MoreVolume - 4-masala yechimi | MathModel.uz

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

4. Bazasi $y = \sqrt {4 - x} $, $x = - 4$ va $x$ o‘qi bilan chegaralangan mintaqa bo‘lgan “silindr”dan kesilgan xanjar hajmini toping. Takozning yuqori va pastki qismlari orasidagi burchak $\frac{\pi }{3}$. "Tsilindr" va takozning eskizini quyidagi rasmga qarang (eskizda ijobiy $x$ o'qi va musbat $y$ o'qi ko'rsatilgan - ular faqat boshqa yo'nalishda).

$Two sketches side by side. The one on the left, labeled Cylinder, shows the solid whose base is the region bounded by $y = \sqrt {4 - x} $, $x = - 4$ and the x-axis, drawn as a tall vertical column standing on that base. The one on the right, labeled Wedge, shows the thin wedge cut out of that column by a plane through the x-axis making an angle of $\frac{\pi }{3}$ with the base. In both sketches the curve $y = \sqrt {4 - x} $ and the line $x = - 4$ are labeled and the angle $\frac{\pi }{3}$ is marked at the near corner.$

Yechim

Kerakli bo'lmasa-da, keling, "tsilindr" ning eskizini qayta tiklaymiz va muammo bayonidan faqat bu safar kesma qanday ko'rinishini ham chizamiz.

$Two sketches of the “cylinder” and the wedge with a typical cross-section drawn in on each and labeled $A\left( x \right)$. The one on the left is the full “cylinder” standing on the region bounded by $y = \sqrt {4 - x} $, $x = - 4$ and the x-axis, and the one on the right is just the wedge cut out of it. On both, the cross-section taken perpendicular to the x-axis is a shaded right triangle, and the curve $y = \sqrt {4 - x} $, the line $x = - 4$ and the angle $\frac{\pi }{3}$ at the near corner are labeled.$

Kesma $x$ o'qiga perpendikulyar bo'lgani uchun biz kesmani $x$ o'qi bo'ylab harakatlantirganda uning maydonini o'zgartiramiz va shuning uchun kesma maydoni $x$ funktsiyasi bo'ladi, i.e. $A\left( x \right)$. Shuni ham yodda tutingki, chizmalarda ko'rsatilganidek, kesma to'g'ri burchakli uchburchak bo'ladi.

2-qadam

Yuqoridagi eskizlar qattiq va ko'ndalang kesimlar 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 mavjud.

$Two sketches. The one on the left is the wedge with the cross-section drawn in for reference and the curve $y = \sqrt {4 - x} $ labeled. The one on the right is just the cross-section itself, a right triangle with the base labeled $b$, the vertical side labeled $h$ and the angle $\frac{\pi }{3}$ at the lower left corner. The dot at the left end of the base marks where the x-axis is and the dot at the right end marks the curve $y = \sqrt {4 - x} $, which is also where the y-axis direction is shown.$

Chapdagi eskiz deyarli biz ko'rgan eskizdir va o'ng tomonda ko'rsatilgan haqiqiy kesma uchun mos yozuvlar nuqtasini berish uchun kiritilgan.

O'ngdagi eskizda ta'kidlanganidek, biz $b$ uchburchak asosini va $h$ uchburchakning balandligini chaqiramiz. Bundan tashqari, taglikning chap tomonidagi nuqta $x$ o'qi kesmada joylashgan joyni va poydevorning o'ng tomonidagi nuqta qattiq jismning chetini (va shuning uchun xanjar) belgilaydigan egri chiziqni ifodalaydi.

Ushbu eskizdan shuni tushunish kerakki, taglikning uzunligi shunchaki $x$ o'qidan egri chiziqgacha bo'lgan masofa yoki,

\[b = \sqrt {4 - x} \]

Xuddi shunday, balandlikni asosiy to'g'ri burchakli trigdan topish mumkin.

\[\tan \left( {\frac{\pi }{3}} \right) = \frac{h}{b}\hspace{0.5in} \Rightarrow \hspace{0.5in}h = b\tan \left( {\frac{\pi }{3}} \right) = \sqrt 3 \sqrt {4 - x} \]

Shunday qilib, $x$ bo'yicha kesmaning maydoni formulasi quyidagicha bo'ladi:

\[A\left( y \right) = \frac{1}{2}bh = \frac{{\sqrt 3 }}{2}{\left( {\sqrt {4 - x} } \right)^2} = \frac{{\sqrt 3 }}{2}\left( {4 - x} \right)\]

3-qadam

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

\[V = \int_{a}^{b}{{A\left( x \right)\,dx}}\]

Muammo bayonotidagi eskizlardan yoki 1-qadamdan biz “birinchi” kesma $x = - 4$ (“silindr”ning orqa uchi) da “oxirgi” kesma $x = 4$ da (“silindr” ning old uchi) paydo bo‘lishini ko‘rishimiz mumkin (bu yerda egri chiziq MATH40 chegarasi bilan kesishadi). integral.

Keyin hajm,

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