Endi, biz sharning bir qismini integrallashayotganimiz sababli, integral uchun sferik koordinatadan foydalanish mantiqiy va chegaralar:

\[\begin{array}{c}0 \le \varphi \le \frac{\pi }{2}\\ 0 \le \theta \le 2\pi \\ 0 \le \rho \le 4\end{array}\]

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

Bu holda bizda sharning to'liq yuqori yarmi bor va shuning uchun \(\theta \) 0 dan \(2\pi \) gacha, \(\varphi \) esa 0 dan \(\frac{\pi }{2}\) gacha bo'ladi.

ushbu limitlarni integralga ulash va sferik koordinatalarga aylantirish,

\[\begin{align*}\iiint\limits_{E}{{10xz + 3\,dV}}& = \int_{0}^{{\frac{\pi }{2}}}{{\int_{0}^{{2\pi }}{{\int_{0}^{4}{{\left[ {10\left( {\rho \sin \varphi \cos \theta } \right)\left( {\rho \cos \varphi } \right) + 3} \right]\left( {{\rho ^2}\sin \varphi } \right)\,d\rho }}\,d\theta }}\,d\varphi }}\\ & = \int_{0}^{{\frac{\pi }{2}}}{{\int_{0}^{{2\pi }}{{\int_{0}^{4}{{10{\rho ^4}{{\sin }^2}\varphi \cos \varphi \cos \theta + 3{\rho ^2}\sin \varphi \,d\rho }}\,d\theta }}\,d\varphi }}\end{align*}\]

\(x\) va \(z\)-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.

\[\begin{align*}\iiint\limits_{E}{{10xz + 3\,dV}} & = \int_{0}^{{\frac{\pi }{2}}}{{\int_{0}^{{2\pi }}{{\left. {\left( {2{\rho ^5}{{\sin }^2}\varphi \cos \varphi \cos \theta + {\rho ^3}\sin \varphi } \right)} \right|_0^4\,d\theta }}\,d\varphi }}\\ & = \int_{0}^{{\frac{\pi }{2}}}{{\int_{0}^{{2\pi }}{{2048{{\sin }^2}\varphi \cos \varphi \cos \theta + 64\sin \varphi \,\,d\theta }}\,d\varphi }}\end{align*}\]

Keling, \(\theta \) integratsiyasini amalga oshiramiz.

\[\begin{align*}\iiint\limits_{E}{{10xz + 3\,dV}} & = \int_{0}^{{\frac{\pi }{2}}}{{\left. {\left( {2048{{\sin }^2}\varphi \cos \varphi \sin \theta + 64\theta \sin \varphi } \right)} \right|_0^{2\pi }\,d\varphi }}\\ & = \int_{0}^{{\frac{\pi }{2}}}{{128\pi \sin \varphi \,d\varphi }}\end{align*}\]

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

\[\iiint\limits_{E}{{10xz + 3\,dV}} = \left. {\left( { - 128\pi \cos \varphi } \right)} \right|_0^{\frac{\pi }{2}} = \require{bbox} \bbox[2pt,border:1px solid black]{{128\pi }}\]

E'tibor bering, bu holda, har bir integralning chegaralari doimiylar bo'lganligi sababli, biz xohlagan tartibda integratsiyani amalga oshirishimiz mumkin edi. Bunday holda, \(\varphi \) ni birinchi yoki ikkinchi marta bajarish "oddiyroq" bo'lishi mumkin edi, chunki bu qolgan integral(lar) uchun integrani sezilarli darajada kamaytiradi.