Asosiy tarkibga o'tish

15.5-bo'lim : Uch karrali integrallar

7. \( \displaystyle \iiint\limits_{E}{{15z\,dV}}\) ni baholang, bunda \(E\) \(z = 2{y^2}\) va \(z = \sqrt {4y} \) bilan chegaralangan \(yz\) tekisligidagi mintaqa oldidagi \(2x + y + z = 4\) va \(4x + 4y + 2z = 20\) orasidagi mintaqadir.

Yechim

Bu hududni tasavvur qilish har doim juda qiyin ko'rinadi. Shunday qilib, keling, quyidagi eskizdan boshlaylik.

A three dimensional sketch showing the two planes that bound the region \(E\), drawn with the \(x\) and \(y\) axes reoriented from the arrangement used in the other three dimensional sketches so that the region is easier to visualize. The large yellow plane is the graph of \(4x + 4y + 2z = 20\) in the first octant and the much smaller green plane, far closer to the origin, is the graph of \(2x + y + z = 4\). The tiny shaded patch right at the origin is the region in the \(yz\)-plane bounded by \(z = 2{y^2}\) and \(z = \sqrt {4y} \), which is sketched separately below because it is so hard to see here.

E'tibor bering, \(x\) va \(y\) o'qlarining "yo'nalishi" biz shu paytgacha qilgan barcha boshqa 3D eskizlardan bu eskiz shaklida farq qiladi. Bu vizualizatsiyaga yordam berish uchun amalga oshiriladi. Qayta yo'nalishsiz chegaralangan hududni tasavvur qilish juda qiyin bo'lar edi.

Har doim o'qlarning yo'nalishi doimo sobit bo'lib qoladi va hech qachon o'zgarmasligini kutmang. Ba'zan yo'nalishni o'zgartirish kerak bo'ladi, shuning uchun biz ma'lum bir sirt yoki hududni tasavvur qilishimiz mumkin.

Xo'sh, biz bu erda ishlayotgan mintaqani tasavvur qilishga harakat qilamiz.

Sariq tekislik 1-banddagi \(4x + 4y + 2z = 20\) grafigidirst oktant (bu erda bizga kerak bo'lgan yagona qism) va yashil tekislik \(2x + y + z = 4\) grafigi. Boshida (ko'rish qiyin) \(yz\) tekisligidagi chegaralangan hududning eskizi muammo bayonida havola qilingan. Yuqoridagi grafikda bu chegaralangan hududni ko'rish qiyin bo'lgani uchun bu erda faqat chegaralangan hududning alohida grafigi mavjud.

A sketch of the two dimensional bounded region in the \(yz\)-plane, with \(y\) on the horizontal axis and \(z\) on the vertical axis. It is the shaded leaf shaped region caught between the red curve labeled \(z = \sqrt {4y} \), which forms the upper left boundary, and the blue curve labeled \(z = 2{y^2}\), which forms the lower right boundary. The two curves meet at the origin and again at \(\left( {1,2} \right)\).

Endi, \(E\) hududini sinab ko‘rish va tasavvur qilish uchun yuqoridagi birinchi eskizdagi ikkita tekislik o‘rtasida pechene xamiri bo‘lagi borligini va \(yz\) tekisligidagi chegaralangan hudud cookie to‘sar ekanligini tasavvur qiling. Biz chegaralangan hududni \(yz\) tekisligidan uzoqlashtiramiz va uni doimo \(yz\) tekisligiga parallel tutamiz.

Chegaralangan hudud tashqariga siljiganida, uning yuqori qismi yashil tekislikka urilganda avval "pechene xamiri" ga aylanadi va chegaralangan hududning pastki qismi yashil tekislikka tegsa, to'liq pechene xamirida bo'ladi. Chegaralangan hudud pechenye xamirini sariq tekislikka yetguncha kesishda davom etadi, chegaralangan hududning tepasi sariq tekislikka urilganda birinchi bo‘lib pechene xamiridan chiqadi. Chegaralangan hududning pastki qismi sariq tekislikka tegsa, chegaralangan hudud cookie xamiridan to'liq chiqib ketadi.

Mana, chegaralangan hudud \(yz\) tekisligidan uzoqlashganda tutadigan yo‘lning eskizi.

A three dimensional sketch of the path that the bounded region traces out as it moves away from the \(yz\)-plane, staying parallel to that plane, with the same reoriented axes as the previous sketch. The large yellow plane \(4x + 4y + 2z = 20\) and the smaller green plane \(2x + y + z = 4\) are both still shown, and the red leaf shaped tube is the full path of the bounded region, running from the origin all the way out to \(x = 5\) regardless of whether it is between the two planes at that point.

E'tibor bering, bu eskiz ikki tekislik o'rtasida bo'ladimi yoki yo'qmi, chegaralangan hududning to'liq yo'lini ko'rsatadi. Yoki ikki tekislik orasidagi pechenye xamiridan mintaqani kesib tashlash yoki qilmaslikdan qat'i nazar, yuqoridan o'xshashlikdan foydalanish. Biz qidirayotgan \(E\) mintaqasi bu yo'lning ikki tekislik o'rtasida joylashgan qismidir.

Bu erda faqat ikkita tekislik orasidagi chegaralangan mintaqaning yo'lining eskizi.

\(E\) mintaqasining yuqori qismi yuqorida ko'rsatilgan qizil qismdir. Biroq, hali ham eskizda bo'lgan ikkita samolyot bilan to'liq hududni tasavvur qilish juda qiyin. Shunday qilib, bu erda eskizdan olib tashlangan ikkita samolyot bilan yana bir eskiz.

A three dimensional sketch of the same setup, but now showing only the portion of the path that lies between the two planes. The red solid starts where the path first passes through the green plane \(2x + y + z = 4\), close to the origin, and stops where it passes through the yellow plane \(4x + 4y + 2z = 20\), and that red solid is the region \(E\) we are after.
A three dimensional sketch of the region \(E\) alone, drawn with the traditional axes and the two planes removed so the solid is easier to see. It is a leaf shaped tube running out along the \(x\)-axis, mostly shaded red, with a sloped yellow face on the front where it follows the plane \(4x + 4y + 2z = 20\) and a sloped rear face where it follows the plane \(2x + y + z = 4\). The same \(x\), \(y\), \(z\) scale as the previous sketches has been kept to help with the visualization.

\(E\) mintaqasi yuqorida ko'rsatilgan qattiqdir. \(E\) ning “old” qismi qiyshaygan va sariq tekislikka ergashadi (shuning uchun buni aniqroq qilish uchun ham sariq rangga bo'yalgan) va \(E\) ning “orqasi” qiya va yashil tekislikka ergashadi.

\(E\) hududining ushbu yakuniy eskizi bilan biz ob'ektni ko'rishga yordam berish uchun "an'anaviy" o'q tizimini ham, "qutili" o'qlar tizimini ham qo'shdik. Shuningdek, biz oldingi tasvirlar kabi \(x\), \(y\), \(z\) shkalasini saqlab qoldik, garchi o'qlar tizimining o'ng tomonida juda ko'p "behuda" bo'sh joy mavjud bo'lsa ham, vizualizatsiyaga yordam beradi.

2-qadam

Xo'sh, bu muammo uchun muammo bayonoti bizga cheklangan hududni aytadi, i.e. \(D\), \(yz\) tekisligida va biz avval \(x\) bilan integratsiya qilishimiz kerak.

Bu shuni anglatadiki, biz \(x\) uchun tekisliklarning har bir tenglamasini echishimiz kerak va biz orqa (yoki yashil) tekislikdan oldingi (yoki sariq) tekislikka integratsiya qilamiz.

Shuning uchun bizda \(x\) uchun quyidagi limitlar mavjud.

\[2 - \frac{1}{2}y - \frac{1}{2}z \le x \le 5 - y - \frac{1}{2}z\]

Esda tutingki, bu turdagi chegaralarni o'rnatishda biz orqadan oldinga o'tamiz va shuning uchun paraboloid pastki chegara bo'ladi, chunki u qattiq jismning "orqasida" joylashgan sirt, \(x = 1\) esa qattiq jismning "old"i va yuqori chegarasi.

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

\[\iiint\limits_{E}{{15z\,dV}} = \iint\limits_{D}{{\left[ {\int_{{2 - \frac{1}{2}y - \frac{1}{2}z}}^{{5 - y - \frac{1}{2}z}}{{15z\,dx}}} \right]\,dA}}\]
3-qadam

Keyinchalik, biz integralni sozlashni tugatishimiz uchun \(D\) uchun limitlar kerak bo'ladi. Yuqorida ta'kidlab o'tilganidek, \(D\) mintaqasi muammo bayonotida berilgan chegaralangan hududdir. Malumot uchun bu yerda biz 1-bosqichda bergan \(D\) eskizining nusxasi.

A sketch of the two dimensional bounded region in the \(yz\)-plane, with \(y\) on the horizontal axis and \(z\) on the vertical axis. It is the shaded leaf shaped region caught between the red curve labeled \(z = \sqrt {4y} \), which forms the upper left boundary, and the blue curve labeled \(z = 2{y^2}\), which forms the lower right boundary. The two curves meet at the origin and again at \(\left( {1,2} \right)\).

Ko'rib turganimizdek, bu mintaqani ikkala tartibda birlashtirish mumkin, ammo tartibdan qat'i nazar, chegaralardan biri kvadratik atama va biri kvadrat ildiz bo'ladi. Shuning uchun, muammo bayonotida berilgan limitlardan foydalanishimiz mumkin.

Shunday qilib, bu erda \(D\) uchun limitlar mavjud.

\[\begin{array}{c}0 \le y \le 1\\ 2{y^2} \le z \le \sqrt {4y} \end{array}\]

ushbu limitlar bizda mavjud integralga ulangan holda,

\[\iiint\limits_{E}{{15z\,dV}} = \int_{0}^{1}{{\int_{{2{y^2}}}^{{\sqrt {4y} }}{{\int_{{2 - \frac{1}{2}y - \frac{1}{2}z}}^{{5 - y - \frac{1}{2}z}}{{15z\,dx}}\,dz}}\,dy}}\]
4-qadam

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

\[\begin{align*}\iiint\limits_{E}{{15z\,dV}} & = \int_{0}^{1}{{\int_{{2{y^2}}}^{{\sqrt {4y} }}{{\left. {\left( {15zx} \right)} \right|_{2 - \frac{1}{2}y - \frac{1}{2}z}^{5 - y - \frac{1}{2}z}\,dz}}\,dy}}\\ & = \int_{0}^{1}{{\int_{{2{y^2}}}^{{\sqrt {4y} }}{{15z\left( {3 - \frac{1}{2}y} \right)\,dz}}\,dy}}\end{align*}\]
5-qadam

Endi \(z\) integratsiyasini bajaramiz.

\[\begin{align*}\iiint\limits_{E}{{15z\,dV}} & = \int_{0}^{1}{{\left. {\frac{{15}}{2}{z^2}\left( {3 - \frac{1}{2}y} \right)} \right|_{2{y^2}}^{\sqrt {4y} }\,dy}}\\ & = \int_{0}^{1}{{15{y^5} - 90{y^4} - 15{y^2} + 90y\,dy}}\end{align*}\]
6-qadam

Va nihoyat, \(y\) integratsiyasini amalga oshiramiz.

\[\iiint\limits_{E}{{15z\,dV}} = \left. {\left( {\frac{5}{2}{y^6} - 18{y^5} - 5{y^3} + 45{y^2}} \right)} \right|_0^1 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{49}}{2}}}\]

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. Limitlar o'rnatilgandan so'ng, integratsiya odatda juda oddiy bo'ladi va shubhasiz, mintaqani tasavvur qilish va ushbu muammo uchun o'rnatilgan chegaralarni olish boshqa ko'plab muammolarga qaraganda qiyinroq bo'lgan.