Asosiy tarkibga o'tish

6.4-bo'lim : Silindrli hajm

7. \(y = {x^2} - 6x + 9\) va \(y = - {x^2} + 6x - 1\) bilan chegaralangan hududni \(x = 8\) chiziq atrofida aylantirish natijasida olingan qattiq jismning hajmini silindrlar 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 downward opening parabola \(y = - {x^2} + 6x - 1\) is the upper boundary and the upward opening parabola \(y = {x^2} - 6x + 9\) is the lower boundary. The two cross at \(\left( {1,4} \right)\) and \(\left( {5,4} \right)\), both marked and labeled, and the lens shaped region between them is shaded in. A vertical dashed line at \(x = 8\) with a small circular arrow on it marks the axis of rotation.

Bizga birozdan keyin kerak bo'ladigan yuqorida ko'rsatilgan kesishish nuqtalarini olish uchun ikkita tenglamani o'rnatish va yechish kifoya.

\[\begin{align*}{x^2} - 6x + 9 & = - {x^2} + 6x - 1\\ 2{x^2} - 12x + 10 & = 0\\ 2\left( {x - 1} \right)\left( {x - 5} \right) & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in} x = 1,\,\,\,x = 5 \hspace{0.25in} \Rightarrow \hspace{0.25in} \left( {1,4} \right)\,\,\,\,\& \,\,\,\,\left( {5,4} \right)\end{align*}\]

Maslahat: Inqilobning qattiq jismining qanday ko'rinishini chizishga yaxshi urinib ko'ring va vakilli silindrda eskiz qiling.

E'tibor bering, buni qilish qiyin bo'lishi mumkin, ayniqsa siz juda ko'p ko'rinadigan odam bo'lmasangiz. Biroq, biz maydon formulasini yozishga borganimizda vakili silindrga ega bo'lish katta yordam berishi mumkin. Bundan tashqari, revolyutsiyaning qattiq jismining eskizisiz vakili tsilindrni olish qiyin bo'lishi mumkin. Shunday qilib, bu eskizlarni olish uchun qo'lingizdan kelganini qiling.

2-qadam

Bu erda inqilobning qattiq qismining eskizi.

A sketch of the solid of revolution obtained by rotating the region about the line \(x = 8\). It is a thick ring or doughnut shaped solid with a hole down the middle around the axis of rotation.

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

A sketch of the solid of revolution with a representative cylinder drawn in and the front half of the solid cut away so that the cylinder can be seen. The cylinder is centered on the dashed line \(x = 8\) and stands inside the ring shaped solid. A sketch of the solid of revolution with a representative cylinder 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 cylinder is centered on the dashed line \(x = 8\) and stands inside the solid.
3-qadam

Endi silindrning sirt maydoni uchun formulani topishimiz kerak. Chunki biz vertikal o'qda joylashgan silindrlardan foydalanamiz (i.e. \(y\) 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, biz kiritishimiz kerak bo'lgan turli miqdorlar bilan vakili silindrning yana bir eskizi.

A sketch of a representative cylinder with all of the quantities needed put in. The cylinder is centered on the dashed line \(x = 8\) with its left edge at some \(x\), and the distance from that edge across to the axis of rotation is labeled radius = \(8 - x\). The vertical distance from the parabola \({x^2} - 6x + 9\) up to the parabola \( - {x^2} + 6x - 1\) is labeled height = \( - 2{x^2} + 12x - 10\), with the two individual distances also marked on the mirrored curves at the right of the sketch.

E'tibor bering, biz barcha balandlikdagi "chiziqlar" ni haqiqiy egri chiziqlarga emas, balki aks ettirilgan egri chiziqlarga qo'yamiz. Bu biz ularni \(y\) o'qiga xalaqit bermaydigan joyga qo'yishimiz uchun qilingan.

Eskizdan biz silindrning \(x = 8\) chizig'ida markazlashtirilganligini va silindrning chap qirrasi qandaydir \(x\) da joylashganligini ko'rishimiz mumkin.

Tsilindrning radiusi aylanish o'qidan silindrning chap chetiga qadar bo'lgan masofadir (i.e. \(8 - x\)).

Tsilindrning yuqori qirrasi qattiq jismning yuqori qismini belgilovchi egri chiziq ustida joylashgan va \(x\) o'qidan \( - {x^2} + 6x - 1\) masofada joylashgan. Tsilindrning pastki qirrasi qattiq jismning pastki qismini belgilovchi egri chiziq ustida joylashgan va u erdan \({x^2} - 6x + 9\) masofada joylashgan. x‑o'qi. Keyin balandlik bu ikkalasining farqidir.

Shunday qilib, silindrning radiusi va kengligi,

\[{\mbox{Radius}} = 8 - x\hspace{0.25in}\hspace{0.25in}{\mbox{Width}} = - {x^2} + 6x - 1 - \left( {{x^2} - 6x + 9} \right) = - 2{x^2} + 12x - 10\]

U holda silindrning maydoni,

\[\begin{align*}A\left( x \right) & = 2\pi \left( {{\mbox{Radius}}} \right)\left( {{\mbox{Height}}} \right)\\ & = 2\pi \left( {8 - x} \right)\left( { - 2{x^2} + 12x - 10} \right) = 2\pi \left( { - 80 + 106x - 28{x^2} + 2{x^3}} \right)\end{align*}\]
4-qadam

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

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

Keyin hajm,

\[V = \int_{1}^{5}{{2\pi \left( { - 80 + 106x - 28{x^2} + 2{x^3}} \right)\,dx}} = 2\left. {\pi \left( { - 80x + 53{x^2} - \frac{{28}}{3}{x^3} + \frac{1}{2}{x^4}} \right)} \right|_1^5 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{640}}{3}\pi }}\]