VolumeWithRings | MathModel.uz

6.3-bo'lim: Halqali ovoz balandligi

Ushbu bo'limda biz inqilobning qattiq hajmini ko'rib chiqamiz. Biz birinchi navbatda inqilobning mustahkamligi nima ekanligini aniqlashimiz kerak. Qattiq inqilobni olish uchun biz $\left[ {a,b} \right]$ oralig'ida $y = f\left( x \right)$ funksiyasidan boshlaymiz.

This is the graph of some unknown function on the domain a<x<b. It is completely in the 1st quadrant and looks like a parabola that opens upwards. The left end of the graph is lower than right end of the graph.

Keyin bu egri chiziqni ma'lum o'q bo'ylab aylantiramiz va inqilobning qattiq yuzasini olamiz. Ushbu munozara uchun keling, egri chiziqni $x$ o'qi atrofida aylantiramiz, garchi u har qanday vertikal yoki gorizontal o'q bo'lishi mumkin. Yuqoridagi egri chiziq uchun buni qilish quyidagi uch o'lchovli hududni beradi.

This is the graph of the solid we get from rotating the graph from above about the x-axis. It looks like an odd vase laying on its side. It has a wide “mouth” on the right side with a narrower “mouth” on the left side and an even narrow middle that corresponds to the vertex of the parabola from the graph.

Keyingi ikki bo'lim davomida biz nima qilishni xohlaymiz - bu ob'ekt hajmini aniqlash.

In Hudud va hajm formulalari Qo'shimchalar bo'limida biz ushbu qattiq jismning hajmi uchun quyidagi formulalarni oldik.

Inqilobning qattiq jismlari uchun hajm formulalari

\[V = \int_{{\,a}}^{{\,b}}{{A\left( x \right)\,dx}}\hspace{0.75in}V = \int_{{\,c}}^{{\,d}}{{A\left( y \right)\,dy}}\]

bu yerda, $A\left( x \right)$ va $A\left( y \right)$ qattiq jismning kesma maydoni funksiyalari. Ko'ndalang kesim maydonini olishning ko'plab usullari mavjud va biz keyingi ikki bo'limda ikkitasini (yoki unga qarashingizga qarab uchtasini) ko'ramiz. $A\left( x \right)$ yoki $A\left( y \right)$ dan foydalanishimiz har bir muammo uchun ishlatiladigan usul va aylanish o'qiga bog'liq bo'ladi.

Ko'ndalang kesim maydonini olishning eng oson usullaridan biri ob'ektni aylanish o'qiga perpendikulyar ravishda kesishdir. Agar ob'ekt qattiq bo'lsa, kesma qattiq disk bo'ladi (yuqoridagi misolimizda bo'lgani kabi) yoki qattiq qismning bir qismini o'yib tashlagan bo'lsak, halqa bo'ladi (biz buni oxir-oqibat ko'ramiz).

Agar biz qattiq diskni oladigan bo'lsak, bu maydon,

\[A = \pi {\left( {{\mbox{radius}}} \right)^2}\]

bu erda radius funktsiyaga va aylanish o'qiga bog'liq bo'ladi.

Agar biz uzuk olgan bo'lsak, bu maydon,

Ringning maydoni

\[A = \pi \left( {{{\left( \begin{array}{c}{\mbox{outer}}\\ {\mbox{radius}}\end{array} \right)}^2} - {{\left( \begin{array}{c}{\mbox{inner}}\\ {\mbox{radius}}\end{array} \right)}^2}} \right)\]

Bu erda yana ikkala radius ham berilgan funktsiyalarga va aylanish o'qiga bog'liq bo'ladi. Shuni ham yodda tutingki, qattiq diskda biz ichki radiusni nol deb hisoblashimiz mumkin va biz qattiq disk uchun to'g'ri formulaga erishamiz va shuning uchun bu foydalanish uchun ancha umumiy formuladir.

Bundan tashqari, ikkala holatda ham maydon $x$ funktsiyasi yoki $y$ funktsiyasi bo'ladimi, biz ko'rib turganimizdek aylanish o'qiga bog'liq bo'ladi.

Ushbu usul ko'pincha deyiladi disklar usuli yoki halqalar usuli.

Keling, bir misol qilaylik.

1-misol $y = {x^2} - 4x + 5$, $x = 1$, $x = 4$ va $x$ o'qi bilan chegaralangan hududni $x$ o'qi atrofida aylantirish natijasida olingan qattiq jismning hajmini aniqlang.

Yechim

Birinchi narsa, chegara hududining eskizini va mintaqani $x$ o'qi atrofida aylantirish natijasida olingan qattiq jismni olishdir. Bizga egri chiziqning mukammal eskizi kerak emas, bizga chegaralangan hudud qanday ko'rinishini his qilish imkonini beradigan narsa kerak, shunda biz qattiq jismning qisqacha eskizini olishimiz mumkin. Shuni yodda tutgan holda shuni ta'kidlashimiz mumkinki, birinchi tenglama faqat $\left( {2,1} \right)$ cho'qqisi bo'lgan paraboladir (siz buni qanday olish kerakligini eslaysiz. cho'qqi parabola to'g'rimi?) va yuqoriga ochiladi va shuning uchun biz uning eskizini chizishga ko'p vaqt sarflashimiz shart emas.

Mana bu ikkala eskiz.

$This is the graph of $y={{x}^{2}}-4x+5$ on the domain 1<x<4. It is completely in the 1st quadrant and it’s vertex is at (2,1).$ This is the graph of the solid we get from rotating the graph from above about the x-axis. It looks like an odd vase laying on its side. It has a wide “mouth” on the right side with a narrower “mouth” on the left side and an even narrow middle that corresponds to the vertex of the parabola from the graph.

Yaxshi, kesmani olish uchun biz har qanday $x$ da qattiqni kesib oldik. Quyida odatiy kesmani ko'rsatadigan bir nechta eskizlar mavjud. O'ngdagi eskizda ob'ektning qopqog'isiz odatiy kesma bilan kesilgan qismi ko'rsatilgan. Chapdagi eskizda biz aylanayotgan egri chiziq, shuningdek, qattiq jismning pastki qismidagi oyna tasviri ko'rsatilgan.

In this graph we cut away the front half of the solid from the previous image and put in a typical disk cross section that is centered on the x-axis. The disk will be described in more detail in the next image

For this image we do away with the solid completely and put in just the bounding curve from the first image in this example and it’s mirrored image around the x-axis. The top and bottom of a typical disk cross section, centered on the x-axis, then touches both of these curves. The radius of the curve is then the distance from the x-axis to the original bounding curve.

Bu holda radius shunchaki $x$ o'qidan egri chiziqgacha bo'lgan masofa va bu yuqorida ko'rsatilganidek, ma'lum $x$ funksiya qiymatidan boshqa narsa emas. Keyin kesma maydoni,

\[A\left( x \right) = \pi {\left( {{x^2} - 4x + 5} \right)^2} = \pi \left( {{x^4} - 8{x^3} + 26{x^2} - 40x + 25} \right)\]

Keyinchalik, integratsiya chegaralarini aniqlashimiz kerak. Chapdan o'ngga ishlaganda birinchi kesma $x = 1$ da, oxirgi kesma esa $x = 4$ da sodir bo'ladi. Bu integratsiya chegaralari.

U holda bu qattiq jismning hajmi,

\[\begin{align*}V &= \int_{{\,a}}^{{\,b}}{{A\left( x \right)\,dx}}\\ & = \pi \int_{{\,1}}^{{\,4}}{{{x^4} - 8{x^3} + 26{x^2} - 40x + 25\,dx}}\\ & = \pi \left. {\left( {\frac{1}{5}{x^5} - 2{x^4} + \frac{{26}}{3}{x^3} - 20{x^2} + 25x} \right)} \right|_1^4\\ & = \frac{{78\pi }}{5}\end{align*}\]

Yuqoridagi misolda ob'ekt qattiq ob'ekt edi, lekin eng qiziqarli narsalar qattiq bo'lmagan narsalardir, shuning uchun ulardan birini ko'rib chiqaylik.

2-misol Birinchi kvadrantda joylashgan $y = \sqrt[3]{x}$ va $\displaystyle y = \frac{x}{4}$ bilan chegaralangan mintaqa qismini y o'qi atrofida aylantirish natijasida olingan qattiq jismning hajmini aniqlang.

Yechim

Birinchidan, chegaralangan hududning grafigini va ob'ektning grafigini olamiz. Esda tutingki, biz chegaralangan hududning faqat birinchi kvadrantda joylashgan qismini xohlaymiz. Uchinchi kvadrantda chegaralangan hududning bir qismi ham bor, lekin biz bu muammo uchun buni xohlamaymiz.

$This is the graphs of $y=\frac{x}{4}$ and $y=\sqrt[3]{x}$ (looks basically like the graph of $\sqrt{x}$) on the domain 0<x<8. This is the bounded region we are after for this example.$ This is the graph of the solid we get from rotating the graph from above about the y-axis. It looks basically like a cone. The inside is hollow and the “walls” of the cone are in the shaped of the bounded region for this example.

This is the graph of the solid we get from rotating the graph from above about the y-axis. It looks basically like a cone. The inside is hollow and the “walls” of the cone are in the shaped of the bounded region for this example.

Ushbu muammoga e'tibor berish kerak bo'lgan bir nechta narsa bor. Birinchidan, biz oxirgi misolda bo'lgani kabi, to'liq ichki qismni emas, balki faqat bu qattiq "devorlar" hajmini qidiramiz.

Keyinchalik, biz aylanish o'qiga perpendikulyar ob'ektni kesib, kesmani olamiz. Ushbu misol uchun kesma halqa bo'ladi (esda tuting, biz faqat devorlarga qaraymiz) va ba'zi $y$ da gorizontal bo'ladi. Bu shuni anglatadiki, halqaning ichki va tashqi radiusi $x$ qiymatlari bo'ladi va shuning uchun biz funktsiyalarimizni $x = f\left( y \right)$ shaklida qayta yozishimiz kerak bo'ladi. Mana, ushbu misol uchun to'g'ri shaklda yozilgan funktsiyalar.

\[\begin{align*}y & = \sqrt[3]{x}\hspace{0.5in}\ \Rightarrow \hspace{0.5in}x = {y^3}\\ y & = \frac{x}{4}\hspace{0.65in} \Rightarrow \hspace{0.5in}x = 4y\end{align*}\]

Mana bu ob'ekt devorlarining chegaralarining bir nechta eskizlari, shuningdek, odatiy halqa. Chapdagi eskiz o'ngdagi rasmga ozgina kontekst berish uchun ob'ektning orqa qismini o'z ichiga oladi.

In this graph we cut away the front half of the solid from the previous image and put in a typical ring cross section that is centered on the y-axis. The ring will be described in more detail in the next image.

For this image we do away with the solid completely and put in just the bounding curves from the first image in this example and their mirrored image around the y-axis. The ring is centered on the y-axis. Its outer radius of the ring is the distance from y-axis to the line (the outer bounded curve) and the inner radius of the ring is the distance from the y-axis to the cube root (the inner bounded curve).

Bu holda ichki radius $y$ o'qidan ichki egri chiziqgacha bo'lgan masofa, tashqi radius esa $y$ o'qidan tashqi egri chiziqgacha bo'lgan masofadir. Bularning ikkalasi ham $x$ masofalari va shuning uchun yuqorida ko'rsatilgandek egri chiziqlar tenglamalari bilan berilgan.

Keyin kesma maydoni,

\[A\left( y \right) = \pi \left( {{{\left( {4y} \right)}^2} - {{\left( {{y^3}} \right)}^2}} \right) = \pi \left( {16{y^2} - {y^6}} \right)\]

Qattiq jismning pastki qismidan yuqoriga qarab ishlaganda biz birinchi kesma $y = 0$ da, oxirgi kesma esa $y = 2$ da bo‘lishini ko‘rishimiz mumkin. Bu integratsiya chegaralari bo'ladi. Keyin hajm,

\[\begin{align*}V &= \int_{{\,c}}^{{\,d}}{{A\left( y \right)\,dy}}\\ & = \pi \int_{{\,0}}^{{\,2}}{{16{y^2} - {y^6}\,dy}}\\ & = \pi \left. {\left( {\frac{{16}}{3}{y^3} - \frac{1}{7}{y^7}} \right)} \right|_0^2\\ & = \frac{{512\pi }}{{21}}\end{align*}\]

Ushbu ikkita misol bilan biz ushbu usul haqida umumiy xulosa chiqarishimiz mumkin. Agar biz gorizontal o'q atrofida aylansak (masalan, $x$ o'qi) u holda kesma maydoni $x$ funktsiyasi bo'ladi. Xuddi shunday, agar biz vertikal o'q atrofida aylansak (masalan, $y$‑o'qi) u holda kesma maydoni $y$ funktsiyasi bo'ladi.

Ushbu bo'limdagi qolgan ikkita misol, biz doimo $x$ yoki $y$ o'qi atrofida aylanish g'oyasiga ko'nikmasligimizga ishonch hosil qiladi.

3-misol $y = {x^2} - 2x$ va $y = x$ bilan chegaralangan hududni $y = 4$ chizig‘i atrofida aylantirish natijasida olingan qattiq jism hajmini aniqlang.

Yechim

Avval chegaralangan hududni va qattiq grafigini olamiz.

$This is the graphs of $y=x$ and $y={{x}^{2}}-2x$ on the domain 0<x<3. The vertex of the parabola is at (1,-1) and the parabola opens upwards intersecting the line at the endpoints of the domain. This is the bounded region we are after for this example.$ This is the graph of the solid we get from rotating the graph from above about the line y=4. It looks basically like a cone. The inside is hollow and the “walls” of the cone are in the shaped of the bounded region for this example. Both ends of this cone are open.

This is the graph of the solid we get from rotating the graph from above about the line y=4. It looks basically like a cone. The inside is hollow and the “walls” of the cone are in the shaped of the bounded region for this example. Both ends of this cone are open.

Shunga qaramay, biz ushbu ob'ektning devorlarining hajmini qidiramiz. Bundan tashqari, biz gorizontal o'q atrofida aylanayotganimiz sababli, kesma maydoni $x$ funktsiyasi bo'lishini bilamiz.

Mana bu ob'ekt devorlarining chegaralarining bir nechta eskizlari, shuningdek, odatiy halqa. Chapdagi eskiz o'ngdagi rasmga ozgina kontekst berish uchun ob'ektning orqa qismini o'z ichiga oladi.

In this graph we cut away the front half of the solid from the previous image and put in a typical ring cross section that is centered on the line y=4. The ring will be described in more detail in the next image.

Endi biz bu erda ichki va tashqi radiusni aniqlashda ehtiyot bo'lishimiz kerak, chunki ular avvalgi ikkita misoldagidek oddiy bo'lmaydi.

Keling, ichki radiusdan boshlaylik, chunki bu biroz aniqroq. Birinchidan, ichki radius $x$ EMAS. $x$ o'qidan halqaning ichki chetigacha bo'lgan masofa $x$, lekin biz radiusni xohlaymiz va bu aylanish o'qidan halqaning ichki chetigacha bo'lgan masofa. Shunday qilib, biz bilamizki, aylanish o'qidan $x$ o'qigacha bo'lgan masofa 4, $x$ o'qidan ichki halqagacha bo'lgan masofa $x$. Ichki radius bu ikkisi orasidagi farq bo'lishi kerak. Yoki,

\[{\mbox{inner radius}} = 4 - x\]

Tashqi radius xuddi shunday ishlaydi. Tashqi radius,

\[{\mbox{outer radius}} = 4 - \left( {{x^2} - 2x} \right) = - {x^2} + 2x + 4\]

E'tibor bering, yuqoridagi eskizdagi odatiy halqaning joylashishini hisobga olsak, tashqi radius formulasi unchalik to'g'ri ko'rinmasligi mumkin, lekin bu aslida to'g'ri. Chizilganidek, halqaning tashqi qirrasi $x$ o'qi ostida joylashgan va bu nuqtada funktsiyaning qiymati manfiy bo'ladi va shuning uchun biz tashqi radius uchun formulada ayirishni amalga oshirganimizda, biz bu masofani 4 ga qo'shishning aniq ta'siriga ega bo'lgan manfiy sonni ayiramiz va bu to'g'ri tashqi radiusni beradi. Xuddi shunday, agar tashqi chet $x$ o'qidan yuqori bo'lsa, funktsiya qiymati ijobiy bo'ladi va shuning uchun biz bu erda halol ayirish qilamiz va bu holda biz to'g'ri radiusni olamiz.

Ushbu holat uchun kesma maydoni,

\[A\left( x \right) = \pi \left( {{{\left( { - {x^2} + 2x + 4} \right)}^2} - {{\left( {4 - x} \right)}^2}} \right) = \pi \left( {{x^4} - 4{x^3} - 5{x^2} + 24x} \right)\]

Birinchi qo'ng'iroq $x = 0$ da, oxirgi qo'ng'iroq esa $x = 3$ da sodir bo'ladi va shuning uchun bu bizning integratsiya chegaralarimizdir. Keyin hajm,

\[\begin{align*}V & = \int_{{\,a}}^{{\,b}}{{A\left( x \right)\,dx}}\\ & = \pi \int_{{\,0}}^{{\,3}}{{{x^4} - 4{x^3} - 5{x^2} + 24x\,dx}}\\ & = \pi \left. {\left( {\frac{1}{5}{x^5} - {x^4} - \frac{5}{3}{x^3} + 12{x^2}} \right)} \right|_0^3\\ & = \frac{{153\pi }}{5}\end{align*}\]

4-misol $y = 2\sqrt {x - 1} $ va $y = x - 1$ bilan chegaralangan hududni $x = - 1$ chiziq atrofida aylantirish natijasida olingan qattiq jism hajmini aniqlang.

Yechim

Oldingi misollardagidek, avval chegaralangan hudud va qattiq jismning grafigini tuzamiz.

$This is the graphs of $y=x-1$ and $y=2\sqrt{x-1}$ on the domain 1<x<5. The square root is always larger than the line and only intersects the line at the endpoints of the domain. This is the bounded region we are after for this example.$ This is the graph of the solid we get from rotating the graph from above about the line x=-1. It looks basically like a cone. The inside is hollow and the “walls” of the cone are in the shaped of the bounded region for this example. Both ends of this cone are open.

This is the graph of the solid we get from rotating the graph from above about the line x=-1. It looks basically like a cone. The inside is hollow and the “walls” of the cone are in the shaped of the bounded region for this example. Both ends of this cone are open.

Keling, e'tibor bering, chunki biz vertikal o'q atrofida aylanamiz va shuning uchun kesma maydoni $y$ funktsiyasi bo'ladi. Bu shuni anglatadiki, biz $y$ nuqtai nazaridan ularni olish uchun funktsiyalarni qayta yozishimiz kerak.

\[\begin{align*}y &= 2\sqrt {x - 1} \hspace{0.5in} \Rightarrow \hspace{0.5in}x = \frac{{{y^2}}}{4} + 1\\ y & = x - 1\hspace{0.75in} \Rightarrow \hspace{0.5in}x = y + 1\end{align*}\]

Mana bu ob'ekt devorlarining chegaralarining bir nechta eskizlari, shuningdek, odatiy halqa. Chapdagi eskiz o'ngdagi rasmga ozgina kontekst berish uchun ob'ektning orqa qismini o'z ichiga oladi.

In this graph we cut away the front half of the solid from the previous image and put in a typical ring cross section that is centered on the line x=-1. The ring will be described in more detail in the next image.

Bu holat uchun ichki va tashqi radius avvalgi misoldan ham o'xshash, ham farq qiladi. Bu misol radiuslar faqat funktsiyalar emasligi ma'nosida o'xshashdir. Ushbu misolda funksiyalar $y$ o'qidan halqalarning chetlarigacha bo'lgan masofalardir. Biroq, halqaning markazi $y$ o'qidan 1 masofada joylashgan. Bu shuni anglatadiki, markazdan qirralargacha bo'lgan masofa aylanish o'qidan $y$ o'qiga (1 masofa) va keyin $y$ o'qidan halqalarning chetiga qadar bo'lgan masofadir.

Demak, radiuslar funksiyalar plyus 1 bo'lib, bu misol oldingi misoldan farq qiladi. Bu erda biz funktsiya qiymatiga masofani qo'shishimiz kerak edi, oldingi misolda biz ushbu masofadan funktsiyani ayirishimiz kerak edi. E'tibor bering, eskizlarsiz ushbu muammolarning radiuslarini olish qiyin bo'lishi mumkin.

Xulosa qilib aytganda, biz ushbu misol uchun ichki va tashqi radius uchun quyidagilarni oldik.

\[\begin{align*}{\mbox{outer radius}} & = y + 1 + 1 = y + 2\\ {\mbox{inner radius}} & = \frac{{{y^2}}}{4} + 1 + 1 = \frac{{{y^2}}}{4} + 2\end{align*}\]

Keyin kesma maydoni,

\[A\left( y \right) = \pi \left( {{{\left( {y + 2} \right)}^2} - {{\left( {\frac{{{y^2}}}{4} + 2} \right)}^2}} \right) = \pi \left( {4y - \frac{{{y^4}}}{{16}}} \right)\]

Birinchi qo'ng'iroq $y = 0$ da, oxirgi halqa esa $y = 4$ da sodir bo'ladi va shuning uchun bu bizning integratsiya chegaralarimiz bo'ladi.

Hajmi,

\[\begin{align*}V & = \int_{{\,c}}^{{\,d}}{{A\left( y \right)\,dy}}\\ & = \pi \int_{{\,0}}^{{\,4}}{{4y - \frac{{{y^4}}}{{16}}\,dy}}\\ & = \pi \left. {\left( {2{y^2} - \frac{1}{{80}}{y^5}} \right)} \right|_0^4\\ & = \frac{{96\pi }}{5}\end{align*}\]