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 sharning 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:

\[\rho \cos \varphi = - \sqrt 3 \rho \sin \varphi \hspace{0.25in} \to \hspace{0.25in} \tan \varphi = - \frac{1}{{\sqrt 3 }} \hspace{0.25in} \to \hspace{0.25in} \varphi = \frac{{5\pi }}{6}\]

Demak, konusning tenglamasi sferik koordinatalar nuqtai nazaridan \(\varphi = \frac{{5\pi }}{6}\) tomonidan berilgan. Biz ishlayotgan hudud konusning ostida joylashganligi sababli, \(\varphi \) \(\frac{{5\pi }}{6}\) dan \(\pi \) gacha bo'lishi kerakligini bilamiz.

Keyin chegaralar,

\[\begin{array}{c}\displaystyle \frac{{5\pi }}{6} \le \varphi \le \pi \\ 0 \le \theta \le 2\pi \\ 0 \le \rho \le 6\end{array}\]

ushbu limitlarni integralga ulash va sferik koordinatalarga aylantirish,

\[\begin{align*}\iiint\limits_{E}{{{x^2}\,dV}} = \int_{{\frac{{5\pi }}{6}}}^{\pi }{{\int_{0}^{{2\pi }}{{\int_{0}^{6}{{{{\left( {\rho \sin \varphi \cos \theta } \right)}^2}\left( {{\rho ^2}\sin \varphi } \right)\,d\rho }}\,d\theta }}\,d\varphi }}\\ & = \int_{{\frac{{5\pi }}{6}}}^{\pi }{{\int_{0}^{{2\pi }}{{\int_{0}^{6}{{{\rho ^4}{{\sin }^3}\varphi {{\cos }^2}\theta \,d\rho }}\,d\theta }}\,d\varphi }}\end{align*}\]

\(z\)-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.

OK, endi biz qilishimiz kerak bo'lgan narsa - bu integralni baholash. Mana \(\rho \) integratsiyasi.

\[\begin{align*}\iiint\limits_{E}{{{x^2}\,dV}} & = \int_{{\frac{{5\pi }}{6}}}^{\pi }{{\int_{0}^{{2\pi }}{{\left. {\left( {\frac{1}{5}{\rho ^5}{{\sin }^3}\varphi {{\cos }^2}\theta } \right)} \right|_0^6\,d\theta }}\,d\varphi }}\\ & = \int_{{\frac{{5\pi }}{6}}}^{\pi }{{\int_{0}^{{2\pi }}{{\frac{{7776}}{5}{{\sin }^3}\varphi {{\cos }^2}\theta \,d\theta }}\,d\varphi }}\end{align*}\]

Keling, \(\theta \) integratsiyasini bajaramiz.

\[\begin{align*}\iiint\limits_{E}{{{x^2}\,dV}} & = \int_{{\frac{{5\pi }}{6}}}^{\pi }{{\int_{0}^{{2\pi }}{{\frac{{7776}}{5}{{\sin }^3}\varphi \left( {\frac{1}{2}} \right)\left( {1 + \cos \left( {2\theta } \right)} \right)\,d\theta }}\,d\varphi }}\\ & = \int_{{\frac{{5\pi }}{6}}}^{\pi }{{\left. {\left( {\frac{{3888}}{5}{{\sin }^3}\varphi \left( {\theta + \frac{1}{2}\sin \left( {2\theta } \right)} \right)} \right)} \right|_0^{2\pi }\,d\varphi }}\\ & = \int_{{\frac{{5\pi }}{6}}}^{\pi }{{\frac{{7776}}{5}\pi {{\sin }^3}\varphi \,d\varphi }}\end{align*}\]

Va nihoyat, biz \(\varphi \) integratsiyasini amalga oshiramiz.

\[\begin{align*}\iiint\limits_{E}{{{x^2}\,dV}} & = \int_{{\frac{{5\pi }}{6}}}^{\pi }{{\frac{{7776}}{5}\pi {{\sin }^2}\varphi \sin \varphi \,d\varphi }}\\ & = \int_{{\frac{{5\pi }}{6}}}^{\pi }{{\frac{{7776}}{5}\pi \left( {1 - {{\cos }^2}\varphi } \right)\sin \varphi \,d\varphi }}\\ & = \left. {\left( {\, - \frac{{7776}}{5}\pi \left( {\cos \varphi - \frac{1}{3}{{\cos }^3}\varphi } \right)\,} \right)} \right|_{\frac{{5\pi }}{6}}^\pi = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{7776}}{5}\pi \left( {\frac{2}{3} - \frac{{3\sqrt 3 }}{8}} \right) = 83.7799}}\end{align*}\]

Oxirgi ikki bosqichda qilgan trig integrallarini qanday qilishni eslaysiz, shunday emasmi? Agar yo'q bo'lsa, siz bosh bo'lishingiz kerak orqaga va Ba'zi Calculus II materiallarini ko'rib chiqing, chunki ular vaqti-vaqti bilan paydo bo'ladi.

Bu shunday emas, bu holda, har bir integralning chegaralari doimiylar bo'lgani uchun biz integratsiyani xohlagan tartibda bajarishimiz mumkin edi.