VolumeWithRings - 2-masala yechimi | MathModel.uz

6.3-bo'lim : Halqali ovoz balandligi

2. $y = 7 - {x^2}$, $x = - 2$, $x = 2$ va $x$ o'qi bilan chegaralangan mintaqani $x$ o'qi atrofida aylantirish natijasida olingan qattiq jismning hajmini aniqlash uchun 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 downward opening parabola $y = 7 - {x^2}$ has its vertex at $\left( {0,7} \right)$ and the region under it between the vertical lines $x = - 2$ and $x = 2$ and above the x-axis is shaded in. A small circular arrow on the x-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 x-axis. It is a rounded, barrel shaped solid lying on its side, widest in the middle at $x = 0$ and with flat circular ends at $x = - 2$ and $x = 2$.$

Mana, vakil diskning bir nechta eskizlari. Chapdagi rasmda qattiq qismning oldingi yarmi kesilgan vakil disk ko'rsatilgan va o'ngdagi rasmda qattiq qismning orqa yarmining "simli ramkasi" bo'lgan vakil disk ko'rsatilgan (i.e. qattiq jismning orqa yarmining chetlarini ifodalovchi egri chiziqlar).

A sketch of the solid of revolution with a representative disk drawn in and the front half of the solid cut away so that the disk can be seen. The disk is centered on the x-axis and stands vertically inside the solid.

A sketch of the solid of revolution with a representative disk 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 disk is centered on the x-axis and stands vertically inside the solid.

3-qadam

Endi biz disk maydoni uchun formulani topishimiz kerak. Biz $x$ o'qida markazlashtirilgan disklardan foydalanayotganimiz sababli, maydon formulasi $x$ nuqtai nazaridan bo'lishi kerakligini bilamiz. Shuning uchun, egri chiziq tenglamasi $x$ (bu holda u allaqachon mavjud) nuqtai nazaridan bo'lishi kerak.

Mana, biz uchun kerak bo'lgan har xil miqdorlar bilan vakil diskning yana bir eskizi.

$A sketch of a representative disk with all of the quantities needed put in. The disk is centered on the x-axis and placed at some $x$, and the distance from the x-axis up to the curve $y = 7 - {x^2}$ is labeled radius = $7 - {x^2}$.$

Eskizdan ko'rinib turibdiki, disk $x$ o'qi markazida joylashgan va ba'zi $x$ da joylashgan. Diskning radiusi - $x$ o'qidan qattiq jismning chetini belgilovchi egri chiziqgacha bo'lgan masofa. Boshqacha aytganda,

\[{\mbox{Radius }} = 7 - {x^2}\]

Diskning maydoni u holda,

\[A\left( x \right) = \pi {\left( {{\mbox{Radius}}} \right)^2} = \pi {\left( {7 - {x^2}} \right)^2} = \pi \left( {49 - 14{x^2} + {x^4}} \right)\]

4-qadam

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

Integral chegaralari uchun biz qattiq jismdagi "birinchi" disk $x = - 2$ da, "oxirgi" disk esa $x = 2$ da paydo bo'lishini ko'rishimiz mumkin. Bizning chegaralarimiz quyidagicha: $ - 2 \le x \le 2$.

Keyin hajm,

\[V = \int_{{ - 2}}^{2}{{\pi \left( {49 - 14{x^2} + {x^4}} \right)\,dx}} = \left. {\pi \left( {49x - \frac{{14}}{3}{x^3} + \frac{1}{5}{x^5}} \right)} \right|_{ - 2}^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{2012}}{{15}}\pi }}\]