VolumeWithCylinder - 2-masala yechimi | MathModel.uz

6.4-bo'lim : Silindrli hajm

2. $\displaystyle y = \frac{1}{x}$, $\displaystyle x = \frac{1}{2}$, $x = 4$ va $x$ o'qi bilan chegaralangan hududni $y$ o'qi 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 curve $y = \frac{1}{x}$ drops from $\left( {\frac{1}{2},2} \right)$ on the left down towards the x-axis on the right and the shaded region is the area under it between the vertical lines $x = \frac{1}{2}$ and $x = 4$ and above the x-axis. A small circular arrow on the y-axis marks it as 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 y-axis. It looks like a wide flat disk or hat brim with a narrow chimney rising out of the middle, since the tall thin part of the region near $x = \frac{1}{2}$ sweeps out the chimney and the long low tail out to $x = 4$ sweeps out the brim.$

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 y-axis and stands inside the solid, with its top edge on the curve that forms the upper surface of 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 y-axis and stands inside the solid.

3-qadam

Endi silindrning sirt maydoni uchun formulani topishimiz kerak. Biz $y$ o'qida joylashgan silindrlardan foydalanayotganimiz sababli, maydon formulasi $x$ nuqtai nazaridan bo'lishi kerakligini bilamiz. Shuning uchun, egri chiziqlar tenglamasi $x$ nuqtai nazaridan bo'lishi kerak (bu holda ular allaqachon mavjud).

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 y-axis with its right edge at some $x$, and the distance from the y-axis across to that edge is labeled radius = $x$. The distance from the x-axis up to the curve $y = \frac{1}{x}$ is labeled height = $\frac{1}{x}$.$

Eskizdan biz tsilindr $y$ o'qining markazida joylashganini va silindrning o'ng qirrasi qandaydir $x$ da joylashganligini ko'rishimiz mumkin.

Tsilindrning radiusi - bu $y$ o'qidan silindrning o'ng chetigacha bo'lgan masofa (i.e. $x$). Tsilindrning balandligi $x$ o'qidan qattiq jismning chetini belgilovchi egri chiziqgacha bo'lgan masofa ($\frac{1}{x}$ masofasi).

Shunday qilib, silindrning radiusi va kengligi,

\[{\mbox{Radius}} = x\hspace{0.25in}\hspace{0.25in}{\mbox{Height}} = \frac{1}{x}\]

U holda silindrning maydoni,

\[A\left( x \right) = 2\pi \left( {{\mbox{Radius}}} \right)\left( {{\mbox{Height}}} \right) = 2\pi \left( x \right)\left( {\frac{1}{x}} \right) = 2\pi \]

Hudud formulasida barcha o'zgaruvchilar bekor qilinishini kutmang. Bu erda bo'lgani kabi, vaqti-vaqti bilan sodir bo'lishi mumkin, lekin u bilan kamdan-kam uchraydi.

4-qadam

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

1-bosqichdagi grafikdan biz qattiq jismdagi "birinchi" silindr $x = \frac{1}{2}$ da, "oxirgi" silindr esa $x = 4$ da paydo bo'lishini ko'rishimiz mumkin. Bizning chegaralarimiz quyidagicha: $\frac{1}{2} \le x \le 4$.

Keyin hajm,

\[V = \int_{{\frac{1}{2}}}^{4}{{2\pi \,dx}} = 2\left. {\pi \left( x \right)} \right|_{\frac{1}{2}}^4 = \require{bbox} \bbox[2pt,border:1px solid black]{{7\pi }}\]