bilamiz.
Ushbu sirt uchun bizga kerak bo'lgan nuqta mahsuloti:
\[\begin{align*}\vec F\left( {x,y,x + 2} \right)\centerdot \vec n & = \left\langle {y\left( {x + 2} \right),x,3{y^2}} \right\rangle \centerdot \frac{{\left\langle {1,0, - 1} \right\rangle }}{{\left\| {\nabla f} \right\|}}\\ & = \frac{1}{{\left\| {\nabla f} \right\|}}\left( {xy + 2y - 3{y^2}} \right)\end{align*}\]Vektor maydonidagi \(z\) ga sirt tenglamasini ulashni unutmang.
U holda integral,
\[\begin{align*}\iint\limits_{{{S_2}}}{{\vec F\centerdot \,d\vec S}} & = \iint\limits_{{{S_2}}}{{\frac{1}{{\left\| {\nabla f} \right\|}}\left( {xy + 2y - 3{y^2}} \right)\,dS}}\\ & = \iint\limits_{D}{{\frac{1}{{\left\| {\nabla f} \right\|}}\left( {xy + 2y - 3{y^2}} \right)\left\| {\nabla f} \right\|\,dA}}\\ & = \iint\limits_{D}{{xy + 2y - 3{y^2}\,dA}}\end{align*}\]E'tibor bering, biz bu integralni qutb koordinatalari bilan tugatishimiz kerak va qutb chegaralari quyidagicha bo'ladi:
\[\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le 2\end{array}\]U holda integral,
\[\begin{align*}\iint\limits_{{{S_2}}}{{\vec F\centerdot \,d\vec S}} & = \iint\limits_{D}{{xy + 2y - 3{y^2}\,dA}}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{\left( {{r^2}\sin \theta \cos \theta + 2r\sin \theta - 3{r^2}{{\sin }^2}\theta } \right)\left( r \right)\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{\frac{1}{2}{r^3}\sin \left( {2\theta } \right) + 2{r^2}\sin \theta - \frac{3}{2}{r^3}\left( {1 - \cos \left( {2\theta } \right)} \right)\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\left. {\frac{1}{8}{r^4}\sin \left( {2\theta } \right) + \frac{2}{3}{r^3}\sin \theta - \frac{3}{8}{r^4}\left( {1 - \cos \left( {2\theta } \right)} \right)\,} \right|_0^2\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{2\sin \left( {2\theta } \right) + \frac{{16}}{3}\sin \theta - 6\left( {1 - \cos \left( {2\theta } \right)} \right)\,d\theta }}\\ & = \left. {\left( { - \cos \left( {2\theta } \right) - \frac{{16}}{3}\cos \theta - 6\left( {\theta - \frac{1}{2}\sin \left( {2\theta } \right)} \right)} \right)} \right|_0^{2\pi } = \underline { - 12\pi } \end{align*}\]4-qadam
Va nihoyat, keling, \({S_3}\) orqali integratsiya qilaylik. Bu holda sirt uchun tenglama \(z - x + 3 = 0\) sifatida yozilishi mumkin va \(D\) \({x^2} + {y^2} \le 4\) diskidir.
\({S_2}\) uchun normal birlik vektor u holda,
\[\vec n = \frac{{\nabla f}}{{\left\| {\nabla f} \right\|}} = \frac{{\left\langle { - 1,0,1} \right\rangle }}{{\left\| {\nabla f} \right\|}}\]Mintaqaning salbiy yo'nalishi bor va shuning uchun yopiq hududga ishora qilishi kerak va shuning uchun yuqoriga yo'naltirilishi kerak (chunki bu silindrning pastki "qopqog'i"). Yuqoridagi normal vektor yuqoriga ishora qiladi (u ijobiy \(z\) komponentiga ega) va bizga kerak bo'lgan normal vektor ham shunday.
Biz gradientning kattaligini hisoblamadik, chunki biz integral bilan ishlashni boshlaganimizda u bekor qilinishini bilamiz.
Ushbu sirt uchun bizga kerak bo'lgan nuqta mahsuloti:
\[\begin{align*}\vec F\left( {x,y,x - 3} \right)\centerdot \vec n & = \left\langle {y\left( {x - 3} \right),x,3{y^2}} \right\rangle \centerdot \frac{{\left\langle { - 1,0,1} \right\rangle }}{{\left\| {\nabla f} \right\|}}\\ & = \frac{1}{{\left\| {\nabla f} \right\|}}\left( { - xy + 3y + 3{y^2}} \right)\end{align*}\]Vektor maydonidagi \(z\) ga sirt tenglamasini ulashni unutmang.
U holda integral,
\[\begin{align*}\iint\limits_{{{S_3}}}{{\vec F\centerdot \,d\vec S}} & = \iint\limits_{{{S_3}}}{{\frac{1}{{\left\| {\nabla f} \right\|}}\left( { - xy + 3y + 3{y^2}} \right)\,dS}}\\ & = \iint\limits_{D}{{\frac{1}{{\left\| {\nabla f} \right\|}}\left( { - xy + 3y + 3{y^2}} \right)\left\| {\nabla f} \right\|\,dA}}\\ & = \iint\limits_{D}{{ - xy + 3y + 3{y^2}\,dA}}\end{align*}\]E'tibor bering, biz bu integralni qutb koordinatalari bilan tugatishimiz kerak va qutb chegaralari quyidagicha bo'ladi:
\[\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le 2\end{array}\]U holda integral,
\[\begin{align*}\iint\limits_{{{S_3}}}{{\vec F\centerdot \,d\vec S}} & = \iint\limits_{D}{{ - xy + 3y + 3{y^2}\,dA}}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{\left( { - {r^2}\sin \theta \cos \theta + 3r\sin \theta + 3{r^2}{{\sin }^2}\theta } \right)\left( r \right)\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{ - \frac{1}{2}{r^3}\sin \left( {2\theta } \right) + 3{r^2}\sin \theta + \frac{3}{2}{r^3}\left( {1 - \cos \left( {2\theta } \right)} \right)\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\left. { - \frac{1}{8}{r^4}\sin \left( {2\theta } \right) + {r^3}\sin \theta + \frac{3}{8}{r^4}\left( {1 - \cos \left( {2\theta } \right)} \right)\,} \right|_0^2\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{ - 2\sin \left( {2\theta } \right) + 8\sin \theta + 6\left( {1 - \cos \left( {2\theta } \right)} \right)\,d\theta }}\\ & = \left. {\left( {\cos \left( {2\theta } \right) - 8\cos \theta + 6\left( {\theta - \frac{1}{2}\sin \left( {2\theta } \right)} \right)} \right)} \right|_0^{2\pi } = \underline {12\pi } \end{align*}\]5-qadam
Endi to'liq sirt bo'yicha integralning qiymatini olish uchun biz qilishimiz kerak bo'lgan narsa yuqoridagi uchta sirt ustidagi har bir integralning qiymatlarini jamlashdir. Shunday qilib,
\[\iint\limits_{S}{{\vec F\centerdot \,d\vec S}} = \left( 0 \right) + \left( { - 12\pi } \right) + \left( {12\pi } \right) = \require{bbox} \bbox[2pt,border:1px solid black]{0}\]Biz har bir alohida integral qiymatlar qayerdan kelganini ko'rsatish uchun qavslar qo'yamiz. Umuman olganda, bular, albatta, kerak emas.