Endi biz integralni silindrsimon koordinatalarga aylantirishimiz kerak. Keling, avval chegaralar bilan shug'ullanaylik.

Biz \(z\) ni birinchi navbatda Kartezian koordinatalarini ishlatish uchun integral o'rnatishda birlashtiramiz va shuning uchun biz uni silindrsimon koordinatalardan foydalanish uchun birinchi navbatda integral o'rnatishda birlashtiramiz. \(z\) chegaralarini silindrsimon koordinatalarga quyidagicha aylantirish oson.

\[{r^2} - 11 \le z \le 9 - 3{r^2}\]

Endi \(x\) va \(y\) chegaralari. Bular asl integraldagi ikkita "tashqi" integraldir va shuning uchun ular \(D\) ni ham aniqlaydilar. Keling, avval \(D\) nima ekanligini aniqlay olamizmi yoki yo'qligini bilib olaylik. Bunga ega bo'lganimizdan so'ng, silindrsimon koordinatalardagi integralimiz uchun \(r\) va \(\theta \) chegaralarini aniqlay olishimiz kerak.

Pastki \(y\) chegarasi \(y = - \sqrt {5 - {x^2}} \) va biz \(D\) radiusi \(\sqrt 5 \) diskining eng ko'p pastki qismi bo'lishini ko'rishimiz mumkin.

\(x\) chegaralaridan biz \(x\) ijobiy bo'lishi kerakligini ko'ramiz va shuning uchun \(D\) \(\sqrt 5 \) radiusi diskining 4-banddagi qismidir.^th kvadrant.

Endi biz \(D\) nima ekanligini bilamiz, shuning uchun integral uchun to'liq chegaralar to'plami:

\[\begin{array}{c}\displaystyle \frac{{3\pi }}{2} \le \theta \le 2\pi \\ 0 \le r \le \sqrt 5 \\ {r^2} - 11 \le z \le 9 - 3{r^2}\end{array}\]

Mayli, integralni silindrsimon koordinatalarga aylantiramiz.

\[\begin{align*}\int_{0}^{{\sqrt 5 }}{{\int_{{ - \sqrt {5 - {x^{\,2}}} }}^{0}{{\int_{{{x^{\,2}} + {y^{\,2}} - 11}}^{{9 - 3{x^{\,2}} - 3{y^{\,2}}}}{{\,\,\,2x - 3y\,\,\,dz}}\,dy}}\,dx}} & = \int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\int_{0}^{{\sqrt 5 }}{{\int_{{{r^{\,2}} - 11}}^{{9 - 3{r^{\,2}}}}{{\,\,\,\left( {2r\cos \theta - 3r\sin \theta } \right)r\,dz}}\,dr}}\,d\theta }}\\ & = \int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\int_{0}^{{\sqrt 5 }}{{\int_{{{r^{\,2}} - 11}}^{{9 - 3{r^{\,2}}}}{{\,\,\,{r^2}\left( {2\cos \theta - 3\sin \theta } \right)\,dz}}\,dr}}\,d\theta }}\end{align*}\]

Shuni unutmangki, integralning kartezian ko'rinishidagi \(dz\,dy\,dz\) asl uchlik integraldagi \(dV\) dan keladi. Bundan tashqari, silindrsimon koordinatalar nuqtai nazaridan bizda \(dV = r\,dz\,dr\,d\theta \) borligini bilamiz va shuning uchun biz \(dz\,dy\,dx = r\,dz\,dr\,d\theta \) ekanligini bilamiz va biz integranda qo'shimcha \(r\)ni olamiz.

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

\[\begin{align*}\int_{0}^{{\sqrt 5 }}{{\int_{{ - \sqrt {5 - {x^{\,2}}} }}^{0}{{\int_{{{x^{\,2}} + {y^{\,2}} - 11}}^{{9 - 3{x^{\,2}} - 3{y^{\,2}}}}{{\,\,\,2x - 3y\,\,\,dz}}\,dy}}\,dx}} & = \int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\int_{0}^{{\sqrt 5 }}{{\left. {\left( {{r^2}\left( {2\cos \theta - 3\sin \theta } \right)z} \right)} \right|_{{r^2} - 11}^{9 - 3{r^2}}\,dr}}\,d\theta }}\\ & = \int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\int_{0}^{{\sqrt 5 }}{{{r^2}\left( {2\cos \theta - 3\sin \theta } \right)\left( {20 - 4{r^2}} \right)\,dr}}\,d\theta }}\\ & = \int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\int_{0}^{{\sqrt 5 }}{{\left( {2\cos \theta - 3\sin \theta } \right)\left( {20{r^2} - 4{r^4}} \right)\,dr}}\,d\theta }}\end{align*}\]

Keling, \(r\) integratsiyasini bajaramiz.

\[\begin{align*}\int_{0}^{{\sqrt 5 }}{{\int_{{ - \sqrt {5 - {x^{\,2}}} }}^{0}{{\int_{{{x^{\,2}} + {y^{\,2}} - 11}}^{{9 - 3{x^{\,2}} - 3{y^{\,2}}}}{{\,\,\,2x - 3y\,\,\,dz}}\,dy}}\,dx}} & = \int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\left. {\left( {\left( {2\cos \theta - 3\sin \theta } \right)\left( {\frac{{20}}{3}{r^3} - \frac{4}{5}{r^5}} \right)} \right)} \right|_0^{\sqrt 5 }\,d\theta }}\\ & = \int_{{\frac{{3\pi }}{2}}}^{{2\pi }}{{\frac{{40}}{3}\sqrt 5 \left( {2\cos \theta - 3\sin \theta } \right)\,d\theta }}\end{align*}\]

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

\[\int_{0}^{{\sqrt 5 }}{{\int_{{ - \sqrt {5 - {x^{\,2}}} }}^{0}{{\int_{{{x^{\,2}} + {y^{\,2}} - 11}}^{{9 - 3{x^{\,2}} - 3{y^{\,2}}}}{{\,\,\,2x - 3y\,\,\,dz}}\,dy}}\,dx}} = \left. {\left( {\frac{{40}}{3}\sqrt 5 \left( {2\sin \theta + 3\cos \theta } \right)} \right)} \right|_{\frac{{3\pi }}{2}}^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{200}}{3}\sqrt 5 }}\]