TICylindricalCoords - 4-masala yechimi | MathModel.uz

15.6-bo'lim : Silindrsimon koordinatalarda uch karrali integrallar

4. $x \le 0$ bilan ${x^2} + {y^2} = 3$ silindr ichidagi $z = 6 - x$ ostida, $z = - \sqrt {4{x^2} + 4{y^2}} $ ustidagi mintaqa hajmini aniqlash uchun uch karrali integraldan foydalaning.

Yechim

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

Mana, $E$ eskizi.

$A three dimensional sketch of the region $E$ drawn with the traditional axes. It is the half of the cylinder ${x^2} + {y^2} = 3$ with $x \le 0$, standing upright, with the orange slanted plane $z = 6 - x$ capping it on top and the cone $z = - \sqrt {4{x^2} + 4{y^2}} $ forming a downward pointing cap at the bottom.$

$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 plane $z = 6 - x$ caps the top of the half cylinder and the cone forms the pointed cap at the bottom.$

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.

$z = 6 - x$ samolyoti silindrdagi yuqori "qopqoq" va $z = - \sqrt {4{x^2} + 4{y^2}} $ konus silindrdagi pastki "qopqoq" hisoblanadi. Muammo bayonotining $x \le 0$ qismi tufayli bizda silindrning faqat yarmi bor.

2-qadam

Bu qattiq jismning hajmi quyidagicha ifodalanadi:

\[V = \iiint\limits_{E}{{dV}}\]

3-qadam

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

\[ - \sqrt {4{x^2} + 4{y^2}} \le z \le 6 - x\]

4-qadam

Ushbu muammo uchun $D$ shunchaki ${x^2} + {y^2} \le \sqrt 3 $ diskining $x \le 0$ bilan bir qismidir. Mana, chegaralar bilan yordam berish uchun $D$ ning qisqacha eskizi.

$A sketch of the region $D$. It is the half of the disk of radius $\sqrt 3 $ centered at the origin with $x \le 0$, so its flat edge lies along the y-axis and the semicircular arc bulges away from it. The inside is shaded orange.$

$D$ aniq diskning bir qismi bo'lgani uchun biz silindrsimon koordinatalardan foydalanamiz. Shunday qilib, bu muammoning silindrsimon koordinatalari.

\[\begin{array}{c}\displaystyle \frac{\pi }{2} \le \theta \le \frac{{3\pi }}{2}\\ 0 \le r \le \sqrt 3 \\ - 2r \le z \le 6 - r\cos \theta \end{array}\]

$z$ chegaralarini silindrsimon koordinatalarga aylantirishni unutmang.

5-qadam

ushbu limitlarni integralga kiritish va silindrsimon koordinatalarga aylantirish,

\[V = \iiint\limits_{E}{{\,dV}} = \int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\int_{0}^{{\sqrt 3 }}{{\int_{{ - 2r}}^{{6 - r\cos \theta }}{{r\,dz}}\,dr}}\,d\theta }}\]

Shuni unutmangki, $dV = r\,dx\,dr\,d\theta $ va shuning uchun biz $dV$ ni silindrsimon koordinatalarga o'tkazishda $r$ni olamiz.

6-qadam

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

\[\begin{align*}V & = \int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\int_{0}^{{\sqrt 3 }}{{\left. {\left( {rz} \right)} \right|_{ - 2r}^{6 - r\cos \theta }\,dr}}\,d\theta }}\\ & = \int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\int_{0}^{{\sqrt 3 }}{{6r - {r^2}\cos \theta + 2{r^2}\,dr}}\,d\theta }}\end{align*}\]

7-qadam

Keling, $r$ integratsiyasini amalga oshiramiz.

\[\begin{align*}V & = \int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{\left. {\left( {3{r^2} - \frac{1}{3}{r^3}\cos \theta + \frac{2}{3}{r^3}} \right)} \right|_0^{\sqrt 3 }\,d\theta }}\\ & = \int_{{\frac{\pi }{2}}}^{{\frac{{3\pi }}{2}}}{{9 - \sqrt 3 \cos \theta + 2\sqrt 3 \,d\theta }}\end{align*}\]

8-qadam

Nihoyat, biz $\theta $ integratsiyasini amalga oshiramiz.

\[V = \left. {\left( {9\theta + 2\sqrt 3 \theta - \sqrt 3 \sin \theta } \right)} \right|_{\frac{\pi }{2}}^{\frac{{3\pi }}{2}} = \require{bbox} \bbox[2pt,border:1px solid black]{{2\sqrt 3 + \left( {9 + 2\sqrt 3 } \right)\pi = 42.6212}}\]