Endi biz integralni sferik koordinatalarga aylantirishimiz kerak. Keling, avval chegaralar haqida qayg'uraylik.

Yuqori \(z\) chegarasidan biz \(z = \sqrt {7 - {x^2} - {y^2}} \) ostida ekanligimizni ko'ramiz (bu faqat \(\sqrt 7 \) radiusli sharning yuqori qismi uchun tenglama).

Pastki \(z\) chegarasidan biz \(z = \sqrt {6{x^2} + 6{y^2}} \) dan yuqori ekanligimizni ko'ramiz (bu shunchaki konusning tenglamasi).

Shunday qilib, biz odatda sferik koordinatalar bilan ishlaganda bo'lgani kabi, biz "muzqaymoq konusi" shaklidagi hududda bo'lganga o'xshaymiz.

Bu bizni quyidagi \(\rho \) chegaralariga olib boradi.

\[0 \le \rho \le \sqrt 7 \]

\(\varphi \) uchun chegaralarni biz ularni konusning tenglamasidan olishimiz mumkin, bu avvalgi bosqichda havola qilingan pastki \(z\) chegarasi. 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:

\[\rho \cos \varphi = \sqrt 6 \rho \sin \varphi \hspace{0.25in} \to \hspace{0.25in} \tan \varphi = \frac{1}{{\sqrt 6 }} \hspace{0.25in} \to \hspace{0.25in}\varphi = {\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 6 }}} \right) = 0.3876\]

Biz ishlayotgan hudud konusning ustida joylashganligi sababli, \(\varphi \) 0 dan 0,3876 gacha bo'lishi kerakligini bilamiz.

Va nihoyat, \(\theta \) chegaralarini olaylik. Malumot uchun bu yerda biz 1-bosqichda topilgan \(x\) va \(y\) chegaralari keltirilgan.

\[\begin{array}{c} - 1 \le x \le 0\\ - \sqrt {1 - {x^2}} \le y \le \sqrt {1 - {x^2}} \end{array}\]

Barcha \(y\) chegaralari bizga shuni aytadiki, asl Dekart koordinatalari integralidan olingan \(D\) mintaqasi radius 1 doirasining bir qismidir. E’tibor bering, bu mantiqiy bo‘lishi kerak, chunki bu ham \(z\) chegaralaridan olingan shar va konusning kesishishidir (buni tekshirish uchun sizga qoldiramiz).

Endi, \(x\) chegaralaridan biz radius 1 doirasining chap tomoniga ega bo'lishimiz kerakligini ko'ramiz va shuning uchun \(\theta \) uchun chegaralar shunday bo'ladi:

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

Integral uchun sferik koordinata chegaralarining to'liq to'plami quyidagicha:

\[\begin{array}{c} \displaystyle 0 \le \varphi \le {\tan ^{ - 1}}\left( {\frac{1}{{\sqrt 6 }}} \right) = 0.3876\\ \displaystyle \frac{\pi }{2} \le \theta \le \frac{{3\pi }}{2}\\0 \le \rho \le \sqrt 7 \end{array}\]

Mayli, integralni sferik koordinatalarga aylantiramiz.

\[\begin{align*}\int_{{ - 1}}^{0}{{\int_{{ - \sqrt {1 - {x^{\,2}}} }}^{{\sqrt {1 - {x^{\,2}}} }}{{\int_{{\sqrt {6{x^{\,2}} + 6{y^{\,2}}} }}^{{\sqrt {7 - {x^{\,2}} - {y^{\,2}}} }}{{\,\,\,18y\,\,\,dz}}\,dy}}\,dx}} & = \int_{0}^{{0.3876}}{{\int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\int_{0}^{{\sqrt 7 }}{{\left( {18\rho \sin \varphi \sin \theta } \right)\left( {{\rho ^2}\sin \varphi } \right)\,d\rho }}\,d\theta }}\,d\varphi }}\\ & = \int_{0}^{{0.3876}}{{\int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\int_{0}^{{\sqrt 7 }}{{18{\rho ^3}{{\sin }^2}\varphi \sin \theta \,d\rho }}\,d\theta }}\,d\varphi }}\end{align*}\]

\(y\) ni sferik koordinatalarga aylantirishni unutmang. Shuni ham unutmangki, \(dz\,dy\,dx\) asl uchlik integraldagi \(dV\) dan keladi. Shuningdek, biz bilamizki, sharsimon koordinatalar nuqtai nazaridan \(dV = {\rho ^2}\sin \varphi \,d\rho \,d\theta \,d\varphi \) va shuning uchun biz o'z navbatida bilamizki,

\[dz\,dy\,dx = {\rho ^2}\sin \varphi \,d\rho \,d\theta \,d\varphi \]

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

\[\begin{align*}\int_{{ - 1}}^{0}{{\int_{{ - \sqrt {1 - {x^{\,2}}} }}^{{\sqrt {1 - {x^{\,2}}} }}{{\int_{{\sqrt {6{x^{\,2}} + 6{y^{\,2}}} }}^{{\sqrt {7 - {x^{\,2}} - {y^{\,2}}} }}{{\,\,\,18y\,\,\,dz}}\,dy}}\,dx}} & = \int_{0}^{{0.3876}}{{\int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\left. {\left( {\frac{9}{2}{\rho ^4}{{\sin }^2}\varphi \sin \theta } \right)} \right|_0^{\sqrt 7 }\,d\theta }}\,d\varphi }}\\ & = \int_{0}^{{0.3876}}{{\int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\frac{{441}}{2}{{\sin }^2}\varphi \sin \theta \,d\theta }}\,d\varphi }}\end{align*}\]

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

\[\begin{align*}\int_{{ - 1}}^{0}{{\int_{{ - \sqrt {1 - {x^{\,2}}} }}^{{\sqrt {1 - {x^{\,2}}} }}{{\int_{{\sqrt {6{x^{\,2}} + 6{y^{\,2}}} }}^{{\sqrt {7 - {x^{\,2}} - {y^{\,2}}} }}{{\,\,\,18y\,\,\,dz}}\,dy}}\,dx}} & = \int_{0}^{{0.3876}}{{\left. {\left( { - \frac{{441}}{2}{{\sin }^2}\varphi \cos \theta } \right)} \right|_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}}\,d\varphi }}\\ & = \int_{0}^{{0.3876}}{{0\,d\varphi }}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{0}\end{align*}\]

Shunday qilib, yuqorida ta'kidlanganidek, biz integrandni nolga tushirganimizdan so'ng, integratsiyani davom ettirish uchun hech qanday sabab yo'q edi, chunki muammoning qolgan qismida javob nolga teng bo'ladi.

Bunday holatlar sodir bo'lganda, bundan hayajonlanmang. Ular vaqti-vaqti bilan bo'ladi va bu shuni anglatadiki, biz boshqacha tarzda integratsiyani biroz tezroq to'xtatamiz.