17.2-bo'lim : Parametrik yuzalar
8. \(z \le 0\) yordamida \({x^2} + {y^2} + {z^2} = 25\) qismining sirt maydonini aniqlang.
Yechim
Biz avval sferani parametrlashimiz kerak va biz bu muammo to'plamida sharni yaratdik, shuning uchun biz bu erda parametrlash haqida batafsil to'xtalmaymiz.
To'liq sfera uchun parametrlash:
\[\vec r\left( {\theta ,\varphi } \right) = \left\langle {5\sin \varphi \cos \theta ,5\sin \varphi \sin \theta ,5\cos \varphi } \right\rangle \]Biz, albatta, to'liq sohani xohlamaymiz. Biz faqat sharning pastki yarmini xohlaymiz, i.e. qismi \(z \le 0\) bilan. Bu biz \(\varphi \)ni \(\frac{1}{2}\pi \le \varphi \le \pi \) bilan cheklashimiz kerakligini anglatadi. Eslatib o'tamiz, \(\varphi \) musbat \(z\) o'qi bilan hosil bo'ladigan burchak nuqtalari va biz faqat \(xy\) tekisligi ostidagi nuqtalarni xohlaganimiz uchun bizga \(\frac{1}{2}\pi \le \varphi \le \pi \) diapazoni kerak bo'ladi.
Biz to'liq pastki yarmini xohlaymiz va shuning uchun biz \(\theta \) diapazoni uchun \(0 \le \theta \le 2\pi \) dan foydalanamiz.
2-qadam
Keyinchalik, biz \({\vec r_\theta } \times {\vec r_\varphi }\) ni hisoblashimiz kerak. Mana o'sha ish.
\[{\vec r_\theta } = \left\langle { - 5\sin \varphi \sin \theta ,5\sin \varphi \cos \theta ,0} \right\rangle \hspace{0.5in}{\vec r_\varphi } = \left\langle {5\cos \varphi \cos \theta ,5\cos \varphi \sin \theta , - 5\sin \varphi } \right\rangle \] \[\begin{align*}{{\vec r}_\theta } \times {{\vec r}_\varphi } & = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{ - 5\sin \varphi \sin \theta }&{5\sin \varphi \cos \theta }&0\\{5\cos \varphi \cos \theta }&{5\cos \varphi \sin \theta }&{ - 5\sin \varphi }\end{array}} \right|\\ & = - 25{\sin ^2}\varphi \cos \theta \,\vec i - 25\sin \varphi \cos \varphi {\sin ^2}\theta \vec k - 25\sin \varphi \cos \varphi {\cos ^2}\theta \vec k - 25{\sin ^2}\varphi \sin \theta \vec j\\ & = - 25{\sin ^2}\varphi \cos \theta \,\vec i - 25{\sin ^2}\varphi \sin \theta \vec j - 25\sin \varphi \cos \varphi \left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)\vec k\\ & = - 25{\sin ^2}\varphi \cos \theta \,\vec i - 25{\sin ^2}\varphi \sin \theta \vec j - 25\sin \varphi \cos \varphi \vec k\end{align*}\]Endi bizga haqiqatan ham kerak bo'lgan narsa,
\[\begin{align*}\left\| {{{\vec r}_\theta } \times {{\vec r}_\varphi }} \right\| & = \sqrt {{{\left( { - 25{{\sin }^2}\varphi \cos \theta } \right)}^2} + {{\left( { - 25{{\sin }^2}\varphi \sin \theta } \right)}^2} + {{\left( { - 25\sin \varphi \cos \varphi } \right)}^2}} \\ & = \sqrt {625{{\sin }^4}\varphi \left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) + 625{{\sin }^2}\varphi {{\cos }^2}\varphi } \\ & = \sqrt {625{{\sin }^2}\varphi \left( {{{\sin }^2}\varphi + {{\cos }^2}\varphi } \right)} \\ & = 25\left| {\sin \varphi } \right|\\ & = 25\sin \varphi \end{align*}\]E'tibor bering, biz mutlaq qiymat satrlarini sinusga tushirishimiz mumkin, chunki biz \(\frac{1}{2}\pi \le \varphi \le \pi \) da sinus ijobiy bo'lishini bilamiz.
3-qadam
U holda sirt maydoni uchun integral,
\[A = \iint\limits_{D}{{25\sin \varphi \,dA}} = \int_{0}^{{2\pi }}{{\int_{{\frac{1}{2}\pi }}^{\pi }{{25\sin \varphi \,d\varphi }}\,d\theta }}\]Yuqoridagi integralda ta'kidlanganidek, \(D\) 1-bosqichda topilgan \(\theta \) va \(\varphi \) diapazonlari xolos.
4-qadam
Endi biz sirt maydonini olish uchun integralni baholashimiz kerak.
\[A = \int_{0}^{{2\pi }}{{\int_{{\frac{1}{2}\pi }}^{\pi }{{25\sin \varphi \,d\varphi }}\,d\theta }} = \int_{0}^{{2\pi }}{{\left. { - 25\cos \varphi } \right|_{\frac{1}{2}\pi }^\pi \,d\theta }} = \int_{0}^{{2\pi }}{{25\,d\theta }} = \require{bbox} \bbox[2pt,border:1px solid black]{{50\pi }}\]