To'g'ri, biz \(yz\)-tekisligi oldida joylashgan tekislikning qismini qidirayotganimiz uchun biz \(x = g\left( {y,z} \right)\) shaklida sirt tenglamasini yozishimiz kerak. Buni qilish yetarlicha oson.

\[x = 1 - y - z\]

Keyinchalik, biz \(D\) nima ekanligini aniqlashimiz kerak. Mana \(S\) sirtining eskizi.

Mana \(D\) hududining eskizi.

E'tibor bering, o'qlar biz \(D\) eskizida ko'rganimizdan boshqacha etiketlangan. Bu eskizni sirtning eskiziga mos kelishini ta'minlash edi. Biz tekislik tenglamasida \(x\) ni nolga tenglashtirib, \(z\) ni yechish orqali gipotenuzaning tenglamasiga keldik. Mana \(y\) va \(z\) diapazonlari.

\[0 \le y \le 1\hspace{0.25in}0 \le z \le 1 - y\]

Endi sirt \(z = g\left( {x,y} \right)\) shaklida bo'lmagani uchun biz yuqoridagi formuladan foydalana olmaymiz. Biroq, yuqorida ta'kidlab o'tilganidek, biz ushbu formulani o'zimizga mos keladiganini olish uchun o'zgartirishimiz mumkin. Mana,

\[\iint\limits_{S}{{f\left( {x,y,z} \right)\,dS}} = \iint\limits_{D}{{f\left( {g\left( {y,z} \right),y,z} \right)\sqrt {1 + {{\left( {\frac{{\partial g}}{{\partial y}}} \right)}^2} + {{\left( {\frac{{\partial g}}{{\partial z}}} \right)}^2}} \,dA}}\]

Formulaga kiritilgan o'zgarishlar biroz aniq o'zgarishlar bo'lishi kerak. Shunday qilib, integral qilaylik.

\[\iint\limits_{S}{{6xy\,dS}} = \iint\limits_{D}{{6\left( {1 - y - z} \right)y\sqrt {1 + {{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2}} \,dA}}\]

E'tibor bering, biz samolyot uchun tekislik tenglamasini kiritdik x integralda. Ushbu nuqtada bizda juda oddiy qo'sh integral bor. Mana o'sha ish.

\[\begin{align*}\iint\limits_{S}{{6xy\,dS}} & = \sqrt 3 \iint\limits_{D}{{6\left( {y - {y^2} - zy} \right)\,dA}}\\ & = 6\sqrt 3 \int_{{\,0}}^{{\,1}}{{\int_{{\,0}}^{{\,1 - y}}{{y - {y^2} - zy\,dz}}\,dy}}\\ & = 6\sqrt 3 \int_{{\,0}}^{{\,1}}{{\left. {\left( {yz - z{y^2} - \frac{1}{2}{z^2}y} \right)} \right|_0^{1 - y}\,dy}}\\ & = 6\sqrt 3 \int_{{\,0}}^{{\,1}}{{\frac{1}{2}y - {y^2} + \frac{1}{2}{y^3}\,dy}}\\ & = 6\sqrt 3 \left. {\left( {\frac{1}{4}{y^2} - \frac{1}{3}{y^3} + \frac{1}{8}{y^4}} \right)} \right|_0^1 = \frac{{\sqrt 3 }}{4}\end{align*}\]

Biz oldingi sfera parametrlarini bergan edik bo'limi. Mana bu soha uchun parametrlash.

\[\vec r\left( {\theta ,\varphi } \right) = 2\sin \varphi \cos \theta \,\vec i + 2\sin \varphi \sin \theta \,\vec j + 2\cos \varphi \,\vec k\]

Biz sharning yuqori yarmida ishlayotganimiz uchun bu erda parametrlar bo'yicha chegaralar mavjud.

\[0 \le \theta \le 2\pi \hspace{0.5in}0 \le \varphi \le \frac{\pi }{2}\]

Keyinchalik, biz \({\vec r_\theta } \times {\vec r_\varphi }\) ni aniqlashimiz kerak. Mana ikkita alohida vektor.

\[\begin{align*}{{\vec r}_\theta }\left( {\theta ,\varphi } \right) & = - 2\sin \varphi \sin \theta \,\vec i + 2\sin \varphi \cos \theta \,\vec j\\ {{\vec r}_\varphi }\left( {\theta ,\varphi } \right) & = 2\cos \varphi \cos \theta \,\vec i + 2\cos \varphi \sin \theta \,\vec j - 2\sin \varphi \,\vec k\end{align*}\]

Endi ko'ndalang mahsulotni olaylik.

\[\begin{align*}{{\vec r}_\theta } \times {{\vec r}_\varphi } & = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{ - 2\sin \varphi \sin \theta }&{2\sin \varphi \cos \theta }&0\\{2\cos \varphi \cos \theta }&{2\cos \varphi \sin \theta }&{ - 2\sin \varphi }\end{array}} \right|\\ & = - 4{\sin ^2}\varphi \cos \theta \,\vec i - 4\sin \varphi \cos \varphi {\sin ^2}\theta \,\vec k - 4{\sin ^2}\varphi \sin \theta \,\vec j - 4\sin \varphi \cos \varphi {\cos ^2}\theta \,\vec k\\ & = - 4{\sin ^2}\varphi \cos \theta \,\vec i - 4{\sin ^2}\varphi \sin \theta \,\vec j - 4\sin \varphi \cos \varphi \left( {{{\sin }^2}\theta \, + {{\cos }^2}\theta } \right)\vec k\\ & = - 4{\sin ^2}\varphi \cos \theta \,\vec i - 4{\sin ^2}\varphi \sin \theta \,\vec j - 4\sin \varphi \cos \varphi \,\vec k\end{align*}\]

Nihoyat, bizga buning kattaligi kerak,

\[\begin{align*}\left\| {{{\vec r}_\theta } \times {{\vec r}_\varphi }} \right\| & = \sqrt {16{{\sin }^4}\varphi {{\cos }^2}\theta + 16{{\sin }^4}\varphi {{\sin }^2}\theta + 16{{\sin }^2}\varphi {{\cos }^2}\varphi } \\ & = \sqrt {16{{\sin }^4}\varphi \left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) + 16{{\sin }^2}\varphi {{\cos }^2}\varphi } \\ & = \sqrt {16{{\sin }^2}\varphi \left( {{{\sin }^2}\varphi + {{\cos }^2}\varphi } \right)} \\ & = 4\sqrt {{{\sin }^2}\varphi } \\ & = 4\left| {\sin \varphi } \right|\\ & = 4\sin \varphi \end{align*}\]

Biz sinusdagi mutlaq qiymat satrlarini tushirishimiz mumkin, chunki sinus biz ishlayotgan \(\varphi \) diapazonida musbat. U holda sirt integrali,

\[\iint\limits_{S}{{z\,dS}} = \iint\limits_{D}{{2\cos \varphi \left( {4\sin \varphi } \right)\,dA}}\]

Shuni unutmangki, biz \(x\), \(y\) va/yoki \(z\)-ni bularga ham ulashimiz kerak, garchi bu holda biz faqat \(z\)-ni ulashimiz kerak edi. Bu erda qo'sh integral uchun baholash.

\[\begin{align*}\iint\limits_{S}{{z\,dS}} & = \int_{{\,0}}^{{\,2\pi }}{{\int_{{\,0}}^{{\,\frac{\pi }{2}}}{{4\sin \left( {2\varphi } \right)\,d\varphi }}\,d\theta }}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\left. {\left( { - 2\cos \left( {2\varphi } \right)} \right)} \right|_0^{\frac{\pi }{2}}\,d\theta }}\\ & = \int_{{\,0}}^{{\,2\pi }}{{4\,d\theta }}\\ & = 8\pi \end{align*}\]

Oldingi tsilindrni parametrlashtirdik bo'limi. Mana bu silindrning parametrizatsiyasi.

\[\vec r\left( {z,\theta } \right) = \sqrt 3 \cos \theta \,\vec i + \sqrt 3 \sin \theta \,\vec j + z\,\vec k\]

Parametrlar diapazonlari,

\[0 \le z \le 6\hspace{0.25in}0 \le \theta \le 2\pi \]

Endi bizga \({\vec r_z} \times {\vec r_\theta }\) kerak. Mana ikkita vektor.

\[\begin{align*}{{\vec r}_z}\left( {z,\theta } \right) & = \,\vec k\\ {{\vec r}_\theta }\left( {z,\theta } \right) & = - \sqrt 3 \sin \theta \,\vec i + \sqrt 3 \cos \theta \,\vec j\end{align*}\]

Mana ko'ndalang mahsulot.

\[\begin{align*}{{\vec r}_z} \times {{\vec r}_\theta } & = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\0&0&1\\{ - \sqrt 3 \sin \theta }&{\sqrt 3 \cos \theta }&0\end{array}} \right|\\ & = - \sqrt 3 \cos \theta \,\vec i - \sqrt 3 \sin \theta \,\vec j\end{align*}\]

Ushbu vektorning kattaligi,

\[\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\| = \sqrt {3{{\cos }^2}\theta + 3{{\sin }^2}\theta } = \sqrt 3 \]

U holda sirt integrali,

\[\begin{align*}\iint\limits_{S}{{y\,dS}} & = \iint\limits_{D}{{\sqrt 3 \sin \theta \left( {\sqrt 3 } \right)\,dA}}\\ & = 3\int_{{\,0}}^{{\,2\pi }}{{\int_{{\,0}}^{6}{{\sin \theta \,dz}}\,d\theta }}\\ & = 3\int_{{\,0}}^{{\,2\pi }}{{6\sin \theta \,d\theta }}\\ & = \left. {\left( { - 18\cos \theta } \right)} \right|_0^{2\pi }\\ & = 0\end{align*}\]

Bu erda kuzatib borishimiz kerak bo'lgan juda ko'p ma'lumotlar mavjud. Birinchidan, biz oldingi misoldagi kabi deyarli bir xil sirtdan foydalanmoqdamiz (integrand boshqacha). Biroq, oldingi misoldan farqli o'laroq, biz bu safar yuzaga yuqori va pastki qismlarni qo'yamiz. Keling, avval sirtning eskizini bilan boshlaylik.

Bu erda ehtiyot bo'lishimiz kerak. Ushbu eskizda haqiqiy sirtning o'zidan ko'proq narsa bor. Biz \({S_1}\) silindrning \(xy\) tekisligidan samolyotga o'tadigan qismi bo'lishiga ruxsat beramiz. Boshqacha qilib aytganda, silindrning yuqori qismi burchak ostida bo'ladi. Biz samolyotning ichida joylashgan qismini chaqiramiz (i.e. silindrdagi qopqoq) \({S_2}\). Nihoyat, silindrning pastki qismi (bu erda ko'rsatilmagan) \(xy\) tekisligidagi \(\sqrt 3 \) radiusli disk bo'lib, \({S_3}\) bilan belgilanadi.

Ushbu integralni bajarish uchun shuni ta'kidlashimiz kerakki, xuddi standart qo'sh integral kabi, agar sirt bo'laklarga bo'lingan bo'lsa, biz ham sirt integralini ajratishimiz mumkin. Shunday qilib, bizning misolimiz uchun bizda bo'ladi,

\[\iint\limits_{S}{{y + z\,dS}} = \iint\limits_{{{S_1}}}{{y + z\,dS}} + \iint\limits_{{{S_2}}}{{y + z\,dS}} + \iint\limits_{{{S_3}}}{{y + z\,dS}}\]

Bu erda biz uchta integralni bajarishimiz kerak. Biroq, biz oldingi misolda birinchisi uchun ishning ko'p qismini bajardik, shuning uchun bundan boshlaylik.

\({S_1}\) : Silindr

Tsilindr va \(\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\|\) parametrlari quyidagicha:

\[\vec r\left( {z,\theta } \right) = \sqrt 3 \cos \theta \,\vec i + \sqrt 3 \sin \theta \,\vec j + z\,\vec k\hspace{0.5in}\left\| {{{\vec r}_z} \times {{\vec r}_\theta }} \right\| = \sqrt 3 \]

Ushbu muammo va avvalgisi o'rtasidagi farq parametrlar bo'yicha cheklovlardir. Mana ular.

\[\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le z \le 4 - y = 4 - \sqrt 3 \sin \theta \end{array}\]

\(z\) uchun yuqori chegara tekislikdir, shuning uchun biz uni ulashimiz mumkin. Biroq, biz tsilindrda bo'lganimiz uchun parametrlashdan \(y\) nima ekanligini bilamiz, shuning uchun uni ham ulashimiz kerak bo'ladi.

Bu erda silindr uchun integral.

\[\begin{align*}\iint\limits_{{{S_1}}}{{y + z\,dS}} & = \iint\limits_{D}{{\left( {\sqrt 3 \sin \theta + z} \right)\left( {\sqrt 3 } \right)\,dA}}\\ & = \sqrt 3 \int_{{\,0}}^{{\,2\pi }}{{\int_{{\,0}}^{{4 - \sqrt 3 \sin \theta }}{{\sqrt 3 \sin \theta + z\,dz}}\,d\theta }}\\ & = \sqrt 3 \int_{{\,0}}^{{\,2\pi }}{{\sqrt 3 \sin \theta \left( {4 - \sqrt 3 \sin \theta } \right) + \frac{1}{2}{{\left( {4 - \sqrt 3 \sin \theta } \right)}^2}\,d\theta }}\\ & = \sqrt 3 \int_{{\,0}}^{{\,2\pi }}{{8 - \frac{3}{2}{{\sin }^2}\theta \,d\theta }}\\ & = \sqrt 3 \int_{{\,0}}^{{\,2\pi }}{{8 - \frac{3}{4}\left( {1 - \cos \left( {2\theta } \right)} \right)d\theta }}\\ & = \sqrt 3 \left. {\left( {\frac{{29}}{4}\theta + \frac{3}{8}\sin \left( {2\theta } \right)} \right)} \right|_0^{2\pi }\\ & = \frac{{29\sqrt 3 \,\pi }}{2}\end{align*}\]

\({S_2}\) : Silindr ustidagi samolyot

Bunday holda, biz hech qanday parametrlashimiz shart emas, chunki u biz ushbu bo'lim boshida bergan formuladan foydalanishga o'rnatilgan. Samolyot \(z = 4 - y\) tomonidan berilganligini unutmang. Shuni ham yodda tutingki, ushbu sirt uchun \(D\) boshlang'ichda markazlashtirilgan \(\sqrt 3 \) radiusli diskdir.

Bu erda samolyot uchun integral.

\[\begin{align*}\iint\limits_{{{S_2}}}{{y + z\,dS}} & = \iint\limits_{D}{{\left( {y + 4 - y} \right)\sqrt {{{\left( 0 \right)}^2} + {{\left( { - 1} \right)}^2} + 1} \,dA}}\\ & = \sqrt 2 \iint\limits_{D}{{4\,dA}}\end{align*}\]

Biz \(z\) uchun ulanishimiz kerakligini unutmang! Endi biz ikkita yo'ldan birida harakat qilishimiz mumkin. Yoki biz integral bilan davom etishimiz mumkin yoki \(\iint\limits_{D}{{dA}}\) \(D\) maydonidan boshqa narsa emasligini eslashimiz mumkin va biz \(D\) \(\sqrt 3 \) radiusli disk ekanligini bilamiz va shuning uchun integralni bajarish uchun hech qanday sabab yo'q.

Mana bu muammo uchun ishning qolgan qismi.

\[\begin{align*}\iint\limits_{{{S_2}}}{{y + z\,dS}} & = 4\sqrt 2 \iint\limits_{D}{{\,dA}}\\ & = 4\sqrt 2 \,\left( {\pi {{\left( {\sqrt 3 } \right)}^2}} \right)\\ & = 12\sqrt 2 \,\pi \end{align*}\]

\({S_3}\) : Tsilindrning pastki qismi

Shunga qaramay, biz pastki qism uchun tenglama \(g\left( {x,y} \right) = 0\) tomonidan berilganligini va \(D\) boshlang'ich markazida joylashgan \(\sqrt 3 \) radiusli disk ekanligini tushunganimizdan so'ng, biz ushbu bo'limda bergan dastlabki formuladan foydalanish uchun o'rnatiladi. Shuningdek, \(z\)-ga ulanishni unutmang.

Mana bu integral uchun ish.

\[\begin{align*}\iint\limits_{{{S_3}}}{{y + z\,dS}} & = \iint\limits_{D}{{\left( {y + 0} \right)\sqrt {{{\left( 0 \right)}^2} + {{\left( 0 \right)}^2} + {{1}}} \,dA}}\\ & = \iint\limits_{D}{{y\,dA}}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{{r^2}\sin \theta \,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\left. {\left( {\frac{1}{3}{r^3}\sin \theta } \right)} \right|_0^{\sqrt 3 }\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\sqrt 3 \sin \theta \,d\theta }}\\ & = \left. { - \sqrt 3 \cos \theta } \right|_0^{2\pi }\\ & = 0\end{align*}\]

Endi biz integralning qiymatini olishimiz mumkin.

\[\begin{align*}\iint\limits_{S}{{y + z\,dS}} & = \iint\limits_{{{S_1}}}{{y + z\,dS}} + \iint\limits_{{{S_2}}}{{y + z\,dS}} + \iint\limits_{{{S_3}}}{{y + z\,dS}}\\ & = \frac{{29\sqrt 3 \,\pi }}{2} + 12\sqrt 2 \,\pi + 0\\ & = \frac{\pi }{2}\left( {29\sqrt 3 + 24\sqrt 2 } \right)\end{align*}\]