17.3-bo'lim : Yuzaki integrallar
4. \( \displaystyle \iint\limits_{S}{{xz\,dS}}\) ni baholang, bu erda \(S\) 3 radiusli sharning \(x \le 0\), \(y \ge 0\) va \(z \ge 0\) qismidir.
Yechim
Keling, ushbu muammoni hal qilishda biz ishlayotgan sirtning qisqacha eskizi bilan boshlaylik.
E'tibor bering, bu muammodagi sirt faqat sharning o'zi. "Qirralar" (o'ng/chapdagi yashil rangli qismlar) eskizda bo'lishiga qaramay, bu sirtning bir qismi emas. Ular eskizda sirtni biroz aniqroq qilishga harakat qilish uchun eskizga kiritilgan. Agar muammo aniq bo'lishi uchun uni qandaydir tarzda belgilab qo'ygan bo'lsa, biz "qirralarni" kiritgan bo'lardik.
2-qadam
Endi bizning sirtimiz shar bo'lgani uchun biz uni parametrlashimiz va sirt integrali uchun quyidagi formuladan foydalanishimiz kerak.
\[\iint\limits_{S}{{f\left( {x,y,z} \right)\,dS}} = \iint\limits_{D}{{f\left( {\vec r\left( {u,v} \right)} \right)\left\| {{{\vec r}_u} \times {{\vec r}_v}} \right\|\,dA}}\]Bu erda parametrlashtirishni amalga oshirishda kerak bo'lganda \(u\) va \(v\) tanlanadi.
Biz oldingi bo'limda sharni qanday parametrlashtirishni ko'rib chiqdik, shuning uchun parametrlash uchun batafsil ma'lumot bermaymiz. Parametrlash - bu,
\[\vec r\left( {\theta ,\varphi } \right) = \left\langle {3\sin \varphi \cos \theta ,3\sin \varphi \sin \theta ,3\cos \varphi } \right\rangle \hspace{0.25in}\,\,\,\,\,\frac{1}{2}\pi \le \theta \le \pi ,\,\,\,\,\,0 \le \varphi \le \frac{1}{2}\pi \]Biz sharning faqat yuqori yarmini olishimizga ishonch hosil qilish uchun \(\varphi \)-da cheklash 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.
\[{\vec r_\theta } = \left\langle { - 3\sin \varphi \sin \theta ,3\sin \varphi \cos \theta ,0} \right\rangle \hspace{0.25in}\hspace{0.25in}{\vec r_\varphi } = \left\langle {3\cos \varphi \cos \theta ,3\cos \varphi \sin \theta , - 3\sin \varphi } \right\rangle \] \[\begin{align*}{{\vec r}_\theta } \times {{\vec r}_\varphi } & = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{ - 3\sin \varphi \sin \theta }&{3\sin \varphi \cos \theta }&0\\{3\cos \varphi \cos \theta }&{3\cos \varphi \sin \theta }&{ - 3\sin \varphi }\end{array}} \right|\\ & = - 9{\sin ^2}\varphi \cos \theta \,\vec i - 9\sin \varphi \cos \varphi {\sin ^2}\theta \vec k - 9\sin \varphi \cos \varphi {\cos ^2}\theta \vec k - 9{\sin ^2}\varphi \sin \theta \vec j\\ & = - 9{\sin ^2}\varphi \cos \theta \,\vec i - 9{\sin ^2}\varphi \sin \theta \vec j - 9\sin \varphi \cos \varphi \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)\vec k\\ & = - 9{\sin ^2}\varphi \cos \theta \,\vec i - 9{\sin ^2}\varphi \sin \theta \vec j - 9\sin \varphi \cos \varphi \vec k\end{align*}\]O'zaro mahsulotning kattaligi,
\[\begin{align*}\left\| {{{\vec r}_\theta } \times {{\vec r}_\varphi }} \right\| & = \sqrt {{{\left( { - 9{{\sin }^2}\varphi \cos \theta } \right)}^2} + {{\left( { - 9{{\sin }^2}\varphi \sin \theta } \right)}^2} + {{\left( { - 9\sin \varphi \cos \varphi } \right)}^2}} \\ & = \sqrt {81{{\sin }^4}\varphi \left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) + 81{{\sin }^2}\varphi {{\cos }^2}\varphi } \\ & = \sqrt {81{{\sin }^2}\varphi \left( {{{\sin }^2}\varphi + {{\cos }^2}\varphi } \right)} \\ & = 9\left| {\sin \varphi } \right|\\ & = 9\sin \varphi \end{align*}\]U holda integral,
\[\iint\limits_{S}{{xz\,dS}} = \iint\limits_{D}{{\left( {3\sin \varphi \cos \theta } \right)\left( {3\cos \varphi } \right)\left( {9\sin \varphi } \right)\,dA}} = \iint\limits_{D}{{81\cos \varphi {{\sin }^2}\varphi \cos \theta \,dA}}\]Yuzaki parametrlashning \(x\) va \(z\) komponentlarini integrandaga ulashni unutmang va \(D\) faqat biz yuqorida parametrlashda qayd etgan \(\theta \) va \(\varphi \) chegaralaridir.
3-qadam
Endi biz qilishimiz kerak bo'lgan yagona narsa - ikki karrali integralni baholash va bu hozir juda qiyin bo'lmasligi kerak.
U holda integral,
\[\begin{align*}\iint\limits_{S}{{xz\,dS}} & = \iint\limits_{D}{{81\cos \varphi {{\sin }^2}\varphi \cos \theta \,dA}}\\ & = \int_{{\frac{1}{2}\pi }}^{\pi }{{\int_{0}^{{\frac{1}{2}\pi }}{{81\cos \varphi {{\sin }^2}\varphi \cos \theta \,d\varphi }}\,d\theta }}\\ & = \int_{{\frac{1}{2}\pi }}^{\pi }{{\left. {\left( {27{{\sin }^3}\varphi \cos \theta \,d\varphi } \right)} \right|_0^{\frac{1}{2}\pi }\,d\theta }}\\ & = \int_{{\frac{1}{2}\pi }}^{\pi }{{27\cos \theta \,d\theta }}\\ & = \left. {\left( {27\sin \theta } \right)} \right|_{\frac{1}{2}\pi }^\pi = \require{bbox} \bbox[2pt,border:1px solid black]{{ - 27}}\end{align*}\]