Endi, biz sharning bir qismini integrallashayotganimiz sababli, integral uchun sferik koordinatadan foydalanish mantiqan to'g'ri keladi va chegaralar:

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

Esda tutingki, \(\varphi \) - bu mintaqani qamrab olganimizda biz aylanayotgan musbat \(z\) o'qidan burchak va \(\theta \) - mintaqani qamrab olganimizda \(z\) o'qi atrofida aylanadigan burchak.

Bu holda bizda sharning faqat \(y \ge 0\) bo'lgan qismi bor va shuning uchun \(\theta \) 0 dan \(\pi \) gacha bo'ladi (esda tutingki, biz \(\theta \)ni ijobiy \(x\) o'qidan o'lchaymiz). Biz \(y \ge 0\) bilan sferaning to'liq yarmini xohlayotganimiz uchun \(\varphi \) 0 dan \(\pi \) gacha bo'lishini bilamiz.

ushbu limitlarni integralga ulash va sferik koordinatalarga aylantirish,

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

\(x\) va \(y\) 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.

Bunday holda, biz integratsiyani hal qilishni osonlashtiradigan juda ko'p soddalashtirishni ham qildik. Iloji bo'lsa, bunday soddalashtirishni unutmang!

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

\[\begin{align*}\iiint\limits_{E}{{{x^2} + {y^2}\,dV}} & = \int_{0}^{\pi }{{\int_{0}^{\pi }{{\left. {\left( {\frac{1}{5}{\rho ^5}{{\sin }^3}\varphi } \right)} \right|_0^2\,d\theta }}\,d\varphi }}\\ & = \int_{0}^{\pi }{{\int_{0}^{\pi }{{\frac{{32}}{5}{{\sin }^3}\varphi \,d\theta }}\,d\varphi }}\end{align*}\]

Keyin \(\theta \) integratsiyasini amalga oshiramiz.

\[\begin{align*}\iiint\limits_{E}{{{x^2} + {y^2}\,dV}} & = \int_{0}^{\pi }{{\left. {\left( {\frac{{32}}{5}\theta {{\sin }^3}\varphi } \right)} \right|_0^\pi \,d\varphi }}\\ & = \int_{0}^{\pi }{{\frac{{32}}{5}\pi {{\sin }^3}\varphi \,d\varphi }}\end{align*}\]

Nihoyat, biz \(\varphi \) integratsiyasini amalga oshiramiz.

\[\begin{align*}\iiint\limits_{E}{{{x^2} + {y^2}\,dV}} & = \int_{0}^{\pi }{{\frac{{32}}{5}\pi {{\sin }^2}\varphi \sin \varphi \,d\varphi }}\\ & = \int_{0}^{\pi }{{\frac{{32}}{5}\pi \left( {1 - {{\cos }^2}\varphi } \right)\sin \varphi \,d\varphi }}\\ & = \left. {\left( { - \frac{{32}}{5}\pi \left( {\cos \varphi - \frac{1}{3}{{\cos }^3}\varphi } \right)} \right)} \right|_0^\pi = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{128}}{{15}}\pi }}\end{align*}\]

Ushbu bosqichda biz qilgan trig integral turlarini qanday qilishni eslaysizmi? 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.

E'tibor bering, bu holda, har bir integralning chegaralari doimiylar bo'lganligi sababli, biz xohlagan tartibda integratsiyani amalga oshirishimiz mumkin edi.