TICylindricalCoords - 3-masala yechimi | MathModel.uz

15.6-bo'lim : Silindrsimon koordinatalarda uch karrali integrallar

3. $\displaystyle \iiint\limits_{E}{{z\,dV}}$ ni baholang, bunda $E$ $x + y + z = 2$ va $x = 0$ ikkita tekislik orasidagi va ${y^2} + {z^2} = 1$ silindrining ichidagi mintaqadir.

Yechim

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

Tsilindr va $x + y + z = 2$ samolyotining qisqacha chizmasi bilan boshlaylik.

$A three dimensional sketch of the cylinder ${y^2} + {z^2} = 1$, drawn in green, together with the red plane $x + y + z = 2$ that slices through it. More of each surface is shown than is strictly needed, so as to make the relationship between the two clear. The plane $x = 0$ is the yz-plane and caps the back of the cylinder, and it is not shown.$

Biz bu yerda ishlayotgan $E$ hududi silindr ichida va muammo bayonida keltirilgan ikkita tekislik o‘rtasida joylashgan. Buning ma'nosi shundaki, ikkita samolyot silindrni yopadi.

Yuqoridagi grafikda biz ikkita sirt o'rtasidagi munosabatni ko'rsatish uchun kerak bo'lganidan ko'proq silindrni va keraklidan ko'proq $x + y + z = 2$ tekisligini joylashtirdik. $x = 0$ samolyoti shunchaki $yz$ tekisligi bo'lib, silindrning orqa qismini "qopqoqlar" bilan qoplaydi va shuning uchun eskizga kiritilmagan.

Endi silindrning $x + y + z = 2$ oldidagi qismidan xalos bo'laylik, chunki u mintaqaning bir qismi emas va $x + y + z = 2$ ning silindr ichida bo'lmagan qismidan xalos bo'laylik. Olingan eskiz $E$ hududidir.

Mana, $E$ eskizi.

$A three dimensional sketch of the region $E$ drawn with the traditional axes. The portion of the cylinder in front of the plane and the portion of the plane outside the cylinder have both been removed, leaving the green tube capped by the red slanted elliptical face of the plane $x + y + z = 2$.$

$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 green tube is the cylinder ${y^2} + {z^2} = 1$ and the red slanted elliptical face is the plane $x + y + z = 2$ capping it.$

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.

2-qadam

Shunday qilib, eskizdan umid qilamanki, $D$ mintaqasi $yz$ tekisligida bo'ladi va shuning uchun biz avval $x$ bilan integratsiya qilishimiz kerak. Bu shuni anglatadiki, $x$ chegaralari,

\[0 \le x \le 2 - y - z\]

Yuqori $x$ chegarasi uchun biz faqat $x$ uchun tekislik tenglamasini echishimiz kerak.

3-qadam

Ushbu muammo uchun $D$ oddiygina ${y^2} + {z^2} \le 1$ diskidir. $D$ $yz$ tekisligida va disk bo'lgani uchun biz silindrsimon koordinatalarning quyidagi "o'zgartirilgan" versiyasidan foydalanishimiz kerak bo'ladi.

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

Bu, shuningdek, birinchi navbatda $x$ ni integratsiyalashimiz 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 diskni $r$ va $\theta $ nuqtai nazaridan osongina tasvirlashimiz mumkin, shuning uchun bu muammoning silindrsimon koordinatalari.

\[\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le 1\\ 0 \le x \le 2 - r\sin \theta - r\cos \theta \end{array}\]

$x$ yuqori chegarasidagi $y$ va $z$ ni silindrsimon koordinataga aylantirishni ham unutmang.

4-qadam

ushbu limitlarni integralga kiritish va silindrsimon koordinatalarga aylantirish,

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

Integrandni o'zgartirilgan silindrsimon koordinatalarimizga aylantirishni unutmang, shuningdek, $dV = r\,dx\,dr\,d\theta $ ni unutmang, shuning uchun biz $dV$ ni silindrsimon koordinatalarga o'tkazishda boshqa $r$ ni tanlaymiz.

5-qadam

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

\[\begin{align*}\iiint\limits_{E}{{z\,dV}} & = \int_{0}^{{2\pi }}{{\int_{0}^{1}{{\left. {\left( {{r^2}\cos \theta \,x} \right)} \right|_0^{2 - r\sin \theta - r\cos \theta }\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{1}{{{r^2}\cos \theta \,\left( {2 - r\sin \theta - r\cos \theta } \right)\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{1}{{2{r^2}\cos \theta - {r^3}\cos \theta \sin \theta - {r^3}{{\cos }^2}\theta \,dr}}\,d\theta }}\end{align*}\]

6-qadam

Keyin $r$ integratsiyasini amalga oshiramiz.

\[\begin{align*}\iiint\limits_{E}{{z\,dV}} & = \int_{0}^{{2\pi }}{{\left. {\left( {\frac{2}{3}{r^3}\cos \theta - \frac{1}{4}{r^4}\cos \theta \sin \theta - \frac{1}{4}{r^4}{{\cos }^2}\theta } \right)} \right|_0^1\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\frac{2}{3}\cos \theta - \frac{1}{4}\cos \theta \sin \theta - \frac{1}{4}{{\cos }^2}\theta \,d\theta }}\end{align*}\]

7-qadam

Va nihoyat, biz $\theta $ integratsiyasini amalga oshiramiz.

\[\begin{align*}\iiint\limits_{E}{{z\,dV}} & = \int_{0}^{{2\pi }}{{\frac{2}{3}\cos \theta - \frac{1}{8}\sin \left( {2\theta } \right) - \frac{1}{8}\left( {1 + \cos \left( {2\theta } \right)} \right)\,d\theta }}\\ & = \left. {\left( {\frac{2}{3}\sin \theta + \frac{1}{{16}}\cos \left( {2\theta } \right) - \frac{1}{8}\left( {\theta + \frac{1}{2}\sin \left( {2\theta } \right)} \right)} \right)} \right|_0^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{\pi }{4}}}\end{align*}\]

Yakuniy integratsiyani amalga oshirishdan oldin integralni soddalashtirishni unutmang. Bu holda biz ikkinchi hadda sinus ikki burchakli va uchinchi hadda kosinus yarim burchak formulasidan foydalanib, bu integrasiyani tez bajarishimiz uchun integrandni soddalashtirdik.