Biz sharning yuqori yarmini olganimiz uchun o'zgaruvchilar uchun chegaralar,

\[\begin{array}{c}0 \le \rho \le 1\\ 0 \le \theta \le 2\pi \\ \displaystyle 0 \le \varphi \le \frac{\pi }{2}\end{array}\]

U holda integral,

\[\begin{align*}\iiint\limits_{E}{{16z\,dV}} & = \int_{0}^{{\,\frac{\pi }{2}}}{{\int_{{\,0}}^{{\,2\pi }}{{\int_{{\,\,0}}^{1}{{{\rho ^2}\sin \varphi \left( {16\rho \cos \varphi } \right)\,d\rho }}\,d\theta }}\,d\varphi }}\\ & = \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{\int_{{\,0}}^{{\,2\pi }}{{\int_{{\,\,0}}^{{\,1}}{{8{\rho ^3}\sin \left( {2\varphi } \right)\,d\rho }}\,d\theta }}\,d\varphi }}\\ & = \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{\int_{{\,0}}^{{\,2\pi }}{{2\sin \left( {2\varphi } \right)\,d\theta }}\,d\varphi }}\\ & = \int_{{\,0}}^{{\,\frac{\pi }{2}}}{{4\pi \sin \left( {2\varphi } \right)\,d\varphi }}\\ & = \left. { - 2\pi \cos \left( {2\varphi } \right)} \right|_0^{\frac{\pi }{2}}\\ & = 4\pi \end{align*}\]

Birinchidan, biz chegaralar haqida g'amxo'rlik qilishimiz kerak. \(E\) hududi asosan teskari muzqaymoq bo'lib, yarmiga bo'lingan, shunda faqat \(x \le 0\) qismi qoladi. Shuning uchun, biz 2 radiusli sharning bir qismida bo'lganimiz uchun, bizda bo'lishi kerak,

\[0 \le \rho \le 2\]

\(\varphi \) uchun biz ehtiyot bo'lishimiz kerak. Muammo bayonotida aytilishicha, konus manfiy \(z\) o'qi bilan \(\frac{\pi }{3}\) burchak hosil qiladi. Biroq, \(\varphi \) ijobiy \(z\) o'qidan o'lchanganini unutmang. Shuning uchun, musbat \(z\) o'qidan o'lchangan holda, konusni "boshlaydigan" birinchi burchak \(\varphi = \frac{{2\pi }}{3}\) bo'ladi va u manfiy \(z\) o'qiga boradi. Shuning uchun biz quyidagi cheklovlarni olamiz \[\varphi \].

\[\frac{{2\pi }}{3} \le \varphi \le \pi \]

Nihoyat, \(\theta \) uchun biz \(x \le 0\) ham aytilganidan foydalanishimiz mumkin. Bu biz \(y\) o'qining chap tomonida ekanligimizni anglatadi va shuning uchun \(\theta \) diapazoni shunday bo'lishi kerak,

\[\frac{\pi }{2} \le \theta \le \frac{{3\pi }}{2}\]

Endi bizda chegaralar bor, biz integralni baholashimiz mumkin.

\[\begin{align*}\iiint\limits_{E}{{z\,x\,dV}} & = \int_{{\frac{{2\pi }}{3}}}^{{\,\pi }}{{\int_{{\,\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\int_{{\,\,0}}^{2}{{\left( {\rho \cos \varphi } \right)\left( {\rho \sin \varphi \cos \theta } \right){\rho ^2}\sin \varphi \,d\rho }}\,d\theta }}\,d\varphi }}\\ & = \int_{{\frac{{2\pi }}{3}}}^{{\,\pi }}{{\int_{{\,\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\int_{{\,\,0}}^{2}{{{\rho ^4}\cos \varphi {{\sin }^2}\varphi \cos \theta \,d\rho }}\,d\theta }}\,d\varphi }}\\ & = \int_{{\frac{{2\pi }}{3}}}^{\pi }{{\int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\frac{{32}}{5}\cos \varphi {{\sin }^2}\varphi \cos \theta \,d\theta }}\,d\varphi }}\\ & = \int_{{\frac{{2\pi }}{3}}}^{\pi }{{ - \frac{{64}}{5}\cos \varphi {{\sin }^2}\varphi \,d\varphi }}\\ & = \left. { - \frac{{64}}{{15}}{{\sin }^3}\varphi } \right|_{\frac{{2\pi }}{3}}^\pi \\ & = \frac{{8\sqrt 3 }}{5}\end{align*}\]

Avval o'zgaruvchilar uchun chegaralarni yozamiz.

\[\begin{array}{c}0 \le y \le 3\\ 0 \le x \le \sqrt {9 - {y^2}} \\ \sqrt {{x^2} + {y^2}} \le z \le \sqrt {18 - {x^2} - {y^2}} \end{array}\]

\(x\) diapazoni bizda 3 radiusli diskning o'ng yarmining boshlang'ich nuqtasida joylashganligini aytadi. Biz \(y\)-ni ijobiy qiymatlar bilan cheklayotganimiz sababli, biz birinchi kvadrantda chorak diskga ega bo'lamiz. Shuning uchun, \(D\) birinchi kvadrantda bo'lganligi sababli, \(E\) birinchi oktantda bo'lishi kerak va bu o'z navbatida bizda \(\theta \) uchun quyidagi diapazon borligini bildiradi (chunki bu \(z\) o'qi atrofidagi burchak).

\[0 \le \theta \le \frac{\pi }{2}\]

Keling, \(z\) diapazoni bizga nimani aytishini ko'rib chiqaylik. Pastki chegara, \(z = \sqrt {{x^2} + {y^2}} \), konusning yuqori yarmidir. Hozircha bu bizga kerak emas, lekin keyinroq kerak bo'ladi. Yuqori chegara, \(z = \sqrt {18 - {x^2} - {y^2}} \), sharning yuqori yarmi,

\[{x^2} + {y^2} + {z^2} = 18\]

va shuning uchun biz endi \(\rho \) uchun quyidagi diapazonga egamiz

\[0 \le \rho \le \sqrt {18} = 3\sqrt 2 \]

Endi bizga faqat \(\varphi \) diapazoni kerak bo'ladi. Buni olishning ikki yo'li bor. Ulardan biri konus va shar kesishgan joydan. Konus uchun tenglamani sharga ulash,

\[\begin{align*}{\left( {\sqrt {{x^2} + {y^2}} } \right)^2} + {z^2} & = 18\\ {z^2} + {z^2} & = 18\\ {z^2} & = 9\\ z & = 3\end{align*}\]

E'tibor bering, biz bu erda \(z\) ni ijobiy deb hisoblashimiz mumkin, chunki biz konusning va/yoki sharning yuqori yarmiga ega ekanligimizni bilamiz. Nihoyat, buni \(z\) konvertatsiyasiga ulang va biz sharda kesishganimizdan beri \(\rho = 3\sqrt 2 \) ekanligini bilishimizdan foydalaning. Bu beradi,

\[\begin{align*}\rho \cos \varphi & = 3\\ 3\sqrt 2 \cos \varphi & = 3\\ \cos \varphi = \frac{1}{{\sqrt 2 }} & = \frac{{\sqrt 2 }}{2}\hspace{0.5in} \Rightarrow \hspace{0.5in}\varphi = \frac{\pi }{4}\end{align*}\]

Shunday qilib, bizda quyidagi diapazon borga o'xshaydi,

\[0 \le \varphi \le \frac{\pi }{4}\]

Ushbu diapazonni olishning boshqa usuli - konusning o'zi. Avval tenglamani silindrsimon koordinatalarga, keyin esa sferik koordinatalarga aylantirib, biz quyidagilarni olamiz:

\[\begin{align*}z& = r\\ \rho \cos \varphi & = \rho \sin \varphi \\ 1 & = \tan \varphi \hspace{0.5in} \Rightarrow \hspace{0.5in}\varphi = \frac{\pi }{4}\end{align*}\]

Shunday qilib, \({\rho ^2} = {x^2} + {y^2} + {z^2}\) ni eslab, integral u holda,

\[\int_{{\,0}}^{{\,3}}{{\int_{{\,0}}^{{\sqrt {9 - {y^2}} }}{{\int_{{\sqrt {{x^2} + {y^2}} }}^{{\sqrt {18 - {x^2} - {y^2}} }}{{{x^2} + {y^2} + {z^2}\,dz}}\,dx}}\,dy}} = \int_{{\,0}}^{{\,{\pi }/{4}\;}}{{\int_{{\,0}}^{{{\pi }/{2}\;}}{{\int_{{\,0}}^{{\,3\sqrt 2 }}{{{\rho ^4}\sin \varphi \,d\rho }}\,d\theta }}\,d\varphi }}\]