Shunday qilib, yuqoridagi eskizdan ma'lum bo'lishi kerakki, biz avval \(z\)ni birlashtirishimiz kerak va shuning uchun biz \(z\) uchun quyidagi limitlarga ega bo'lamiz.

\[2{x^2} + 2{y^2} - 7 \le z \le 1\]

Ushbu muammo uchun \(D\) 1-qadamda chizilgan hududni “qoplaydigan” diskdir. Biz disk tenglamasini muammo bayonidagi ikkita tenglamani teng qilib, biroz qayta yozish orqali aniqlashimiz mumkin.

\[2{x^2} + 2{y^2} - 7 = 1\hspace{0.25in} \to \hspace{0.25in}2{x^2} + 2{y^2} = 8\hspace{0.25in} \to \hspace{0.25in}{x^2} + {y^2} = 4\]

Shunday qilib, \(D\) \({x^2} + {y^2} \le 4\) diskidir va bu integral uchun silindrsimon koordinatalardan foydalanishimiz kerakligi aniq bo'lishi kerak.

Mana bu muammoning silindrsimon koordinatalari.

\[\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le 2\\ 2{r^2} - 7 \le z \le 1\end{array}\]

2-bosqichdagi \(z\) chegaralarini silindrsimon koordinatalarga aylantirishni ham unutmang.

ushbu limitlarni integralga kiritish va silindrsimon koordinatalarga aylantirish,

\[\begin{align*}\iiint\limits_{E}{{4xy\,dV}} & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{\int_{{2{r^2} - 7}}^{1}{{4\left( {r\cos \theta } \right)\left( {r\sin \theta } \right)r\,dz}}\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{\int_{{2{r^2} - 7}}^{1}{{4{r^3}\cos \theta \sin \theta \,dz}}\,dr}}\,d\theta }}\end{align*}\]

\(x\) va \(y\)-ni silindrsimon koordinatalarga aylantirishni unutmang, shuningdek, \(dV = r\,dz\,dr\,d\theta \)-ni unutmang, shuning uchun \(dV\)-ni silindrsimon koordinatalarga o'tkazishda biz boshqa \(r\)-ni tanlaymiz.

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

\[\begin{align*}\iiint\limits_{E}{{4xy\,dV}} & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{\left. {\left( {4{r^3}\cos \theta \sin \theta z} \right)} \right|_{2{r^2} - 7}^1\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{4{r^3}\left( {8 - 2{r^2}} \right)\cos \theta \sin \theta \,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{2}{{\left( {32{r^3} - 8{r^5}} \right)\cos \theta \sin \theta \,dr}}\,d\theta }}\end{align*}\]

Keyin \(r\) integratsiyasini amalga oshiramiz.

\[\begin{align*}\iiint\limits_{E}{{4xy\,dV}} & = \int_{0}^{{2\pi }}{{\left. {\left( {8{r^4} - \frac{4}{3}{r^6}} \right)\cos \theta \sin \theta } \right|_0^2\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\frac{{128}}{3}\cos \theta \sin \theta \,d\theta }}\end{align*}\]

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

\[\iiint\limits_{E}{{4xy\,dV}} = \int_{0}^{{2\pi }}{{\frac{{64}}{3}\sin \left( {2\theta } \right)\,d\theta }} = \left. { - \frac{{32}}{3}\cos \left( {2\theta } \right)} \right|_0^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{0}\]

E'tibor bering, biz integrallashdan biroz oldin integralni soddalashtirish uchun sinus uchun ikki burchakli formuladan foydalanganmiz. Agar xohlasak, biz ushbu bosqich uchun ikkita almashtirishdan birini amalga oshirishimiz mumkin edi (va biz, albatta, bir xil javobni olamiz).