MoreVolume - 3-masala yechimi | MathModel.uz

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

3. Bazasi $x = 2 - {y^2}$ va $x = {y^2} - 2$ bilan chegaralangan mintaqa bo‘lgan va ko‘ndalang kesimlari asosi $y$ o‘qiga perpendikulyar bo‘lgan teng yonli uchburchaklar, asosi va uzunligi teng bo‘lgan ikki tomoni orasidagi burchak $\frac{\pi }{4}$ 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 region between the two parabolas $x = {y^2} - 2$ and $x = 2 - {y^2}$, drawn in perspective, and a shaded isosceles triangle stands vertically on the base with its base running across the region. The two base angles of the triangle are each labeled $\frac{\pi }{4}$, the distance out along the base is labeled $x$ and half the width 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 the pointed, lens shaped region between the two parabolas, with the two boundary curves colored orange and green, and the cross-sections taken perpendicular to the y-axis are isosceles triangles, so the solid rises to a ridge along the middle. 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, biz 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 top one is the figure from the problem statement, an isosceles triangle with both base angles labeled $\frac{\pi }{4}$ standing on the base of the solid, with the left boundary curve colored orange and the right one green. The bottom one looks straight down at the base from above and shows the region between $x = {y^2} - 2$ on the left in orange and $x = 2 - {y^2}$ on the right in green, with the cross-section itself the horizontal red line across the region.$

Yuqoridagi eskiz haqiqatdan ham muammo bayonida keltirilgan grafik bo'lib, u pastdagi eskiz bilan mos yozuvlar uchun kiritilgan. Pastki qismdagi eskiz to'g'ridan-to'g'ri yuqoridan kesmaga qaraydi va qizil chiziq bilan ko'rsatilgan.

Keling, faqat kesmaning qisqacha eskizini olamiz va $b$ uchburchak asosining uzunligini va $h$ uchburchakning balandligini chaqiramiz.

$A sketch of a single cross-section, which is an isosceles triangle with both base angles labeled $\frac{\pi }{4}$ and a dashed line from the top vertex down to the base labeled $h$. Along the bottom a black dot marks where the y-axis passes through, splitting the base into two pieces each labeled $\frac{b}{2}$, and the two ends of the base are marked with dots and labeled $x = {y^2} - 2$ on the left and $x = 2 - {y^2}$ on the right.$

Endi, pastki qismida biz $y$ o'qining kesmada joylashgan joyini qora nuqta bilan belgiladik va to'q sariq va yashil nuqtalar qattiq jismning pastki qismining chap va o'ng tomonlarini belgilaydigan chap va o'ng egri chiziqlarni ifodalaydi. Bundan tashqari, ko'rishimiz mumkinki, kesma qandaydir $y$ da joylashgan bo'lsa, yashil nuqta $y$ o'qidan $2 - {y^2}$ masofasida bo'lishi kerak. Xuddi shunday, to'q sariq nuqta $y$ o'qidan $ - \left( {{y^2} - 2} \right) = 2 - {y^2}$ masofasida bo'lishi kerak (esda tuting, biz bu erda masofa ijobiy bo'lishini xohlaymiz va shuning uchun ijobiy masofani olish uchun funktsiyaga minus belgisini qo'shamiz).

Endi biz uchburchakning asosi bilan berilganligini ko'rishimiz mumkin:

\[\frac{b}{2} = 2 - {y^2}\hspace{0.5in} \Rightarrow \hspace{0.5in}b = 2\left( {2 - {y^2}} \right)\]

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

\[\tan \left( {\frac{\pi }{4}} \right) = \frac{h}{{{}^{b}/{}_{2}}}\hspace{0.5in} \Rightarrow \hspace{0.5in}h = \frac{b}{2}\tan \left( {\frac{\pi }{4}} \right) = 2 - {y^2}\]

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

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

3-qadam

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

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

$x = 0$ ni qattiq jismning asosining chap va o‘ng tomonlarini aniqlovchi tenglamalardan biriga qo‘yish orqali (ular $y$ o‘qida kesishganligi sababli) biz “birinchi” kesma $y = - \sqrt 2 $ da, “oxirgi” kesma esa $y = \sqrt 2 $ chegarasida paydo bo‘lishini ko‘rishimiz mumkin.

Keyin hajm,

\[V = \int_{{ - \sqrt 2 }}^{{\sqrt 2 }}{{4 - 4{y^2} + {y^4}\,dy}} = \left. {\left( {4y - \frac{4}{3}{y^3} + \frac{1}{5}{y^5}} \right)} \right|_{ - \sqrt 2 }^{\sqrt 2 } = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{64\sqrt 2 }}{{15}}}}\]