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15.6-bo'lim: Silindrsimon koordinatalarda uch karrali integrallar

Ushbu bo'limda biz silindrsimon koordinatalarda to'liq bajarilgan uchlik integrallarni ko'rib chiqmoqchimiz. Eslab qoling silindrsimon koordinatalar haqiqatan ham qutb koordinatalarini uch o'lchamga kengaytirishdan boshqa narsa emas. Quyida silindrsimon koordinatalarni aylantirish formulalari keltirilgan.

\[x = r\cos \theta \hspace{0.25in}y = r\sin \theta \hspace{0.25in}z = z\]

Silindrsimon koordinatalarda integralni bajarish uchun biz \(dV\) silindrsimon koordinatalar nuqtai nazaridan qanday bo'lishini bilishimiz kerak. Biz ko'rsata olamiz O'zgaruvchilarning o'zgarishi ushbu bobning bo'limi,

Silindrsimon koordinatalarda \(dV\)

\[dV = r\;dz\,dr\,d\theta \]

Biz birlashayotgan \(E\) mintaqasi,

\[\begin{align*}E & = \left\{ {\left( {x,y,z} \right)|\left( {x,y} \right) \in D,\,\,\,{u_1}\left( {x,y} \right) \le z \le {u_2}\left( {x,y} \right)} \right\}\\ & = \left\{ {\left( {r,\theta ,z} \right)|\alpha \le \theta \le \beta ,\,\,{h_1}\left( \theta \right) \le r \le {h_2}\left( \theta \right),\,\,\,{u_1}\left( {r\cos \theta ,r\sin \theta } \right) \le z \le {u_2}\left( {r\cos \theta ,r\sin \theta } \right)} \right\}\end{align*}\]

E'tibor bering, biz buni faqat \(D\) \(xy\) tekisligida joylashgan \(E\) uchun berganmiz. Agar \(D\) \(yz\) tekisligida yoki kerak bo'lganda \(xz\) tekislikda bo'lsa, biz buni mos ravishda o'zgartirishimiz mumkin.

Silindrsimon koordinatalar nuqtai nazaridan uch karrali integral,

Silindrsimon koordinatalarda uch karrali integrallar

\[\iiint\limits_{E}{{f\left( {x,y,z} \right)\,dV}} = \int_{{\,\alpha }}^{{\,\beta }}{{\int_{{{h_1}\left( \theta \right)}}^{{{h_2}\left( \theta \right)}}{{\int_{{{u_1}\left( {r\cos \theta ,r\sin \theta } \right)}}^{{{u_2}\left( {r\cos \theta ,r\sin \theta } \right)}}{{r\,f\left( {r\cos \theta ,r\sin \theta ,z} \right)\,dz}}\,dr}}\,d\theta }}\]

\(r\)-ga qo'shishni unutmang va barcha \(x\) va \(y\)-lar ham silindrsimon koordinatalarga aylantirilishiga ishonch hosil qiling.

Keling, misolni ko'rib chiqaylik.

1-misol \( \displaystyle \iiint\limits_{E}{{y\,dV}}\) ni baholang, bu erda \(E\) \(z = x + 2\) tekisligi ostida \(xy\) tekisligi ustidagi va \({x^2} + {y^2} = 1\) va \({x^2} + {y^2} = 4\) tsilindrlari orasida joylashgan mintaqadir.
Yechim

Haqiqatan ham, konvertatsiya qilish va keyin integralni baholashdan tashqari, bu bilan juda ko'p ish yo'q.

Biz \(z\) diapazonini silindrsimon koordinatalar bo'yicha olishdan boshlaymiz.

\[0 \le z \le x + 2\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}0 \le z \le r\cos \theta + 2\]

Esda tutingki, biz \(xy\) tekisligidan yuqoridamiz va shuning uchun biz \(z = 0\) tekisligidan yuqoridamiz.

Keyinchalik, \(D\) mintaqasi \(xy\) tekisligidagi \({x^2} + {y^2} = 1\) va \({x^2} + {y^2} = 4\) ikkita doiralari orasidagi mintaqadir va shuning uchun uning diapazonlari:

\[0 \le \theta \le 2\pi \hspace{0.25in}\hspace{0.25in}1 \le r \le 2\]

Mana integral.

\[\begin{align*}\iiint\limits_{E}{{y\,dV}} & = \int_{0}^{{2\pi }}{{\int_{1}^{2}{{\int_{0}^{{r\cos \theta + 2}}{{\left( {r\sin \theta } \right)r\,dz}}\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{1}^{2}{{{r^2}\sin \theta \left( {r\cos \theta + 2} \right)\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{1}^{2}{{\frac{1}{2}{r^3}\sin \left( {2\theta } \right) + 2{r^2}\sin \theta \,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\left. {\left( {\frac{1}{8}{r^4}\sin \left( {2\theta } \right) + \frac{2}{3}{r^3}\sin \theta } \right)} \right|_1^2\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\frac{{15}}{8}\sin \left( {2\theta } \right) + \frac{{14}}{3}\sin \theta \,d\theta }}\\ & = \left. {\left( { - \frac{{15}}{{16}}\cos \left( {2\theta } \right) - \frac{{14}}{3}\cos \theta } \right)} \right|_0^{2\pi }\\ & = 0\end{align*}\]

Xuddi qutbli koordinatalarni o'z ichiga olgan qo'sh integral bilan qilganimiz kabi, biz \(x\), \(y\) va \(z\) nuqtai nazaridan takrorlangan integraldan boshlashimiz va uni silindrsimon koordinatalarga aylantirishimiz mumkin.

2-misol \(\displaystyle \int_{{\, - 1}}^{{\,1}}{{\int_{{\,0}}^{{\,\sqrt {1 - {y^2}} }}{{\int_{{\,{x^2} + {y^2}}}^{{\,\sqrt {{x^2} + {y^2}} }}{{xyz\,dz}}\,dx}}\,dy}}\) ni silindrsimon koordinatalarda integralga aylantiring.
Yechim

Mana bu takrorlangan integraldan o'zgaruvchilar diapazonlari.

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

Birinchi ikkita tengsizlik \(D\) mintaqasini belgilaydi va \(x\) ning yuqori va pastki chegaralari \(x = \sqrt {1 - {y^2}} \) va \(x = 0\) bo‘lgani uchun biz bosh markazida joylashgan 1 radiusli o‘ng yarim doiraning kamida bir qismini olganligini bilamiz. \(y\) diapazoni \( - 1 \le y \le 1\) bo'lgani uchun biz 1 radiusli diskning to'liq o'ng yarmiga ega ekanligini bilamiz. Shunday qilib, silindrsimon koordinatalarda \(D\) diapazonlari:

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

Endi faqat \(z\) diapazonining chegaralarini o'zgartirish qoladi, ammo bu juda yomon emas.

\[{r^2} \le z \le r\]

Yon eslatmada e'tibor bering, bu erda pastki chegara elliptik paraboloid va yuqori chegara konusdir. Shuning uchun \(E\) bu ikki sirt orasidagi hududning bir qismidir.

Integral bu,

\[\begin{align*}\int_{{\, - 1}}^{{\,1}}{{\int_{{\,0}}^{{\,\sqrt {1 - {y^2}} }}{{\int_{{\,{x^2} + {y^2}}}^{{\,\sqrt {{x^2} + {y^2}} }}{{xyz\,dz}}\,dx}}\,dy}} & = \int_{{\, - {\pi }/{2}\;}}^{{\,{\pi }/{2}\;}}{{\int_{{\,0}}^{{\,1}}{{\int_{{\,{r^2}}}^{{\,r}}{{r\left( {r\cos \theta } \right)\left( {r\sin \theta } \right)z\,dz}}\,dr}}\,d\theta }}\\ & = \int_{{\, - {\pi }/{2}\;}}^{{\,{\pi }/{2}\;}}{{\int_{{\,0}}^{{\,1}}{{\int_{{\,{r^2}}}^{{\,r}}{{z{r^3}\cos \theta \sin \theta \,dz}}\,dr}}\,d\theta }}\end{align*}\]