TripleIntegrals - 5-masala yechimi | MathModel.uz

15.5-bo'lim : Uch karrali integrallar

5. $ \displaystyle \iiint\limits_{E}{{12y - 8x\,dV}}$ ni baholang, bunda $E$ $y = 10 - 2z$ orqasidagi va $xz$ tekisligidagi mintaqa oldidagi $z = 2x$, $z = 5$ va $x = 0$ bilan chegaralangan.

Yechim

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

$A three dimensional sketch of the region $E$ drawn with the traditional axes. The yellow surface across the top is the portion of the plane $y = 10 - 2z$ that lies in front of the region in the $xz$-plane, and the dark red surface below it is the side of the solid that is visible from this angle. The solid is widest where it meets the $xz$-plane, in which $z$ runs up to 5, and it narrows to a point out at $y = 10$ on the $y$-axis.$

$A three dimensional sketch of the same wedge shaped region, this time drawn with a set of “box” axes to make the surfaces easier to visualize. The yellow surface is the plane $y = 10 - 2z$ and the dark red surface below it is the visible side of the solid, which tapers to a point at $y = 10$.$

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.

Mintaqaning yuqori qismi (sariq rangli sirt) muammo bayonotida berilgan $xz$ tekisligi mintaqasi oldida joylashgan $y = 10 - 2z$ qismining grafigi.

Qizil sirt $z = 2x$ tomonidan aniqlangan tekislikdir. Bu $xz$-tekisligini $z = 2x$ chizig'ida kesib o'tuvchi va yuqorida $y = 10 - 2z$ yuzasi bilan chegaralangan tekislikdir.

2-qadam

$xz$ tekisligidagi $z = 2x$, $z = 5$ va $x = 0$ bilan chegaralangan, muammo bayonida havola qilingan mintaqa $D$ hududidir va shuning uchun biz bilamizki, bu muammoni hal qilish uchun avval $y$ ni birlashtirishimiz kerak (chunki MATH36X da $xz$-samolyot).

Shuning uchun bizda $y$ uchun quyidagi limitlar mavjud.

\[0 \le y \le 10 - 2z\]

ushbu limitlar bilan biz uch karra integralni ham hech bo'lmaganda qisman quyidagicha o'rnatishimiz mumkin.

\[\iiint\limits_{E}{{12y - 8x\,dV}} = \iint\limits_{D}{{\left[ {\int_{0}^{{10 - 2z}}{{12y - 8x\,dy}}} \right]\,dA}}\]

3-qadam

Keyinchalik, biz integralni sozlashni tugatishimiz uchun $D$ uchun limitlar kerak bo'ladi. Ushbu muammo uchun $D$ yuqorida ta'kidlaganimizdek, muammo bayonotidagi $xz$ tekisligidagi mintaqadir.

Mana $D$ ning qisqacha eskizi.

$A sketch of the two dimensional region $D$ in the $xz$-plane. It is the shaded triangle bounded above by the red horizontal line labeled $z = 5$, bounded on the right by the blue line labeled $z = 2x$, or $x = \frac{1}{2}z$, and bounded on the left by the $z$-axis, i.e. $x = 0$. The triangle has vertices at the origin, $\left( {0,5} \right)$ and $\left( {\frac{5}{2},5} \right)$.$

Bu mintaqani o'ylashning yana bir usuli shundaki, u 1-bosqichda chizilgan qattiq jismning "orqasi" va biz 1-bosqichda chizilgan qattiq narsa bu mintaqaning orqasida. Boshqacha qilib aytganda, musbat y o'qi to'g'ridan-to'g'ri sahifaga qaytadi, salbiy $y$ o'qi esa to'g'ridan-to'g'ri sahifadan chiqadi.

$D$ eskizida ta'kidlanganidek, biz quyidagi chegaralar to'plamidan birini osongina aniqlashimiz mumkin.

\[\begin{array}{*{20}{c}}\begin{aligned}& \,\,\,0 \le x \le \frac{5}{2}\\ & 2x \le z \le 5\end{aligned}&{\,\,\,\,\,\,\,\,\,\,\,{\mbox{OR}}\,\,\,\,\,\,\,\,\,\,}&\begin{aligned} & \,\, 0 \le z \le 5\\ & 0 \le x \le \frac{1}{2}z\end{aligned}\end{array}\]

Integrand ulardan biri boshqasidan osonroq bo'lishini anglatmaydi, shuning uchun biz 2 dan foydalanamiz^nd Boshqa sabablarga ko'ra chegaralar to'plami ikkalasining pastki chegaralari nolga teng bo'lib, integratsiya jarayonida narsalarni biroz chiroyli qilishi mumkin.

ushbu limitlar bizda mavjud integralga ulangan holda,

\[\iiint\limits_{E}{{12y - 8x\,dV}} = \int_{0}^{5}{{\int_{0}^{{\frac{1}{2}z}}{{\int_{0}^{{10 - 2z}}{{12y - 8x\,dy}}\,dx}}\,dz}}\]

4-qadam

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

\[\begin{align*}\iiint\limits_{E}{{12y - 8x\,dV}} & = \int_{0}^{5}{{\int_{0}^{{\frac{1}{2}z}}{{\left. {\left( {6{y^2} - 8xy} \right)} \right|_0^{10 - 2z}\,dx}}\,dz}}\\ & = \int_{0}^{5}{{\int_{0}^{{\frac{1}{2}z}}{{6{{\left( {10 - 2z} \right)}^2} - 8x\left( {10 - 2z} \right)\,dx}}\,dz}}\end{align*}\]

E'tibor bering, biz bu erda hech qanday soddalashtirmadik, chunki biz biron bir soddalashtirishni amalga oshirishimiz kerakligi hali aniq emas. Keyingi integratsiya $x$ bilan bog'liq va birinchi atama integratsiyaga tegishli bo'lsa, konstanta hisoblanadi va ikkinchi atama $x$ bilan ko'paytiriladi va shuning uchun $x$ integratsiyasi uchun uni shu shaklda qoldirish osonroq bo'ladi.

5-qadam

Endi $x$ integratsiyasini bajaramiz.

\[\begin{align*}\iiint\limits_{E}{{12y - 8x\,dV}} & = \int_{0}^{5}{{\left. {\left[ {6{{\left( {10 - 2z} \right)}^2}x - 4{x^2}\left( {10 - 2z} \right)} \right]} \right|_0^{\frac{1}{2}z}\,dz}}\\ & = \int_{0}^{5}{{3z{{\left( {10 - 2z} \right)}^2} - {z^2}\left( {10 - 2z} \right)\,dz}}\\ & = \int_{0}^{5}{{14{z^3} - 130{z^2} + 300z\,dz}}\end{align*}\]

6-qadam

Va nihoyat, $z$ integratsiyasini amalga oshiramiz.

\[\iiint\limits_{E}{{12y - 8x\,dV}} = \left. {\left( {\frac{7}{2}{z^4} - \frac{{130}}{3}{z^3} + 150{z^2}} \right)} \right|_0^5 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{3125}}{6}}}\]

Shunday qilib, biz barcha limitlarni o'rnatganimizdan so'ng, bu muammo uchun integratsiya unchalik yomon emas edi. Ko'pincha bu muammolar bilan shunday bo'ladi. O'rnatilgan integrallar uchun chegaralarni olish muammoning eng qiyin qismi bo'ladi, lekin har doim emas. Ular o'rnatilgach, integratsiya odatda juda oddiy.