Area_Volume_Formulas | MathModel.uz

Ilova A.6: Maydon va hajm formulalari

Ushbu bo'limda biz ikkita egri chiziq orasidagi maydonni va aylanish jismining hajmini olish uchun ishlatiladigan formulalarni olamiz.

Ikki egri chiziq orasidagi maydon

Biz $\left[ {a,b} \right]$ oralig'ida $y = f\left( x \right)$ va $y = g\left( x \right)$ orasidagi maydonni aniqlash formulasidan boshlaymiz. Shuningdek, biz $f\left( x \right) \ge g\left( x \right)$ ni $\left[ {a,b} \right]$ da deb hisoblaymiz.

Endi biz buni ko'rib chiqqanimizdek davom etamiz Hudud muammosi Integrallar bo'limida. Biz avval intervalni $n$ har birining uzunligi bo'lgan teng subintervallarga ajratamiz,

\[\Delta x = \frac{{b - a}}{n}\]

Keyin, har bir kichik intervalda, $x_i^*$ nuqtani tanlang va keyin har bir intervalda to'rtburchaklardan quyidagi tarzda foydalanishimiz mumkin.

$This is the graph of two functions, f(x) and g(x), on the domain a<x<b. The graph of f(x) is always larger than the graph of g(x). Along the x-axis are points labeled $x_{0}$, $x_{1}$, $x_{2}$ on the left end $x_{i-1}$,$x_{i}$ in the middle and $x_{n-1}$, $x_{n}$ on the right end. Between each pair of points is another set of points labeled $x_{1}^{*}$, $x_{2}^{*}$, $x_{3}^{*}$ on the left $x_{i}^{*}$ in the middle and $x_{n}^{*}$ on the right. Each of the “*” points are used to get the height of a rectangle above it. So, there are two rectangles on the left one in the middle and one on the right that are used to represent the area between the two curves. The bottom of each rectangle is on g(x) and the top is on f(x).$

Ushbu to'rtburchaklarning har birining balandligi quyidagicha ifodalanadi:

\[f\left( {x_i^*} \right) - g\left( {x_i^*} \right)\]

va har bir to'rtburchakning maydoni u holda,

\[\left( {f\left( {x_i^*} \right) - g\left( {x_i^*} \right)} \right)\Delta x\]

Shunday qilib, ikkita egri chiziq orasidagi maydon quyidagicha taxmin qilinadi:

\[A \approx \sum\limits_{i = 1}^n {\left( {f\left( {x_i^*} \right) - g\left( {x_i^*} \right)} \right)\Delta x} \]

Aniq maydon,

\[A = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {\left( {f\left( {x_i^*} \right) - g\left( {x_i^*} \right)} \right)\Delta x} \]

Endi eslatib o'tamiz aniq integralning ta'rifi bu boshqa narsa emas,

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

Yuqoridagi formula ikkita funktsiya $y = f\left( x \right)$ va $y = g\left( x \right)$ shaklida bo'lganda ishlaydi. Biroq, barcha funktsiyalar bu shaklda emas.

Ba'zan biz $\left[ {c,d} \right]$ oralig'ida $x = f\left( y \right)$ va $x = g\left( y \right)$ o'rtasidagi ko'rinishdagi funktsiyalar bilan ishlashga majbur bo'lamiz ($y$ qiymatlari oralig'i ...). Bu sodir bo'lganda, hosila bir xil bo'ladi. Avval biz $\left[ {c,d} \right]$ da $f\left( y \right) \ge g\left( y \right)$ deb faraz qilishdan boshlaymiz. Keyin biz intervalni teng subintervallarga bo'lishimiz va bu intervallarning har birida to'rtburchaklar qurishimiz mumkin. Mana bu holatning eskizi.

$This is the graph of two functions, f(y) and g(y), on the domain c<y<d. The graph of f(y) is always larger than the graph of g(y). Along the y-axis are points labeled $y_{0}$, $y_{1}$, $y_{2}$ on the left end $y_{i-1}$,$y_{i}$ in the middle and $y_{n-1}$, $y_{n}$ on the right end. Between each pair of points is another set of points labeled $y_{1}^{*}$, $y_{2}^{*}$, $y_{3}^{*}$ on the left $y_{i}^{*}$ in the middle and $xy_{n}^{*}$ on the right. Each of the “*” points are used to get the width of a rectangle along it. So, there are two rectangles on the bottom, one in the middle and one on the top that are used to represent the area between the two curves. The left side of each rectangle is on g(x) and the right side is on f(x).$

Yuqoridagi ishlardan so'ng, biz hudud uchun quyidagilarga erishamiz:

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

Shunday qilib, funktsiyalar qanday shaklda bo'lishidan qat'i nazar, biz asosan bir xil formuladan foydalanamiz.

Solid of Revolution uchun jildlar

Buning formulasini olishdan oldin, ehtimol, birinchi navbatda, inqilobning qattiqligi nima ekanligini aniqlab olishimiz 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 hosila uchun keling, egri chiziqni $x$ o'qi atrofida aylantiramiz. Buni bajarish 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.

Biz ushbu ob'ektning ichki qismining hajmini aniqlamoqchimiz. Buning uchun biz ikkita egri chiziq orasidagi maydon uchun qilganimizdek harakat qilamiz. Biz avval intervalni $n$ kenglik subintervallariga ajratamiz,

\[\Delta x = \frac{{b - a}}{n}\]

Keyin har bir subintervaldan $x_i^*$ nuqtani tanlaymiz.

Endi, ikkita egri chiziq orasidagi maydonda biz har bir subintervalda to'rtburchaklar yordamida maydonni yaqinlashtirdik. Hajmlar uchun biz maydonni taxmin qilish uchun har bir subintervalda disklardan foydalanamiz. Har bir diskning yuzining maydoni $A\left( {x_i^*} \right)$ tomonidan berilgan va har bir diskning hajmi

\[{V_i} = A\left( {x_i^*} \right)\Delta x\]

Mana buning eskizi,

This is a sketch of the solid approximated by a series of 12 disks centered on the x-axis and whose edges are on the graph of the function defining the boundary of the solid.

Mintaqaning hajmini quyidagicha taxmin qilish mumkin:

\[V \approx \sum\limits_{i = 1}^n {A\left( {x_i^*} \right)\Delta x} \]

Shunda aniq hajm,

\[\begin{align*}V & = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n {A\left( {x_i^*} \right)\Delta x} \\ & = \int_{{\,a}}^{{\,b}}{{A\left( x \right)\,dx}}\end{align*}\]

Shunday qilib, bu holda hajm har qanday $x$, $A\left( x \right)$ da kesma maydonining integrali bo'ladi. Shuni ham yodda tutingki, bu holda kesma maydoni aylana bo'lib, biz uzoqroqqa borib, buning uchun formulani ham olishimiz mumkin. Biroq, yuqoridagi formula umumiyroq va kesma olishning har qanday usuli uchun ishlaydi, shuning uchun biz uni avvalgidek qoldiramiz.

Biz ushbu formuladan haqiqatda foydalanadigan bo'limlarda, shuningdek, $x$ o'rniga $y$ funktsiyasi bo'lgan kesma maydonini beradigan kesmani yaratish usullari mavjudligini ko'ramiz. Bunday hollarda formula quyidagicha bo'ladi:

\[V = \int_{{\,c}}^{{\,d}}{{A\left( y \right)\,dy}},\,\,\,\hspace{0.25in}c \le y \le d\]

Bu holda biz $x$ o'qi atrofida egri chiziqni aylantirishni ko'rib chiqdik, ammo biz egri chiziqni $y$ o'qi atrofida osongina aylantirishimiz mumkin edi. Aslida, biz egri chiziqni istalgan vertikal yoki gorizontal o'q atrofida aylantirishimiz mumkin va bularning barchasida quyidagi formulalardan birini yoki ikkalasini ishlatishimiz mumkin.

\[V = \int_{{\,a}}^{{\,b}}{{A\left( x \right)\,dx}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}V = \int_{{\,c}}^{{\,d}}{{A\left( y \right)\,dy}}\]