TripleIntegrals | MathModel.uz

15.5-bo'lim: Uch karrali integrallar

Endi biz ikki o'lchovli mintaqada qanday qilib integratsiya qilishni bilganimizdan so'ng, biz uch o'lchovli mintaqada integratsiyaga o'tishimiz kerak. Biz ikki o'lchovli hududni integrallash uchun qo'shaloq integraldan foydalandik va shuning uchun biz uni ishlatishimiz ajablanarli emas. uch karra integral uch o'lchovli mintaqada integratsiya qilish. Umumiy uch karrali integrallarning yozuvi:

\[\iiint\limits_{E}{{f\left( {x,y,z} \right)\,dV}}\]

Keling, qutiga integratsiya qilishdan boshlaylik,

\[B = \left[ {a,b} \right] \times \left[ {c,d} \right] \times \left[ {r,s} \right]\]

E'tibor bering, ushbu belgidan foydalanganda biz $x$ ning birinchisini, $y$ ning ikkinchisini va $z$ ning uchinchisini keltiramiz.

Bu holda uch karrali integral,

\[\iiint\limits_{B}{{f\left( {x,y,z} \right)\,dV}} = \int_{{\,r}}^{{\,s}}{{\int_{{\,c}}^{{\,d}}{{\int_{{\,a}}^{{\,b}}{{f\left( {x,y,z} \right)dx}}\,dy}}\,dz}}\]

E'tibor bering, biz avval $x$, so'ngra $y$ va nihoyat $z$ ni birlashtirdik, lekin aslida bu tartibda integrallar uchun hech qanday sabab yo'q. Integralni bajarish uchun 6 xil mumkin bo'lgan buyruqlar mavjud va siz integralni qaysi tartibda bajarishingiz funktsiyaga bog'liq bo'ladi va siz o'ylagan tartib eng oson bo'ladi. Buyurtmadan qat'i nazar, biz bir xil javobni olamiz.

Keling, ushbu turdagi uchlik integralga tezkor misol keltiraylik.

1-misol Quyidagi integralni baholang. \[\iiint\limits_{B}{{8xyz\,dV}} \hspace{0.5in} B = \left[ {2,3} \right] \times \left[ {1,2} \right] \times \left[ {0,1} \right]\]

Yechim

Buyurtma muhim emasligini ta'kidlash uchun, keling, yuqorida sanab o'tilgandan boshqa tartibni ishlataylik. Biz integralni quyidagi tartibda bajaramiz.

\[\begin{align*}\iiint\limits_{B}{{8xyz\,dV}} & = \int_{{\,1}}^{{\,2}}{{\int_{{\,2}}^{{\,3}}{{\int_{{\,0}}^{1}{{8xyz\,dz}}\,dx}}\,dy}}\\ & = \int_{{\,1}}^{{\,2}}{{\int_{{\,2}}^{{\,3}}{{\left. {4xy{z^2}} \right|_0^1\,dx}}\,dy}}\\ & = \int_{{\,1}}^{{\,2}}{{\int_{{\,2}}^{{\,3}}{{4xy\,dx}}\,dy}}\\ & = \int_{{\,1}}^{{\,2}}{{\left. {2{x^2}y} \right|_2^3\,dy}}\\ & = \int_{{\,1}}^{{\,2}}{{10y\,dy}} = 15\end{align*}\]

Ko'proq umumiy hududlarga o'tishdan oldin, keling, uchlik integral haqida chiroyli geometrik talqinni olaylik, shunda biz undan keyingi misollarda foydalanishimiz mumkin.

Uch karra integral sifatida mintaqa hajmi

$E$ uch o'lchovli hududining hajmi integral bilan berilgan,

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

Keling, umumiy uch o'lchovli hududlarga o'taylik. Umumiy mintaqa uchun bizda uch xil imkoniyat bor. Bu erda birinchi imkoniyatning eskizi.

$This is a sketch of a generic solid that is defined in such a way that the top of the solid can always be given by a function defined as $z=u_{2}\left(x,y\right)$ and that the bottom of the solid can always be given by a function defined as $z=u_{1}\left(x,y\right)$. The sides of the solid is just a vertical surface that connects the top and bottom of the surface. For this this particular graph the top and bottom are shown as basically flat surfaces but they do not need to be flat. Below the solid is a general region in the xy-plane denoted as D. This is the region that the points (x,y) come from to sketch the top/bottom surface of the solid.$

Bunday holda biz $E$ mintaqasini quyidagicha aniqlaymiz,

\[E = \left\{ {\left( {x,y,z} \right)|\left( {x,y} \right) \in D,\,\,\,{u_1}\left( {x,y} \right) \le z \le {u_2}\left( {x,y} \right)} \right\}\]

bu erda $\left( {x,y} \right) \in D$ - bu $\left( {x,y} \right)$ nuqtasi $xy$ tekisligidan $D$ hududida joylashganligini bildiruvchi belgi. Bu holda biz uch karrali integralni quyidagicha baholaymiz:

\[\iiint\limits_{E}{{f\left( {x,y,z} \right)\,dV}} = \iint\limits_{D}{{\left[ {\int_{{\,{u_1}\left( {x,y} \right)}}^{{\,{u_2}\left( {x,y} \right)}}{{f\left( {x,y,z} \right)\,dz}}} \right]\,dA}}\]

bu erda qo'sh integralni biz oldingi ikki bo'limda ko'rgan usullarning har qandayida baholash mumkin. Boshqacha qilib aytadigan bo'lsak, biz $x$ ga nisbatan birinchi bo'lib integratsiya qilishimiz mumkin, biz $y$ ga nisbatan avval integratsiya qilishimiz yoki kerak bo'lganda qutb koordinatalaridan foydalanishimiz mumkin.

2-misol $\displaystyle \iiint\limits_{E}{{2x\,dV}}$ ni baholang, bu erda $E$ birinchi oktantda joylashgan $2x + 3y + z = 6$ tekisligi ostidagi mintaqadir.

Yechim

Biz birinchi navbatda aniqlashimiz kerak oktant. Ikki o'lchovli koordinatalar tizimini to'rtta kvadrantga bo'lish mumkin bo'lgani kabi, uch o'lchovli koordinatalar tizimini sakkiz oktantga bo'lish mumkin. Birinchi oktant bu oktant bo'lib, unda uchta koordinatasi ham ijobiy bo'ladi.

Mana birinchi oktantdagi samolyotning eskizi.

Fazoviy koordinatalar sistemasi va egri chiziq grafigi

This is a boxed 3D coordinate system. The z-axis is right vertical edge of the box, the x-axis is the top right edge of the box and the y-axis is the bottom front edge of the box. This is a sketch of the portion of the plane that is in the 1st octant. It is a triangle with vertices at (0,0,6), (3,0,0) and (0,2,0).

Endi $xy$ tekisligida $D$ mintaqasini aniqlashimiz kerak. Biz yuqoridan ob'ektga to'g'ridan-to'g'ri qaragandek qilib, mintaqaning vizualizatsiyasini olishimiz mumkin. Biz ko'rgan narsa $xy$ tekisligidagi $D$ mintaqasi bo'ladi. Shunday qilib, $D$ uchlari $\left( {0,0} \right)$, $\left( {3,0} \right)$ va $\left( {0,2} \right)$ bo'lgan uchburchak bo'ladi. Mana $D$ eskizi.

Endi bizga integratsiya chegaralari kerak. Samolyot ostida va birinchi oktantda bo'lganimiz sababli (biz $z = 0$ tekisligidan yuqoridamiz) $z$ uchun quyidagi cheklovlar mavjud.

\[0 \le z \le 6 - 2x - 3y\]

Quyidagi ikkita tengsizliklar to‘plamidan birini qo‘llagan holda $D$ bo‘yicha qo‘sh integralni integrallashimiz mumkin.

\[\begin{matrix} \begin{aligned} & \,\,\,\,\,\,0\le x\le 3 \\ & 0\le y\le -\frac{2}{3}x+2 \\ \end{aligned} & \hspace{0.5in} & \begin{aligned} & 0\le x\le -\frac{3}{2}y+3 \\ & \,\,\,\,\,\,0\le y\le 2 \\ \end{aligned} \\ \end{matrix}\]

Hech birining boshqasidan ustunligi yo'qligi sababli biz birinchisidan foydalanamiz. U holda integral,

\[\begin{align*}\iiint\limits_{E}{{2x\,dV}} &= \iint\limits_{D}{{\left[ {\int_{{\,0}}^{{6 - 2x - 3y}}{{2x\,dz}}} \right]\,dA}}\\ & = \iint\limits_{D}{{\left. {2xz} \right|_0^{6 - 2x - 3y}\,dA}}\\ & = \int_{{\,0}}^{{\,3}}{{\int_{0}^{{ - \frac{2}{3}x + 2}}{{2x\left( {6 - 2x - 3y} \right)\,dy}}\,dx}}\\ & = \int_{0}^{3}{{\left. {\left( {12xy - 4{x^2}y - 3x{y^2}} \right)} \right|_0^{ - \frac{2}{3}x + 2}\,dx}}\\ & = \int_{0}^{3}{{\frac{4}{3}{x^3} - 8{x^2} + 12x\,dx}}\\ & = \left. {\left( {\frac{1}{3}{x^4} - \frac{8}{3}{x^3} + 6{x^2}} \right)} \right|_0^3\\ & = 9\end{align*}\]

Keling, uchlik integrallar uchun duch kelishimiz mumkin bo'lgan ikkinchi uch o'lchovli mintaqaga o'tamiz. Mana bu hududning eskizi.

Ushbu imkoniyat uchun biz $E$ mintaqasini quyidagicha aniqlaymiz,

\[E = \left\{ {\left( {x,y,z} \right)|\left( {y,z} \right) \in D,\,\,\,{u_1}\left( {y,z} \right) \le x \le {u_2}\left( {y,z} \right)} \right\}\]

Shunday qilib, $D$ mintaqasi $yz$ tekisligidagi mintaqa bo'ladi. Mana bu integrallarni qanday baholaymiz.

\[\iiint\limits_{E}{{f\left( {x,y,z} \right)\,dV}} = \iint\limits_{D}{{\left[ {\int_{{\,{u_1}\left( {y,z} \right)}}^{{\,{u_2}\left( {y,z} \right)}}{{f\left( {x,y,z} \right)\,dx}}} \right]\,dA}}\]

Birinchi imkoniyatda bo'lgani kabi bizda $yz$ tekisligida qo'sh integralni bajarish uchun ikkita variant, shuningdek, kerak bo'lganda qutb koordinatalaridan foydalanish imkoniyati mavjud.

3-misol $x + y + z = 8$ tekisligi orqasida va $\displaystyle z = \frac{3}{2}\,\,\sqrt y $ va $\displaystyle z = \frac{3}{4}y$ bilan chegaralangan $yz$-tekisligida mintaqa oldida joylashgan hudud hajmini aniqlang.

Yechim

Bu holatda bizga $D$ berildi va shuning uchun biz buni topish uchun chindan ham ishlashimiz shart emas. Bu erda $D$ hududining eskizi, shuningdek, samolyotning qisqacha eskizi va samolyotning yonidan prognoz qilingan $D$ ni aniqlovchi egri chiziqlari, shuning uchun biz ishlayotgan mintaqa qanday ko'rinishi haqida tasavvurga ega bo'lishimiz mumkin.

$This is the 2D graph on a yz-axis system with domain 0<y<4 of $z=\frac{3}{2} \sqrt{y}$ and $z=\frac{3}{4}y$. In this domain the graph of $z=\frac{3}{2} \sqrt{y}$ is always larger than the graph of $z=\frac{3}{4}y$. The area between the two functions has been shaded in.$ Fazoviy koordinatalar sistemasi va egri chiziq grafigi

Endi yuqoridagi mintaqaning grafigi hammasi yaxshi, lekin u bizga mintaqa nima ekanligini ko'rsatmaydi. Shunday qilib, bu erda mintaqaning o'zi eskizi.

Bu erda har bir o'zgaruvchi uchun chegaralar mavjud.

\[\begin{array}{c}0 \le y \le 4\\ \displaystyle \frac{3}{4}y \le z \le \frac{3}{2}\sqrt y \\ 0 \le x \le 8 - y - z\end{array}\]

Keyin hajm,

\[\begin{align*}V &= \iiint\limits_{E}{{\,dV}} = \iint\limits_{D}{{\left[ {\int_{0}^{{8 - y - z}}{{\,dx}}} \right]\,dA}}\\ & = \int_{0}^{4}{{\int_{{{{3y}}/{4}\;}}^{{{{3\sqrt y }}/{2}\;}}{{8 - y - z\,dz}}\,dy}}\\ & = \int_{0}^{4}{{\left. {\left( {8z - yz - \frac{1}{2}{z^2}} \right)} \right|_{\frac{{3y}}{4}}^{\frac{{3\sqrt y }}{2}}\,dy}}\\ & = \int_{0}^{4}{{12{y^{\frac{1}{2}}} - \frac{{57}}{8}y - \frac{3}{2}{y^{\frac{3}{2}}} + \frac{{33}}{{32}}{y^2}\,dy}}\\ & = \left. {\left( {8{y^{\frac{3}{2}}} - \frac{{57}}{{16}}{y^2} - \frac{3}{5}{y^{\frac{5}{2}}} + \frac{{11}}{{32}}{y^3}} \right)} \right|_0^4 = \frac{{49}}{5}\end{align*}\]

Endi biz uch karrali integrallar uchun uch o'lchamli mumkin bo'lgan uchinchi (va oxirgi) mintaqani ko'rib chiqishimiz kerak. Mana bu hududning eskizi.

Ushbu yakuniy holatda $E$ quyidagicha aniqlanadi:

\[E = \left\{ {\left( {x,y,z} \right)|\left( {x,z} \right) \in D,\,\,\,{u_1}\left( {x,z} \right) \le y \le {u_2}\left( {x,z} \right)} \right\}\]

va bu erda $D$ mintaqasi $xz$ tekisligidagi mintaqa bo'ladi. Mana bu integrallarni qanday baholaymiz.

\[\iiint\limits_{E}{{f\left( {x,y,z} \right)\,dV}} = \iint\limits_{D}{{\left[ {\int_{{\,{u_1}\left( {x,z} \right)}}^{{\,{u_2}\left( {x,z} \right)}}{{f\left( {x,y,z} \right)\,dy}}} \right]\,dA}}\]

Bu erda biz $D$ ni $xz$ tekisligiga integratsiya qilish uchun ikkita mumkin bo'lgan buyurtmalardan birini ishlatishimiz mumkin yoki kerak bo'lganda qutb koordinatalaridan foydalanishimiz mumkin.

4-misol $\displaystyle \iiint\limits_{E}{{\sqrt {3{x^2} + 3{z^2}} \,dV}}$ ni baholang, bunda $E$ $y = 2{x^2} + 2{z^2}$ va $y = 8$ tekisligi bilan chegaralangan qattiq jismdir.

Yechim

Mana qattiq $E$ ning eskizi.

This is a boxed 3D coordinate system. The z-axis is left vertical edge of the box, the x-axis is the bottom right edge of the box and the y-axis is the top front edge of the box. This is the sketch of the vaguely cup shaped solid from the elliptic paraboloid given in the problem statement. It starts at the origin, is centered on the y-axis and opens up in the positive y direction. The solid is capped at y=8 by a disk.

$xz$ tekisligidagi $D$ mintaqasini ushbu qattiq jismning oldida “tikish” orqali topish mumkin va biz $D$ $xz$ tekisligida disk bo‘lishini ko‘rishimiz mumkin. Bu disk qattiq jismning old qismidan keladi va biz elliptik paraboloid va tekislikni teng qilib, disk tenglamasini aniqlashimiz mumkin.

\[2{x^2} + 2{z^2} = 8\hspace{0.25in} \Rightarrow \hspace{0.25in}{x^2} + {z^2} = 4\]

Bu mintaqa, shuningdek, integral, ikkalasi ham qutb koordinatalari kabi narsalarni ishlatishimiz kerakligini taklif qiladi. Biroq, biz $xz$ tekisligidamiz va biz faqat $xy$ tekisligida qutb koordinatalarini ko'rdik. Bu muammo emas. Biz ularni har doim quyidagi ta'rif bilan $xz$ tekisligiga "tarjima qilishimiz" mumkin.

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

Mintaqada $y$ mavjud emasligi sababli biz $z$ ga barcha formulalarda $y$ o'rnini egallashiga ruxsat beramiz. E'tibor bering, bu ta'riflar formulaga ham olib keladi,

\[{x^2} + {z^2} = {r^2}\]

Buning yordamida biz ushbu integral uchun kerak bo'ladigan o'zgaruvchilar chegaralariga erishishimiz mumkin.

\[\begin{array}{c}2{x^2} + 2{z^2} \le y \le 8\\ 0 \le r \le 2\\ 0 \le \theta \le 2\pi \end{array}\]

U holda integral,

\[\begin{align*}\iiint\limits_{E}{{\sqrt {3{x^2} + 3{z^2}} \,dV}} & = \iint\limits_{D}{{\left[ {\int_{{2{x^2} + 2{z^2}}}^{{\,8}}{{\sqrt {3{x^2} + 3{z^2}} \,dy}}} \right]\,dA}}\\ & = \iint\limits_{D}{{\left. {\left( {y\sqrt {3{x^2} + 3{z^2}} } \right)} \right|_{2{x^2} + 2{z^2}}^8\,dA}}\\ & = \iint\limits_{D}{{\sqrt {3\left( {{x^2} + {z^2}} \right)} \left( {8 - \left( {2{x^2} + 2{z^2}} \right)} \right)\,dA}}\end{align*}\]

Endi biz qutb koordinatalarida qo'sh integralni bajarmoqchi bo'lganimiz sababli, keling, hamma narsani qutb koordinatalariga aylantiramiz. Integral bu,

\[\begin{align*}\sqrt {3\left( {{x^2} + {z^2}} \right)} \left( {8 - \left( {2{x^2} + 2{z^2}} \right)} \right) &= \sqrt {3{r^2}} \left( {8 - 2{r^2}} \right)\\ & = \sqrt 3 \,\,r\left( {8 - 2{r^2}} \right)\\ & = \sqrt 3 \left( {8r - 2{r^3}} \right)\end{align*}\]

U holda integral,

\[\begin{align*}\iiint\limits_{E}{{\sqrt {3{x^2} + 3{z^2}} \,dV}} & = \iint\limits_{D}{{\sqrt 3 \,\,\left( {8r - 2{r^3}} \right)dA}}\\ & = \sqrt 3 \int_{{\,0}}^{{\,2\pi }}{{\int_{{\,0}}^{{\,2}}{{\left( {8r - 2{r^3}} \right)r\,dr}}\,d\theta }}\\ & = \sqrt 3 \int_{{\,0}}^{{\,2\pi }}{{\left. {\left( {\frac{8}{3}{r^3} - \frac{2}{5}{r^5}} \right)} \right|_0^2\,d\theta }}\\ & = \sqrt 3 \int_{{\,0}}^{{\,2\pi }}{{\frac{{128}}{{15}}\,d\theta }}\\ & = \frac{{256\sqrt 3 \,\pi }}{{15}}\end{align*}\]