Asosiy tarkibga o'tish

6.4-bo'lim : Silindrli hajm

8. \(\displaystyle y = \frac{{{{\bf{e}}^{\frac{1}{2}x}}}}{{x + 2}}\), \(\displaystyle y = 5 - \frac{1}{4}x\), \(x = - 1\) va \(x = 6\) bilan chegaralangan hududni \(x = - 2\) chizig‘i 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 line \(y = 5 - \frac{1}{4}x\) is the upper boundary and the curve \(y = \frac{{{{\bf{e}}^{\frac{1}{2}x}}}}{{x + 2}}\) is the lower boundary, and the region between them runs from the vertical line \(x = - 1\) on the left to \(x = 6\) on the right. It is shaded in and gets narrower towards the right end. A vertical dashed line at \(x = - 2\) 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 \(x = - 2\). It is a wide, shallow bowl shaped solid with a narrow tube running up 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 \(x = - 2\) and stands 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 \(x = - 2\) 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 \(x\) jihatidan 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 \(x = - 2\) with its right edge at some \(x\), and the distance from the axis of rotation across to that edge is labeled radius = \(2 + x\), made up of a piece of length 2 and a piece of length \(x\). The vertical distance from the curve \(\frac{{{{\bf{e}}^{\frac{1}{2}x}}}}{{x + 2}}\) up to the line \(5 - \frac{1}{4}x\) is labeled height = \(5 - \frac{1}{4}x - \frac{{{{\bf{e}}^{\frac{1}{2}x}}}}{{x + 2}}\), with the two individual distances also marked.

Eskizdan biz silindrni \(x = - 2\) chizig'ida markazlashganini va silindrning o'ng qirrasi qandaydir \(x\) da joylashganligini ko'rishimiz mumkin.

Bu erda radiusga biroz ehtiyot bo'lishimiz kerak, chunki silindrning o'ng qirrasi qattiq jismga kesishgan joyiga qarab \(y\) o'qining chap va o'ng tomonida bo'lishi mumkin.

Yuqoridagi eskizda ko'rsatilganidek, silindrning o'ng qirrasi \(y\) o'qining o'ng tomonidagi ob'ektga kesilgan bo'lsa, u holda radius aylanish o'qining \(y\) o'qigacha bo'lgan masofasi (2 masofa) va \(y\) o'qidan \(y\) o'qining o'ng tomonidagi masofaga (\(y\) o'qining o'ng tomoniga bo'lgan masofa) radiusdir. Shuning uchun, bu holda, radius \(2 + x\).

Boshqa tomondan, agar silindrning o'ng qirrasi uning chap tomonidagi qattiq qismga kesilsa y‑o'qi bo'lsa, radius aylanish o'qidan \(y\) o'qigacha bo'lgan masofa (2 masofa) minus silindrning o'ng chetidan \(y\) o'qigacha bo'lgan masofa bo'ladi. Biroq, bu holda, o'ng qirrani belgilaydigan \(x\) qiymati manfiy qiymatdir va shuning uchun silindrning o'ng chetining \(y\) o'qiga masofasi \( - x\) bo'lishi kerak. Buni ijobiy miqdorga aylantirish uchun minus belgisi kerak. Shuning uchun bu holda radius \(2 - \left( { - x} \right) = 2 + x\), birinchi holatda bo'lgani kabi.

Tsilindrning yuqori qirrasi qattiq jismning yuqori qismini belgilaydigan egri chiziqda joylashgan va \(x\) o'qidan \(5 - \frac{1}{4}x\) masofasida joylashgan. Tsilindrning pastki qirrasi qattiq jismning pastki qismini belgilovchi egri chiziq ustida joylashgan va \(x\) o'qidan \(\frac{{{{\bf{e}}^{\frac{1}{2}x}}}}{{x + 2}}\) masofasi. Keyin balandlik bu ikkalasining farqidir.

Shunday qilib, silindrning radiusi va kengligi,

\[{\mbox{Radius}} = 2 + x\hspace{0.25in}\hspace{0.25in}{\mbox{Width}} = 5 - \frac{1}{4}x - \frac{{{{\bf{e}}^{\frac{1}{2}x}}}}{{x + 2}}\]

U holda silindrning maydoni,

\[A\left( x \right) = 2\pi \left( {{\mbox{Radius}}} \right)\left( {{\mbox{Height}}} \right) = 2\pi \left( {2 + x} \right)\left( {5 - \frac{1}{4}x - \frac{{{{\bf{e}}^{\frac{1}{2}x}}}}{{x + 2}}} \right) = 2\pi \left( {10 + \frac{9}{2}x - \frac{1}{4}{x^2} - {{\bf{e}}^{\frac{1}{2}x}}} \right)\]
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 \(x = 6\) da paydo bo'lishini ko'rishimiz mumkin. Bizning chegaralarimiz quyidagicha: \( - 1 \le x \le 6\).

Keyin hajm,

\[\begin{align*}V & = \int_{{ - 1}}^{6}{{2\pi \left( {10 + \frac{9}{2}x - \frac{1}{4}{x^2} - {{\bf{e}}^{\frac{1}{2}x}}} \right)\,dx}}\\ & = 2\left. {\pi \left( {10x + \frac{9}{4}{x^2} - \frac{1}{{12}}{x^3} - 2{{\bf{e}}^{\frac{1}{2}x}}} \right)} \right|_{ - 1}^6 = \require{bbox} \bbox[2pt,border:1px solid black]{{2\pi \left( {\frac{{392}}{3} + 2{{\bf{e}}^{ - \,\,\frac{1}{2}}} - 2{{\bf{e}}^3}} \right)}}\end{align*}\]