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6.3-bo'lim : Halqali ovoz balandligi

6. \(y = 10 - 6x + {x^2}\), \(y = - 10 + 6x - {x^2}\), \(x = 1\) va \(x = 5\) bilan chegaralangan hududni \(y = 8\) chizig‘i atrofida aylantirish natijasida olingan qattiq jism hajmini 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 upward opening parabola \(y = 10 - 6x + {x^2}\) is the upper boundary and the downward opening parabola \(y = - 10 + 6x - {x^2}\) is the lower boundary, and the region between them runs from the vertical line \(x = 1\) to \(x = 5\). It is shaded in and pinches in towards the middle. A horizontal dashed line at \(y = 8\) with a small circular arrow on it marks the axis of rotation.
2-qadam

Bu erda inqilobning qattiq qismining eskizi.

A sketch of the solid of revolution obtained by rotating the region about the line \(y = 8\). It is a solid that is wide at both ends and pinched in around the middle, rather like a spool, with a hole running through it along the axis of rotation.

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 \(y = 8\) and stands vertically inside the 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 \(y = 8\) and stands vertically inside the solid.
3-qadam

Endi halqaning maydoni uchun formulani topishimiz kerak. Chunki biz gorizontal o'qda joylashgan halqalardan foydalanamiz (i.e. \(x\) o'qiga parallel) biz bilamizki, maydon formulasi \(x\) nuqtai nazaridan bo'lishi kerak. Shuning uchun, egri chiziqlar tenglamalari \(x\) (bu holda ular allaqachon mavjud) jihatidan 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 \(y = 8\) and placed at some \(x\). The inner radius is labeled i.r. = \( - 2 + 6x - {x^2}\), the distance from the axis of rotation down to the curve \(y = 10 - 6x + {x^2}\), and the outer radius is labeled o.r. = \(18 - 6x + {x^2}\), the distance from the axis of rotation down to the curve \(y = - 10 + 6x - {x^2}\), with the distance 8 from the axis down to the x-axis also marked.

Eskizdan biz halqaning \(y = 8\) chizig'ida markazlashtirilganligini va qandaydir \(x\)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 \(x\) o'qidan 8 masofada ekanligiga e'tibor bering. Keyinchalik, qattiq jismning ichki chetini belgilovchi egri chiziq \(x\) o'qidan \(y = 10 - 6x + {x^2}\) masofasidir. Ichki radius bu ikki masofa orasidagi farq yoki,

\[{\mbox{Inner Radius}} = 8 - \left( {10 - 6x + {x^2}} \right) = - 2 + 6x - {x^2}\]

Tashqi radius ham xuddi shunday tarzda hisoblanadi. Bu aylanish o'qidan \(x\) o'qigacha bo'lgan masofa (8 masofa) va keyin u qattiq jismning tashqi chetini belgilovchi egri chiziqqa yetguncha \(x\) o'qi ostida davom etadi. Shunday qilib, biz bu ikki masofani qo'shishimiz kerak, lekin ehtiyot bo'lishimiz kerak, chunki "pastki" funktsiya aslida manfiy qiymatdir va shuning uchun pastki funktsiyadagi nuqtaning \(x\) o'qidan masofasi aslida: eskizda ko'rsatilganidek, \( - \left( { - 10 + 6x - {x^2}} \right)\). Tenglama oldidagi manfiy funktsiyaning manfiy qiymati ijobiy miqdorga aylantirilishini ta'minlaydi (bizning masofamiz uchun kerak). Tashqi radius bu ikki masofaning yig'indisi yoki,

\[{\mbox{Outer Radius}} = 8 - \left( { - 10 + 6x - {x^2}} \right) = 18 - 6x + {x^2}\]

U holda halqaning maydoni,

\[\begin{align*}A\left( x \right) & = \pi \left[ {{{\left( {{\mbox{Outer Radius}}} \right)}^2} - {{\left( {{\mbox{Inner Radius}}} \right)}^2}} \right]\\ & = \pi \left[ {{{\left( {18 - 6x + {x^2}} \right)}^2} - {{\left( { - 2 + 6x - {x^2}} \right)}^2}} \right] = \pi \left( {320 - 192x + 32{x^2}} \right)\end{align*}\]
4-qadam

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

1-bosqichdagi grafikdan biz qattiq jismdagi "birinchi" halqa \(x = 1\) da, "oxirgi" halqa esa \(x = 5\) da paydo bo'lishini ko'rishimiz mumkin. Bizning chegaralarimiz quyidagicha: \(1 \le x \le 5\).

Keyin hajm,

\[V = \int_{1}^{5}{{\pi \left( {320 - 192x + 32{x^2}} \right)\,dx}} = \left. {\pi \left( {320x - 96{x^2} + \frac{{32}}{3}{x^3}} \right)} \right|_1^5 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{896}}{3}\pi }}\]