Asosiy tarkibga o'tish

6.3-bo'lim : Halqali ovoz balandligi

7. \(x = {y^2} - 4\) va \(x = 6 - 3y\) bilan chegaralangan hududni \(x = 24\) chizig‘i atrofida aylantirish natijasida olingan qattiq jism hajmini aniqlash uchun disklar/halqalar usulidan foydalaning.

Yechim

Muammoni biror joyda boshlashimiz kerak, shuning uchun "oddiy" ni boshlaylik.

Chegaralangan mintaqaning qanday ko'rinishini bilish ushbu turdagi muammolarning ko'pchiligiga yordam beradi, chunki biz maydon formulasini o'rnatish uchun barcha egri chiziqlar bir-biriga qanday bog'liqligini bilishimiz kerak va bizga grafik ko'pincha yordam beradigan integral uchun limitlar kerak bo'ladi.

Bu erda aylanish o'qi bilan chegaralangan hududning eskizi ko'rsatilgan.

A sketch of the bounded region with the axis of rotation shown. The parabola \(x = {y^2} - 4\) opens to the right with vertex at \(\left( { - 4,0} \right)\) and the line \(x = 6 - 3y\) falls to the right. The two cross at \(\left( {0,2} \right)\) and \(\left( {21, - 5} \right)\), both marked and labeled, and the region between them is shaded in. A vertical dashed line at \(x = 24\) with a small circular arrow on it marks the axis of rotation.

Eskizda ko'rsatilgan kesishish nuqtalarini olish uchun biz qilishimiz kerak bo'lgan yagona narsa ikkita tenglamani tenglashtirish va yechish (bular bizga birozdan keyin kerak bo'ladi).

\[\begin{align*}{y^2} - 4 & = 6 - 3y\\ {y^2} + 3y - 10 & = 0\\ \left( {y + 5} \right)\left( {y - 2} \right) & = 0\hspace{0.25in}\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\,\,\,y = - 5,\,\,\,\,y = 2\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\left( {21, - 5} \right)\,\,\,\& \,\,\,\left( {0,2} \right)\end{align*}\]
2-qadam

Bu erda inqilobning qattiq qismining eskizi.

A sketch of the solid of revolution obtained by rotating the region about the line \(x = 24\). It is a wide, shallow bowl or saucer shaped solid centered on the axis of rotation, open at the top and rounded underneath.

Mana vakillik halqasining bir nechta eskizlari. Chapdagi rasmda qattiq qismning old yarmi kesilgan va o'ngdagi rasmda qattiq qismning orqa yarmining "simli ramkasi" bo'lgan vakili halqa ko'rsatilgan (i.e. qattiq jismning orqa yarmining chetlarini ifodalovchi egri chiziqlar).

A sketch of the solid of revolution with a representative ring drawn in and the front half of the solid cut away so that the ring can be seen. The ring is centered on the dashed line \(x = 24\) and lies horizontally inside the saucer shaped solid. A sketch of the solid of revolution with a representative ring drawn in and only a “wire frame” of the back half of the solid shown, so just the curves that form the edges of the solid are drawn. The ring is centered on the dashed line \(x = 24\) and lies horizontally inside the solid.
3-qadam

Endi halqaning maydoni uchun formulani topishimiz kerak. Chunki biz vertikal o'qda joylashgan halqalardan foydalanamiz (i.e. \(y\) o'qiga parallel) biz bilamizki, maydon formulasi \(y\) nuqtai nazaridan bo'lishi kerak. Shuning uchun, egri chiziqlar tenglamasi \(y\) (bu holda ular allaqachon mavjud) nuqtai nazaridan bo'lishi kerak.

Mana vakillik uzukining yana bir eskizi, unda biz uchun kerak bo'lgan turli miqdorlar mavjud.

A sketch of a representative ring with all of the quantities needed put in. The ring is centered on the dashed line \(x = 24\) and placed at some \(y\). The inner radius is labeled i.r. = \(18 + 3y\), the distance from the axis of rotation in to the line \(x = 6 - 3y\), and the outer radius is labeled o.r. = \(28 - {y^2}\), the distance from the axis of rotation out to the parabola \(x = {y^2} - 4\), with the distance 24 from the axis to the y-axis also marked.

Eskizdan biz halqaning \(x = 24\) chizig'ida markazlashtirilganligini va qandaydir \(y\)da joylashganini ko'rishimiz mumkin.

Keyin halqaning ichki radiusi aylanish o'qidan qattiq jismning ichki chetini belgilovchi egri chiziqgacha bo'lgan masofadir. Buning uchun formulani aniqlash uchun birinchi navbatda aylanish o'qi \(y\) o'qidan 24 masofada ekanligiga e'tibor bering. Keyinchalik, qattiq jismning ichki chetini belgilovchi egri chiziq \(y\) o'qidan \(x = 6 - 3y\) masofasidir. Ichki radius bu ikki masofa orasidagi farq yoki,

\[{\mbox{Inner Radius}} = 24 - \left( {6 - 3y} \right) = 18 + 3y\]

Tashqi radius ham xuddi shunday tarzda hisoblanadi, lekin biroz qiyinroq. Bu holda qattiq jismning tashqi chetini belgilovchi egri chiziq \(y\) o'qining chap va o'ng tomonida paydo bo'ladi.

Keling, avval yuqoridagi eskizda ko'rsatilganidek, ishni ko'rib chiqaylik. Bu holda qattiq jismning tashqi chetini aniqlovchi funksiyaning qiymati \(y\) o'qining chap tomonida joylashgan va shuning uchun manfiy qiymatga ega. Ushbu nuqtaning \(y\) o'qidan masofasi \( - \left( {{y^2} - 4} \right)\) bo'ladi, bu erda minus belgisi manfiy funktsiya qiymatini masofa uchun kerak bo'lgan ijobiy qiymatga aylantiradi. Bu holat uchun tashqi radius aylanish o'qining \(y\) o'qigacha bo'lgan masofasi (masofa 24) va tashqi chetini \(y\) o'qigacha aniqlaydigan egri chiziqning masofasining yig'indisidir (biz buni yuqorida topdik).

Agar qattiq jismning tashqi chetini belgilovchi egri chiziq \(y\) o'qining o'ng tomonida bo'lsa, u ijobiy qiymatga ega bo'ladi va shuning uchun egri chiziq va \(y\) o'qidagi nuqtalar masofasi faqat \({y^2} - 4\) bo'ladi. Bu holda bizga minus belgisi kerak emas, chunki funktsiya qiymati allaqachon ijobiy, bu bizga masofa uchun kerak. Bu holda tashqi radius keyin aylanish o'qidan \(y\) o'qigacha bo'lgan masofa (24 masofa) minus bu yangi masofa.

Yaxshiyamki, har ikkala holatda ham tashqi radius shunday bo'ladi,

\[{\mbox{Outer Radius}} = 24 - \left( {{y^2} - 4} \right) = 28 - {y^2}\]

Shuni esda tutingki, chetni belgilovchi egri chiziq ijobiy va salbiy qiymatlarga ega bo'lsa, radiusning yakuniy tenglamasi (muammoga qarab ichki yoki tashqi) bir xil bo'ladi. Siz faqat ko'rib chiqmoqchi bo'lgan ishni o'rnatishda ehtiyot bo'lishingiz kerak. Agar siz birinchi holatni to'g'ri o'rnatgan bo'lsangiz, ikkinchisini qilishingiz shart emas, chunki formula bir xil bo'ladi.

U holda halqaning maydoni,

\[\begin{align*}A\left( x \right) & = \pi \left[ {{{\left( {{\mbox{Outer Radius}}} \right)}^2} - {{\left( {{\mbox{Inner Radius}}} \right)}^2}} \right]\\ & = \pi \left[ {{{\left( {28 - {y^2}} \right)}^2} - {{\left( {18 + 3y} \right)}^2}} \right] = \pi \left( {460 - 108y - 65{y^2} + {y^4}} \right)\end{align*}\]
4-qadam

Yakuniy bosqich - bu hajm uchun integralni o'rnatish va uni baholash.

1-bosqichda topilgan ikkita egri chiziqning kesishish nuqtalaridan biz qattiq jismdagi “birinchi” halqa \(y = - 5\) da, “oxirgi” halqa esa \(y = 2\) da paydo bo‘lishini ko‘rishimiz mumkin. Bizning chegaralarimiz quyidagicha: \( - 5 \le y \le 2\).

Keyin hajm,

\[V = \int_{{ - 5}}^{2}{{\pi \left( {460 - 108y - 65{y^2} + {y^4}} \right)\,dy}} = \left. {\pi \left( {460y - 54{y^2} - \frac{{65}}{3}{y^3} + \frac{1}{5}{y^5}} \right)} \right|_{ - 5}^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{31556}}{{15}}\pi }}\]