15.7-bo'lim : Sferik koordinatalarda uch karrali integrallar
3. \( \displaystyle \iiint\limits_{E}{{3z\,dV}}\) ni baholang, bunda \(E\) \({x^2} + {y^2} + {z^2} = 1\) va \(z = \sqrt {{x^2} + {y^2}} \) ichidagi mintaqadir.
Yechim
Mayli, keling, \(E\) hududining qisqacha chizmasidan boshlaylik, shunda biz nima bilan shug'ullanayotganimizni his qilishimiz mumkin.
Biz "an'anaviy" o'qlar to'plami bilan bir qatorda "quti" o'qlari to'plami bilan eskizlarni berdik, bu sirt va mintaqani tasavvur qilishda yordam beradi.
Bunday holda, biz radius 1 sferasi ostidagi mintaqa bilan ishlaymiz (eskizlardagi to'q sariq sirt) ichida joylashgan (i.e. yuqorida) konus (eskizlardagi yashil sirt).
2-qadam
Endi, biz sharning bir qismini integrallashayotganimiz sababli, integral uchun sferik koordinatadan foydalanish mantiqiy.
\(\rho \) va \(\theta \) uchun chegaralar juda aniq bo'lishi kerak, chunki ular faqat shar radiusiga va \(z\) o'qi atrofida aylanish orqali biz qancha sfera olishimizga to'g'ri keladi.
\(\varphi \) uchun chegaralar hech qanday tarzda aniq berilmagan, ammo biz ularni konusning tenglamasidan olishimiz mumkin. Birinchidan, biz bilamizki, silindrsimon koordinatalar nuqtai nazaridan, \(\sqrt {{x^2} + {y^2}} = r\) va biz bilamizki, sharsimon koordinatalar nuqtai nazaridan, \(r = \rho \sin \varphi \). Shuning uchun, agar biz konusning tenglamasini sferik koordinatalarga aylantirsak, biz quyidagilarni olamiz:
\[\rho \cos \varphi = \rho \sin \varphi \hspace{0.25in} \to \hspace{0.25in} \tan \varphi = 1 \hspace{0.25in} \to \hspace{0.25in} \varphi = \frac{\pi }{4}\]Demak, konusning tenglamasi sferik koordinatalar nuqtai nazaridan \(\varphi = \frac{\pi }{4}\) tomonidan berilgan. Biz ishlayotgan hudud konusning ustida joylashganligi sababli, \(\varphi \) 0 dan \(\frac{\pi }{4}\) gacha bo'lishi kerakligini bilamiz.
Keyin chegaralar,
\[\begin{array}{c} \displaystyle 0 \le \varphi \le \frac{\pi }{4}\\ 0 \le \theta \le 2\pi \\ 0 \le \rho \le 1\end{array}\]3-qadam
ushbu limitlarni integralga ulash va sferik koordinatalarga aylantirish,
\[\begin{align*}\iiint\limits_{E}{{3z\,dV}} & = \int_{0}^{{\frac{\pi }{4}}}{{\int_{0}^{{2\pi }}{{\int_{0}^{1}{{\left( {3\rho \cos \varphi } \right)\left( {{\rho ^2}\sin \varphi } \right)\,d\rho }}\,d\theta }}\,d\varphi }}\\ & = \int_{0}^{{\frac{\pi }{4}}}{{\int_{0}^{{2\pi }}{{\int_{0}^{1}{{3{\rho ^3}\cos \varphi \sin \varphi \,d\rho }}\,d\theta }}\,d\varphi }}\end{align*}\]\(z\)-ni sferik koordinatalarga aylantirishni unutmang, shuningdek, \(dV = {\rho ^2}\sin \varphi \,d\rho \,d\theta \,d\varphi \)-ni unutmang, shuning uchun biz \(dV\)-ni sferik koordinatalarga o'tkazishda bir nechta qo'shimcha shartlarni tanlaymiz.
4-qadam
OK, endi biz qilishimiz kerak bo'lgan narsa - bu integralni baholash. Mana \(\rho \) integratsiyasi.
\[\begin{align*}\iiint\limits_{E}{{3z\,dV}} & = \int_{0}^{{\frac{\pi }{4}}}{{\int_{0}^{{2\pi }}{{\left. {\left( {\frac{3}{4}{\rho ^4}\cos \varphi \sin \varphi } \right)} \right|_0^1\,d\theta }}\,d\varphi }}\\ & = \int_{0}^{{\frac{\pi }{4}}}{{\int_{0}^{{2\pi }}{{\frac{3}{4}\cos \varphi \sin \varphi \,d\theta }}\,d\varphi }}\end{align*}\]5-qadam
Keling, \(\theta \) integratsiyasini bajaramiz.
\[\begin{align*}\iiint\limits_{E}{{3z\,dV}} & = \int_{0}^{{\frac{\pi }{4}}}{{\left. {\left( {\frac{3}{4}\theta \cos \varphi \sin \varphi } \right)} \right|_0^{2\pi }\,d\varphi }}\\ & = \int_{0}^{{\frac{\pi }{4}}}{{\,\frac{3}{2}\pi \cos \varphi \sin \varphi \,d\varphi }}\end{align*}\]6-qadam
Va nihoyat, biz \(\varphi \) integratsiyasini amalga oshiramiz.
\[\begin{align*}\iiint\limits_{E}{{3z\,dV}} & = \int_{0}^{{\frac{\pi }{4}}}{{\,\frac{3}{4}\pi \sin \left( {2\varphi } \right)\,d\varphi }}\\ & = \left. {\left( {\, - \frac{3}{8}\pi \cos \left( {2\varphi } \right)\,} \right)} \right|_0^{\frac{\pi }{4}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{3}{8}\pi }}\end{align*}\]Biz integralni tezda bajara oladigan narsaga kamaytirish uchun sinus uchun ikki burchakli formuladan foydalandik.
Bu shunday emas, bu holda, har bir integralning chegaralari doimiylar bo'lgani uchun biz integratsiyani xohlagan tartibda bajarishimiz mumkin edi.