17.3-bo'lim : Yuzaki integrallar
6. \( \displaystyle \iint\limits_{S}{{x - z\,dS}}\) ni baholang, bunda \(S\) - \({x^2} + {y^2} = 4\), \(z = x - 3\) 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.
\({S_1}:{\mbox{ Cylinder given by }}{x^2} + {y^2} = 4\,\,\,\left( {i.e{\mbox{ the walls of the solid}}} \right)\)
\({S_2}:{\mbox{ Plane given by }}z = x + 2{\rm{ }}\left( {i.e{\mbox{ the top cap of the cylinder}}} \right)\)
\({S_3}:{\mbox{ Plane given by }}z = x - 3{\rm{ }}\left( {i.e{\mbox{ the bottom cap of the cylinder}}} \right)\)
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.
\[{\vec r_z} = \left\langle {0,0,1} \right\rangle \hspace{0.25in}\hspace{0.25in}{\vec r_\theta } = \left\langle { - 2\sin \theta ,2\cos \theta ,0} \right\rangle \] \[{\vec r_z} \times {\vec r_\theta } = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\0&0&1\\{ - 2\sin \theta }&{2\cos \theta }&0\end{array}} \right| = - 2\cos \theta \vec i - 2\sin \theta \vec j\]Keyinchalik, bizga ko'ndalang mahsulotning kattaligi kerak bo'ladi, shuning uchun bu erda.
\[\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\| = \sqrt {4{{\cos }^2}\theta + 4{{\sin }^2}\theta } = 2\]Bu holda integral,
\[\iint\limits_{{{S_1}}}{{x - z\,dS}} = \iint\limits_{D}{{\left[ {2\cos \theta - z} \right]\left( 2 \right)\,dA}} = \iint\limits_{D}{{4\cos \theta - 2z\,dA}}\]Sirt parametrlarini integralga kiritishni unutmang va o'zaro faoliyat mahsulotning kattaligini qo'shishni unutmang!
Endi, bu sirt uchun \(D\) biz yuqorida bergan \(z\) va \(\theta \) limitlaridan boshqa narsa emas.
Endi bu sirt uchun integralni hisoblaymiz.
\[\begin{align*}\iint\limits_{{{S_1}}}{{x - z\,dS}} & = \int_{0}^{{2\pi }}{{\int_{{2\cos \theta - 3}}^{{2\cos \theta + 2}}{{4\cos \theta - 2z\,dz}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\left. {\left( {4z\cos \theta - {z^2}} \right)} \right|_{2\cos \theta - 3}^{2\cos \theta + 2}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{4\cos \theta \left[ {2\cos \theta + 2 - \left( {2\cos \theta - 3} \right)} \right] - \left[ {{{\left( {2\cos \theta + 2} \right)}^2} - {{\left( {2\cos \theta - 3} \right)}^2}} \right]\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{5\,d\theta }} = \underline {10\pi } \end{align*}\]Oddiylashtirishni unutmang! \(z\) integratsiyasidan so'ng biz ushbu muammoni ko'rganimizdek, integral haqiqatan ham chigal ko'rindi, ammo juda oddiy soddalashtirilgandan so'ng u nihoyatda sodda integrandga qisqardi.
3-qadam
Keyinchalik biz \({S_2}\)-ga g'amxo'rlik qilamiz. Bu holda sirt uchun tenglama oddiygina \(z = x + 2\), \(D\) esa \({x^2} + {y^2} \le 4\) diskidir.
Bu holda integral,
\[\iint\limits_{{{S_2}}}{{x - z\,dS}} = \iint\limits_{D}{{\left[ {x - \left( {x + 2} \right)} \right]\sqrt {{{\left( 1 \right)}^2} + {{\left( 0 \right)}^2} + 1} \,dA}} = \iint\limits_{D}{{ - 2\sqrt 2 \,dA}} = - 2\sqrt 2 \iint\limits_{D}{{dA}}\]OK, bu holda biz integralni baholashni amalga oshirishimiz shart emas, chunki biz buni bilamiz,
\[\iint\limits_{D}{{dA}} = {\mbox{Area of }}D\]va bu holda \(D\) shunchaki disk va biz uning maydonini hech qanday baholamasdan tezda aniqlay olamiz.
Shunday qilib, bu sirt uchun integrali faqat,
\[\iint\limits_{{{S_2}}}{{x - z\,dS}} = - 2\sqrt 2 \left( {{\mbox{Area of }}D} \right) = - 2\sqrt 2 \left[ {{{\left( 2 \right)}^2}\pi } \right] = \underline { - 8\sqrt 2 \pi } \]4-qadam
Va nihoyat, keling, \({S_3}\) orqali integratsiya qilaylik. Bu holda sirt uchun tenglama oddiygina \(z = x - 3\), \(D\) esa \({x^2} + {y^2} \le 4\) diskidir.
Bu holda integral,
\[\begin{align*}\iint\limits_{{{S_3}}}{{x - z\,dS}} & = \iint\limits_{D}{{\left[ {x - \left( {x - 3} \right)} \right]\sqrt {{{\left( 1 \right)}^2} + {{\left( 0 \right)}^2} + 1} \,dA}}\\ & = \iint\limits_{D}{{3\sqrt 2 \,dA}} = 3\sqrt 2 \iint\limits_{D}{{dA}} = 3\sqrt 2 \left( {4\pi } \right) = \underline {12\sqrt 2 \pi } \end{align*}\]Shunday qilib, bu holda integral 3-bosqichdagi integralga o'xshash bo'ldi va shuning uchun biz bu erda hech qanday tushuntirish bermadik.
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,
\[\iint\limits_{S}{{x - z\,dS}} = \left( {10\pi } \right) + \left( { - 8\sqrt 2 \pi } \right) + \left( {12\sqrt 2 \pi } \right) = \left( {10 + 4\sqrt 2 } \right)\pi = \require{bbox} \bbox[2pt,border:1px solid black]{{49.1875}}\]Biz har bir alohida integral qiymatlar qayerdan kelganini ko'rsatish uchun qavslar qo'yamiz. Umuman olganda, bular, albatta, kerak emas.