Asosiy tarkibga o'tish

6.4-bo'lim : Silindrli hajm

6. \(x = {y^2} - 4\) va \(x = 6 - 3y\) bilan chegaralangan hududni \(y = - 8\) 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 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 \(\left( {21, - 5} \right)\), which is marked and labeled, and the region between them is shaded in. A horizontal dashed line at \(y = - 8\) 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.

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

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, vakillik silindriga ega bo'lish, biz maydon formulasini yozishga borganimizda katta yordam berishi mumkin. Bundan tashqari, revolyutsiyaning qattiq jismining eskizisiz vakili silindrni olish qiyin bo'lishi mumkin. Shunday qilib, ushbu 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 \(y = - 8\). It is a cone shaped solid lying on its side, wide and rounded at the left end and tapering down towards the right end where it meets 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 \(y = - 8\) and lies inside the 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 \(y = - 8\) and lies inside the solid.
3-qadam

Endi silindrning sirt maydoni uchun formulani topishimiz kerak. Chunki biz gorizontal o'qda joylashgan silindrlardan foydalanamiz (i.e. \(x\) o'qiga parallel) biz bilamizki, maydon formulasi \(y\) jihatidan bo'lishi kerak. Shuning uchun, egri chiziqlar tenglamalari \(y\) (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 \(y = - 8\) with its upper edge at some \(y\), and the distance from the axis of rotation up to that edge is labeled radius = \(8 + y\). The horizontal distance from the parabola \(x = {y^2} - 4\) across to the line \(x = 6 - 3y\) is labeled width = \(10 - 3y - {y^2}\), with the two individual distances also marked.

Eskizdan biz silindrni \(y = - 8\) chizig'ida markazlashganini va silindrning yuqori qirrasi qandaydir \(y\) da joylashganligini ko'rishimiz mumkin.

Tsilindrning radiusi bu muammo uchun biroz qiyin.

Birinchidan, aylanish o'qi \(x\) o'qidan 8 ga pastroq masofada ekanligiga e'tibor bering. Keyinchalik, silindrning yuqori qirrasi ma'lum bir \(y\) da, ammo \(y\) silindrni chizgan nuqtada manfiy bo'lgani uchun, bu yuqori chetning \(x\) o'qi ostidagi masofasi aslida \( - y\) ekanligini anglatadi. Buni masofa uchun zarur bo'lgan ijobiy miqdorga aylantirish uchun minus kerak. Bu silindrning radiusi bu ikki masofaning farqiga teng yoki,

\[8 - \left( { - y} \right) = 8 + y\]

Endi shuni yodda tutingki, silindrning yuqori qirrasi \(x\) o'qidan yuqoriga ko'tarilganda, \(x\) o'qi ustidagi yuqori chetining masofasi faqat \(y\) bo'ladi. Bu safar \(y\) ijobiy bo'lgani uchun bizga minus belgisi kerak emas (va aslida buni xohlamaymiz, chunki bu masofani manfiy miqdorga aylantiradi). Keyin radius aylanish o'qining \(x\) o'qidan masofasi (hali 8 masofa) va \(x\) o'qi ustidagi yuqori chetining masofasi (bu \(y\)) yoki,

\[8 + y\]

Ikkala holatda ham biz bir xil radiusni olamiz.

Tsilindrning o'ng qirrasi qattiq jismning o'ng qismini belgilaydigan egri chiziqda joylashgan va \(y\) o'qidan \(6 - 3y\) masofasi. Tsilindrning chap qirrasi qattiq jismning chap qismini belgilovchi egri chiziqda joylashgan va \(y\) o'qidan \({y^2} - 4\) masofasida joylashgan. Kenglik bu ikkalasining farqidir.

Shunday qilib, silindrning radiusi va kengligi,

\[{\mbox{Radius}} = 8 + y\hspace{0.25in}\hspace{0.25in}{\mbox{Width}} = 6 - 3y - \left( {{y^2} - 4} \right) = 10 - 3y - {y^2}\]

U holda silindrning maydoni,

\[A\left( y \right) = 2\pi \left( {{\mbox{Radius}}} \right)\left( {{\mbox{Height}}} \right) = 2\pi \left( {8 + y} \right)\left( {10 - 3y - {y^2}} \right) = 2\pi \left( {80 - 14y - 11{y^2} - {y^3}} \right)\]
4-qadam

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

1-bosqichdagi grafikdan biz qattiq jismdagi "birinchi" silindr \(y = - 5\) da, "oxirgi" silindr esa \(y = 2\) da paydo bo'lishini ko'rishimiz mumkin. Bizning chegaralarimiz quyidagicha: \( - 5 \le y \le 2\).

Keyin hajm,

\[V = \int_{{ - 5}}^{2}{{2\pi \left( {80 - 14y - 11{y^2} - {y^3}} \right)\,dy}} = 2\left. {\pi \left( {80y - 7{y^2} - \frac{{11}}{3}{y^3} - \frac{1}{4}{y^4}} \right)} \right|_{ - 5}^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{4459}}{6}\pi }}\]