TICylindricalCoords - 2-masala yechimi | MathModel.uz

15.6-bo'lim : Silindrsimon koordinatalarda uch karrali integrallar

2. $\displaystyle \iiint\limits_{E}{{{{\bf{e}}^{ - {x^{\,2}} - {z^{\,2}}}}\,dV}}$ ni baholang, bunda $E$ ${x^2} + {z^2} = 4$ va ${x^2} + {z^2} = 9$ ikkita tsilindr orasidagi mintaqadir $1 \le y \le 5$ va $z \le 0$.

Yechim

Mayli, keling, $E$ hududining qisqacha chizmasidan boshlaylik, shunda biz nima bilan shug'ullanayotganimizni his qilishimiz mumkin.

Biz bilamizki, ${x^2} + {z^2} = 4$ va ${x^2} + {z^2} = 9$ mos ravishda $y$ o'qida joylashgan 2 va 3 radiusli silindrlardir. $1 \le y \le 5$ diapazoni bizga faqat $y$ ning ushbu diapazonida silindrlarga ega bo'lishini aytadi. Va nihoyat, $z \le 0$ bizga silindrlarning har birining faqat pastki yarmiga ega bo'lishini aytadi.

Mana, $E$ eskizi.

$A three dimensional sketch of the region $E$ drawn with the traditional axes. It is the region between the two cylinders of radius 2 and 3 centered on the y-axis, restricted to $1 \le y \le 5$ and $z \le 0$, so it is a half pipe shaped trough. The orange surface is the outer cylinder, the blue surface is the inner cylinder and the tan ring at the front is the portion of the plane $y = 5$ that caps it. The cap at $y = 1$ is not visible from this angle.$

$A three dimensional sketch of the same region $E$, this time drawn with a set of “box” axes to make the surface and region easier to visualize. The orange and blue surfaces are the outer and inner cylinders and the tan ring caps the front at $y = 5$.$

Biz "an'anaviy" o'qlar to'plami bilan bir qatorda "quti" o'qlari to'plami bilan eskizlarni berdik, bu sirt va mintaqani tasavvur qilishda yordam beradi.

$E$ ning "old" qismi $y = 5$ samolyotining old qismini "yopib turadigan" qismi va eskizdagi to'q sariq halqadir. $E$ ning "orqa" qismi $y = 1$ tekisligining mintaqaning orqa qismini qoplaydigan va eksa tizimining yo'nalishi tufayli eskizda ko'rsatilmagan qismidir.

2-qadam

Shunday qilib, yuqoridagi eskizdan ko'rinib turibdiki, $D$ mintaqasi $xz$ tekisligiga qaytadi va shuning uchun biz avval $y$ bilan integratsiya qilishimiz kerak. Bu holda bu yanada osonroq, chunki yuzalarning old va orqa qismlari mos ravishda $y = 5$ va $y = 1$ tekisliklaridir. Bu shuni anglatadiki, $y$ chegaralari,

\[1 \le y \le 5\]

3-qadam

Keling, ushbu muammo uchun $D$ haqida o'ylab ko'raylik. Agar biz ob'ektga $y$ o'qi bo'ylab qarasak, quyida ko'rsatilgandek 2 va 3 radiusli halqaning pastki yarmini ko'ramiz.

$A sketch of the region $D$ as seen looking at $E$ from along the y-axis. It is the lower half of the ring between the circles of radius 2 and 3 centered at the origin, shaded orange with the inner circle drawn in blue. Note that the x-axis has been drawn with the positive values on the left and the negative values on the right, to match what you would actually see looking along the y-axis.$

E'tibor bering, $x$ o'qi yo'nalishi standart yo'nalishdan $y$ o'qi bo'ylab $E$ga qaraganingizda ko'rgan narsangizga aniqroq mos kelish uchun o'zgartiriladi. Boshqacha qilib aytganda, ijobiy $x$ qiymatlari chap tomonda va salbiy $x$ qiymatlari o'ng tomonda. $x$ o'qining yo'nalishi o'zgarmaydi, biz hali ham $xz$ tekisligida joylashgan halqaning bir qismini ko'ramiz va bu aslida bu muammo uchun $D$.

Chunki $D$ $xz$ tekisligida va u halqaning bir qismi bo‘lib, biz silindrsimon koordinatalarning quyidagi “o‘zgartirilgan” versiyasidan foydalanishimiz kerakligini bildiradi.

\[\begin{align*}x & = r\cos \theta \\ y & = y\\ z & = r\sin \theta \end{align*}\]

Bu, shuningdek, birinchi navbatda $y$ ni integratsiya qilishimiz kerakligi (2-bosqichda aniqlaganimizdek) va silindrsimon koordinatalar bilan integratsiyaning birinchi o'zgaruvchisi har doim "erkin" o'zgaruvchidir (i.e. trig funktsiyalari bilan bog'liq emas).

Shunday qilib, biz halqani $r$ va $\theta $ nuqtai nazaridan osongina tasvirlashimiz mumkin, shuning uchun bu muammoning silindrsimon koordinatalari.

\[\begin{array}{c}\pi \le \theta \le 2\pi \\ \,2 \le r \le 3\\ \,1 \le y \le 5\end{array}\]

4-qadam

ushbu limitlarni integralga kiritish va silindrsimon koordinatalarga aylantirish,

\[\iiint\limits_{E}{{{{\bf{e}}^{ - {x^{\,2}} - {z^{\,2}}}}\,dV}} = \int_{\pi }^{{2\pi }}{{\int_{2}^{3}{{\int_{1}^{5}{{r\,{{\bf{e}}^{ - \,\,{r^{\,2}}}}\,dy}}\,dr}}\,d\theta }}\]

O'zgartirilgan silindrsimon koordinatalarimiz ostida ${x^2} + {z^2} = {r^2}$-ni unutmang, shuningdek, $dV = r\,dy\,dr\,d\theta $-ni unutmang, shuning uchun biz $dV$-ni silindrsimon koordinatalarga o'tkazishda boshqa $r$-ni tanlaymiz (bu $r$ integratsiyasida juda foydali bo'ladi).

5-qadam

OK, endi biz qilishimiz kerak bo'lgan narsa - bu integralni baholash. Mana $y$ integratsiyasi.

\[\begin{align*}\iiint\limits_{E}{{{{\bf{e}}^{ - {x^{\,2}} - {z^{\,2}}}}\,dV}} & = \int_{\pi }^{{2\pi }}{{\int_{2}^{3}{{\left. {r\,{{\bf{e}}^{ - \,\,{r^{\,2}}}}y} \right|_1^5\,dr}}\,d\theta }}\\ & = \int_{\pi }^{{2\pi }}{{\int_{2}^{3}{{4r\,{{\bf{e}}^{ - \,\,{r^{\,2}}}}\,dr}}\,d\theta }}\end{align*}\]

6-qadam

Keling, $r$ integratsiyasini amalga oshiramiz.

\[\begin{align*}\iiint\limits_{E}{{{{\bf{e}}^{ - {x^{\,2}} - {z^{\,2}}}}\,dV}} & = \int_{\pi }^{{2\pi }}{{\left. {\left( { - 2\,{{\bf{e}}^{ - \,\,{r^{\,2}}}}} \right)} \right|_2^3\,d\theta }}\\ & = \int_{\pi }^{{2\pi }}{{2\left( {{{\bf{e}}^{ - 4}} - {{\bf{e}}^{ - 9}}} \right)\,d\theta }}\end{align*}\]

7-qadam

Nihoyat, biz $\theta $ integratsiyasini amalga oshiramiz.

\[\iiint\limits_{E}{{{{\bf{e}}^{ - {x^{\,2}} - {z^{\,2}}}}\,dV}} = \left. {2\left( {{{\bf{e}}^{ - 4}} - {{\bf{e}}^{ - 9}}} \right)\theta} \right|_\pi ^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{2\pi \left( {{{\bf{e}}^{ - 4}} - {{\bf{e}}^{ - 9}}} \right) = 0.1143}}\]

Buning eng qiyin qismi, ehtimol, $E$ eskizi edi. Buni va integral uchun $D$ ni qanday olish kerakligini ko'rganingizdan so'ng, muammoning qolgan qismi juda oddiy edi.