Asosiy tarkibga o'tish

6.4-bo'lim : Silindrli hajm

4. \(y = 4x\) va \(y = {x^3}\) bilan chegaralangan hududni \(x\) o'qi atrofida aylantirish natijasida olingan qattiq jismning hajmini silindrlar usulidan foydalaning. Ushbu muammo uchun \(x \ge 0\) deb hisoblang.

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 \(y = 4x\) is the upper boundary and the curve \(y = {x^3}\) is the lower boundary, and the two meet at the origin and again at \(\left( {2,8} \right)\), which is marked and labeled. The region between them is shaded in and a small circular arrow on the x-axis marks it as the axis of rotation.

Grafikda ko'rsatilgan kesishish nuqtalarini olish uchun bizga birozdan keyin kerak bo'ladigan tenglamalarni bir-biriga tenglashtirish va yechish kerak.

\[\begin{align*}{x^3} & = 4x\\ {x^3} - 4x & = 0\\ x\left( {{x^2} - 4} \right) & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in} x = 0,\,\,\,\,\,\,x = \pm 2\hspace{0.25in} \Rightarrow \hspace{0.25in}\left( {0,0} \right)\,\,\,\,\& \,\,\,\,\left( {2,8} \right)\end{align*}\]

E'tibor bering, muammo bayonotida \(x \ge 0\) deb taxmin qilingan va shuning uchun biz \(x = - 2\) kesishish nuqtasidan foydalanmaymiz.

Maslahat: Inqilobning qattiq jismining qanday ko'rinishini chizishga yaxshi urinib ko'ring va vakilli silindrda eskiz qiling.

E'tibor bering, buni qilish qiyin bo'lishi mumkin, ayniqsa siz juda ko'p ko'rinadigan odam bo'lmasangiz. Biroq, vakillik silindriga ega bo'lish, biz maydon formulasini yozishga borganimizda katta yordam berishi mumkin. Bundan tashqari, revolyutsiyaning qattiq jismining eskizisiz vakili silindrni olish qiyin bo'lishi mumkin. Shunday qilib, ushbu eskizlarni olish uchun qo'lingizdan kelganini qiling.

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, football shaped solid that comes to a point at the origin on the left and is widest near \(x = 2\), with a cone shaped hollow scooped out of the left end.

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 x-axis and lies 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 x-axis and lies inside the solid.
3-qadam

Endi silindrning sirt maydoni uchun formulani topishimiz kerak. Biz \(x\) o'qida joylashgan silindrlardan foydalanayotganimiz sababli, maydon formulasi \(y\) nuqtai nazaridan bo'lishi kerakligini bilamiz. Shuning uchun biz egri chiziqlar tenglamalarini \(y\) ko'rinishida qayta yozishimiz kerak bo'ladi.

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 x-axis with its upper edge at some \(y\), and the distance from the x-axis up to that edge is labeled radius = \(y\). The two curves have been rewritten in terms of \(y\) as \(x = \frac{1}{4}y\) and \(x = \sqrt[3]{y}\), and the horizontal distance between them is labeled width = \({y^{\frac{1}{3}}} - \frac{1}{4}y\), with the two individual distances also marked.

Eskizdan biz silindrning \(x\) o'qida joylashganligini va silindrning yuqori qirrasi qandaydir \(y\) da joylashganligini ko'rishimiz mumkin.

Tsilindrning radiusi - bu \(x\) o'qidan silindrning yuqori chetigacha bo'lgan masofa (i.e. \(y\)).

Tsilindrning o'ng qirrasi qattiq jismning o'ng qismini belgilovchi egri chiziqda joylashgan va \(y\) o'qidan \({y^{\frac{1}{3}}}\) masofasi. Tsilindrning chap qirrasi qattiq jismning chap qismini belgilovchi egri chiziqda joylashgan va \(y\) o'qidan \(\frac{1}{4}y\) masofada joylashgan. Keyin balandlik bu ikkalasining farqidir.

Shunday qilib, silindrning radiusi va kengligi,

\[{\mbox{Radius}} = y\hspace{0.25in}\hspace{0.25in}{\mbox{Width}} = {y^{\frac{1}{3}}} - \frac{1}{4}y\]

U holda silindrning maydoni,

\[A\left( y \right) = 2\pi \left( {{\mbox{Radius}}} \right)\left( {{\mbox{Height}}} \right) = 2\pi \left( y \right)\left( {{y^{\frac{1}{3}}} - \frac{1}{4}y} \right) = 2\pi \left( {{y^{\frac{4}{3}}} - \frac{1}{4}{y^2}} \right)\]
4-qadam

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

1-bosqichdagi grafikdan biz qattiq jismdagi "birinchi" silindr \(y = 0\) da, "oxirgi" silindr esa \(y = 8\) da paydo bo'lishini ko'rishimiz mumkin. Bizning chegaralarimiz quyidagicha: \(0 \le y \le 8\).

Keyin hajm,

\[V = \int_{0}^{8}{{2\pi \left( {{y^{\frac{4}{3}}} - \frac{1}{4}{y^2}} \right)\,dy}} = 2\left. {\pi \left( {\frac{3}{7}{y^{\frac{7}{3}}} - \frac{1}{{12}}{y^3}} \right)} \right|_0^8 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{512}}{{21}}\pi }}\]