Asosiy tarkibga o'tish

6.5-bo'lim : Ko'proq ovoz balandligi muammolari

2. Asosi \(r\) radiusli disk va kesmalari kvadrat bo‘lgan qattiq jismning hajmini toping. Ko'ndalang kesimlarning eskizini ko'rish uchun quyidagi rasmga qarang.

A sketch showing a typical cross-section of the solid. The base of the solid is the disk \({x^2} + {y^2} = {r^2}\), drawn in perspective as an ellipse, and a shaded square stands vertically on the disk with its base running across the disk. The distance from the center of the disk out along the base is labeled \(x\) and half the width of the square is labeled \(y\).
Yechim

Bu erda uch xil burchakdagi qattiq jismning bir nechta eskizlari. Malumot uchun musbat \(x\) o'qi va musbat \(y\) o'qi ko'rsatilgan.

Three sketches of the solid from different angles with a typical cross-section shown on each and labeled \(A\left( y \right)\). The base of the solid is a disk and the cross-sections taken perpendicular to the y-axis are squares, so the solid looks like a rounded, dome shaped block that is flat on the bottom. For reference the positive x-axis and positive y-axis are shown on each sketch.

Kesma \(y\) o'qiga perpendikulyar bo'lgani uchun biz kesmani \(y\) o'qi bo'ylab harakatlantirganda uning maydonini o'zgartiramiz va shuning uchun kesma maydoni \(y\) funktsiyasi bo'ladi, i.e. \(A\left( y \right)\).

2-qadam

Yuqoridagi eskizlar qattiq jismning qanday ko'rinishini his qilish uchun yoqimli bo'lsa-da, bizga kerak bo'lgan narsa shunchaki kesmaning eskizidir. Shunday qilib, bu erda tasavvurlar maydonining bir nechta eskizlari.

Two sketches of the cross-sectional area. The one on the left is the figure from the problem statement, a square standing on the disk with the left half of the boundary circle colored orange and the right half colored green. The one on the right looks straight down at the base from above and shows the circle \({x^2} + {y^2} = {r^2}\) with the left half labeled \(x = - \sqrt {{r^2} - {y^2}} \) in orange and the right half labeled \(x = \sqrt {{r^2} - {y^2}} \) in green, and the cross-section itself is the horizontal red line across the circle.

Chapdagi eskiz haqiqatan ham muammo bayonida berilgan grafik bo'lib, biz o'ng/chap tomonlarni rangga bo'yatganimiz uchun u o'ngdagi eskizga mos keladi. O'ngdagi eskiz to'g'ridan-to'g'ri yuqoridan kesmaga qaraydi va qizil chiziq bilan ko'rsatilgan.

Keling, faqat kesmaning qisqacha eskizini olamiz va har bir kvadratning yon tomonining uzunligini \(s\) deb ataymiz.

A sketch of a single cross-section, which is a square with each side labeled \(s\). Along the bottom of the square a black dot marks where the y-axis passes through, splitting the base into two pieces each labeled \(\frac{s}{2}\), and the two ends of the base are marked with dots and labeled \(x = - \sqrt {{r^2} - {y^2}} \) on the left and \(x = \sqrt {{r^2} - {y^2}} \) on the right.

Endi pastki qismida biz \(y\) o'qining kesmada joylashgan joyini qora nuqta bilan belgilab oldik va to'q sariq va yashil nuqtalar aylananing chap va o'ng qismlarini ko'rsatadi. Bundan tashqari, ko'rishimiz mumkinki, kesma qandaydir \(y\) da joylashgan bo'lsa, yashil nuqta \(y\) o'qidan \(\sqrt {{r^2} - {y^2}} \) masofasida bo'lishi kerak. Xuddi shunday, to'q sariq nuqta ham \(y\) o'qidan \(\sqrt {{r^2} - {y^2}} \) masofasida bo'lishi kerak (esda tuting, biz bu erda masofa ijobiy bo'lishini xohlaymiz va shuning uchun ijobiy masofani olish uchun funktsiyadan minus belgisini tushiramiz).

Endi biz kvadratning maydoni shunchaki \({s^2}\) ekanligini bilamiz va yuqoridagi muhokamadan shuni ko'ramizki,

\[\frac{s}{2} = \sqrt {{r^2} - {y^2}} \hspace{0.5in} \Rightarrow \hspace{0.5in}s = 2\sqrt {{r^2} - {y^2}} \]

Shunday qilib, \(y\) nuqtai nazaridan kesmaning maydoni formulasi:

\[A\left( y \right) = {s^2} = {\left( {2\sqrt {{r^2} - {y^2}} } \right)^2} = 4\left( {{r^2} - {y^2}} \right)\]
3-qadam

Nihoyat, bizga hajmning o'zi kerak. Quyidagi integralni baholash orqali hajm topilishini bilamiz.

\[V = \int_{c}^{d}{{A\left( y \right)\,dy}}\]

Bizda allaqachon 2-bosqichdan \(A\left( y \right)\) formulasi bor va 1-bosqichdagi eskizlardan biz “birinchi” kesma \(y = - r\) da, “oxirgi” kesma esa \(y = r\) da sodir bo‘lishini ko‘rishimiz mumkin, shuning uchun bu integral uchun chegaralardir.

Keyin hajm,

\[V = \int_{{ - r}}^{r}{{4\left( {{r^2} - {y^2}} \right)\,dy}} = \left. {4\left( {y{r^2} - \frac{1}{3}{y^3}} \right)} \right|_{ - r}^r = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{16}}{3}{r^3}}}\]

\(r\) integral va maydon formulasidan hayajonlanmang. Bu shunchaki doimiy. Haqiqatan ham o'zgaruvchan yagona harf - \(y\).