DIGeneralRegion | MathModel.uz

15.3-bo'lim: Umumiy hududlar bo'yicha qo'sh integrallar

Oldingida biz to'rtburchaklar mintaqalar ustidagi qo'sh integrallarni ko'rib chiqdik. Muammo shundaki, aksariyat mintaqalar to'rtburchaklar emas, shuning uchun biz quyidagi qo'sh integralni ko'rib chiqishimiz kerak,

\[\iint\limits_{D}{{f\left( {x,y} \right)\,dA}}\]

bu erda $D$ har qanday mintaqadir.

Biz ko'rib chiqishimiz kerak bo'lgan ikki turdagi mintaqalar mavjud. Mana ularning ikkalasining eskizi.

$This is the 2D graph of two unknown functions on the domain a<x<b that are in the 1st quadrant. The graph of $y=g_{2}(x)$ is always over the graph of $y=g_{1}(x)$. The two graphs come to a point at x=b. This is done only to show that it is possible for the graph graphs to touch as long as the graph of $g_{2}(x)$ always remains above the graph of $g_{1}(x)$ everywhere else. The area between the two functions has been shaded in.$ $This is the 2D graph of two unknown functions on the domain $c<y<d$ that are in the 1st quadrant. The graph of $x=h_{2}(y)$ is always to the right of the graph of $x=h_{1}(y)$. The area between the two functions has been shaded in.$

Biz tez-tez ishlatamiz quruvchi belgisini o'rnatish bu hududlarni tavsiflash uchun. Mana 1-holda mintaqaning ta'rifi

\[D = \left\{ {\left( {x,y} \right)|a \le x \le b,\,\,{g_1}\left( x \right) \le y \le {g_2}\left( x \right)} \right\}\]

va bu erda 2-holatdagi mintaqa uchun ta'rif.

\[D = \left\{ {\left( {x,y} \right)|{h_1}\left( y \right) \le x \le {h_2}\left( y \right),\,c \le y \le d} \right\}\]

Bu belgi haqiqatan ham biz $\left( {x,y} \right)$ nuqtalaridan foydalanamiz, deyishning oddiy usuli, bunda ikkala koordinata ham berilgan ikkita tengsizlikni qondiradi.

Bu ikkala holat uchun qo'sh integral takrorlangan integrallar nuqtai nazaridan quyidagicha aniqlanadi.

$D = \left\{ {\left( {x,y} \right)|a \le x \le b,\,\,{g_1}\left( x \right) \le y \le {g_2}\left( x \right)} \right\}$ integral sifatida belgilangan 1-holatda,

\[\iint\limits_{D}{{f\left( {x,y} \right)\,dA}} = \int_{{\,a}}^{{\,b}}{{\int_{{{g_{\,1}}\left( x \right)}}^{{{g_{\,2}}\left( x \right)}}{{f\left( {x,y} \right)\,dy}}\,dx}}\]

$D = \left\{ {\left( {x,y} \right)|{h_1}\left( y \right) \le x \le {h_2}\left( y \right),\,c \le y \le d} \right\}$ integral sifatida belgilangan 2-holatda,

\[\iint\limits_{D}{{f\left( {x,y} \right)\,dA}} = \int_{{\,c}}^{{\,d}}{{\int_{{h{\,_1}\left( y \right)}}^{{{h_{\,2}}\left( y \right)}}{{f\left( {x,y} \right)\,dx}}\,dy}}\]

Mana, ba'zi misollar keltirishdan oldin ko'rib chiqishimiz kerak bo'lgan qo'sh integralning ba'zi xususiyatlari. E'tibor bering, bu uchta xususiyat haqiqatan ham qo'sh integrallarga kengaytirilgan yagona integrallar xususiyatlarining kengaytmalari.

Ikki tomonlama integralning xossalari

$\displaystyle \iint\limits_{D}{{f\left( {x,y} \right) + g\left( {x,y} \right)\,dA}} = \iint\limits_{D}{{f\left( {x,y} \right)\,dA}} + \iint\limits_{D}{{g\left( {x,y} \right)\,dA}}$
$\displaystyle \iint\limits_{D}{{cf\left( {x,y} \right)\,dA}} = c\iint\limits_{D}{{f\left( {x,y} \right)\,dA}}$, bu erda $c$ har qanday doimiydir.
Agar $D$ mintaqasini ${D_1}$ va ${D_2}$ ikkita alohida mintaqaga bo'lish mumkin bo'lsa, integral quyidagicha yozilishi mumkin. \[\iint\limits_{D}{{f\left( {x,y} \right)\,dA}} = \iint\limits_{{{D_1}}}{{f\left( {x,y} \right)\,dA}} + \iint\limits_{{{D_2}}}{{f\left( {x,y} \right)\,dA}}\]

Keling, umumiy mintaqalar bo'yicha qo'sh integrallarning ba'zi misollarini ko'rib chiqaylik.

1-misol Quyidagi integrallarning har birini berilgan $D$ mintaqasi bo‘yicha baholang.

$\displaystyle \iint\limits_{D}{{{{\bf{e}}^{\frac{x}{y}}}\,dA}}$, $D = \left\{ {\left( {x,y} \right)|1 \le y \le 2,\,\,y \le x \le {y^3}} \right\}$
$\displaystyle \iint\limits_{D}{{4xy - {y^3}\,dA}}$, $D$ $y = \sqrt x $ va $y = {x^3}$ bilan chegaralangan mintaqadir.
$\displaystyle \iint\limits_{D}{{6{x^2} - 40y\,dA}}$, $D$ uchburchakdir $\left( {0,3} \right)$, $\left( {1,1} \right)$ va $\left( {5,3} \right)$ uchlari.

a $\displaystyle \iint\limits_{D}{{{{\bf{e}}^{\frac{x}{y}}}\,dA}}$, $D = \left\{ {\left( {x,y} \right)|1 \le y \le 2,\,\,y \le x \le {y^3}} \right\}$ Yechim

Xo'sh, bu birinchisi yuqoridagi formuladan foydalanish uchun o'rnatilgan, shuning uchun buni qilaylik.

\[\begin{align*}\iint\limits_{D}{{{{\bf{e}}^{\frac{x}{y}}}\,dA}} & = \int_{{\,1}}^{{\,2}}{{\int_{{\,y}}^{{{y^3}}}{{{{\bf{e}}^{\frac{x}{y}}}\,dx}}\,dy}} = \int_{{\,1}}^{{\,2}}{{\left. {y\,{{\bf{e}}^{\frac{x}{y}}}} \right|_y^{{y^3}}\,dy}}\\ & = \int_{{\,1}}^{{\,2}}{{y\,{{\bf{e}}^{{y^2}}} - y{{\bf{e}}^1}\,dy}}\\ & = \left. {\left( {\frac{1}{2}{{\bf{e}}^{{y^2}}} - \frac{1}{2}{y^2}{{\bf{e}}^1}} \right)} \right|_1^2 = \frac{1}{2}{{\bf{e}}^4} - 2{{\bf{e}}^1}\end{align*}\]

b $\displaystyle \iint\limits_{D}{{4xy - {y^3}\,dA}}$, $D$ $y = \sqrt x $ va $y = {x^3}$ bilan chegaralangan mintaqadir. Yechim

Bu holda biz integralni bajarishimiz kerak bo'lgan $x$ va $y$ uchun ikkita tengsizlikni aniqlashimiz kerak. Buni qilishning eng yaxshi usuli - bu ikkita egri chiziqning grafigi. Mana eskiz.

$This is the 2D graph on the domain 0<x<1 of $y=\sqrt{x}$ and $y=x^{3}$. In this domain the graph of $y=\sqrt{x}$ is always larger than the graph of $y=x^{3}$. The area between the two functions has been shaded in.$

Shunday qilib, eskizdan biz ikkita tengsizlik ekanligini ko'rishimiz mumkin,

\[0 \le x \le 1\hspace{0.25in}{x^3} \le y \le \sqrt x \]

Endi biz integralni qila olamiz,

\[\begin{align*}\iint\limits_{D}{{4xy - {y^3}\,dA}} & = \int_{{\,0}}^{{\,1}}{{\int_{{\,{x^3}}}^{{\,\sqrt x }}{{4xy - {y^3}\,dy}}\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{\left. {\left( {2x{y^2} - \frac{1}{4}{y^4}} \right)} \right|_{{x^3}}^{\sqrt x }\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{\frac{7}{4}{x^2} - 2{x^7} + \frac{1}{4}{x^{12}}\,dx}}\\ & = \left. {\left( {\frac{7}{{12}}{x^3} - \frac{1}{4}{x^8} + \frac{1}{{52}}{x^{13}}} \right)} \right|_0^1 = \frac{{55}}{{156}}\end{align*}\]

c $\displaystyle \iint\limits_{D}{{6{x^2} - 40y\,dA}}$, $D$ uchburchakdir $\left( {0,3} \right)$, $\left( {1,1} \right)$ va $\left( {5,3} \right)$ uchlari. Yechim

Bu safar mintaqa haqida kamroq ma'lumot oldik. Keling, buni uchburchakning eskizini chizishdan boshlaylik.

Har bir chekkada ikkita nuqta borligi sababli, har bir chekka uchun tenglamalarni olish oson va shuning uchun tenglamalarni tekshirishni sizga qoldiramiz.

Endi bu hududni tasvirlashning ikkita usuli bor. Agar biz $x$ funksiyalaridan foydalansak, rasmda ko'rsatilganidek, mintaqani ikki xil bo'lakka ajratishimiz kerak bo'ladi, chunki pastki funktsiya $x$ qiymatiga qarab farq qiladi. Bunday holda mintaqa $D = {D_1} \cup {D_2}$ tomonidan beriladi, bu erda,

\[\begin{align*}{D_1} & = \left\{ {\left( {x,y} \right)|0 \le x \le 1,\,\,\, - 2x + 3 \le y \le 3} \right\}\\ {D_2} & = \left\{ {\left( {x,y} \right)|1 \le x \le 5,\,\,\,\frac{1}{2}x + \frac{1}{2} \le y \le 3} \right\}\end{align*}\]

E'tibor bering, $ \cup $ "birlashma" belgisidir va shunchaki $D$ ikkita mintaqani tarash orqali oladigan mintaqa ekanligini anglatadi. Agar buni qilsak, har bir mintaqa uchun ikkita alohida integral qilishimiz kerak bo'ladi.

Bunga yo'l qo'ymaslik uchun biz narsalarni aylantirib, $x$ uchun ikkita tenglamani hal qilishimiz mumkin,

\[\begin{align*}y & = - 2x + 3\hspace{0.5in} \Rightarrow \hspace{0.5in}x = - \frac{1}{2}y + \frac{3}{2}\\ y & = \frac{1}{2}x + \frac{1}{2}\hspace{0.5in} \Rightarrow \hspace{0.5in}x = 2y - 1\end{align*}\]

Agar biz buni qilsak, biz bir xil funktsiya har doim o'ngda va bir xil funktsiya har doim chap tomonda ekanligini ko'rishimiz mumkin, shuning uchun mintaqa,

\[D = \left\{ {\left( {x,y} \right)|\,\, - \frac{1}{2}y + \frac{3}{2} \le x \le 2y - 1,\,\,\,1 \le y \le 3} \right\}\]

Mintaqani bu shaklda yozish, biz boshqacha qilishimiz kerak bo'lgan ikkita integral o'rniga bitta integralni bajarishni anglatadi.

Ikkala yo'l ham bir xil javob berishi kerak va shuning uchun biz mintaqani bo'lish eslatmalarida misol olishimiz mumkin, ikkala integralni ham bajaramiz.

Yechim 1

\[\begin{align*}\iint\limits_{D}{{6{x^2} - 40y\,dA}} & = \iint\limits_{{{D_1}}}{{6{x^2} - 40y\,dA}} + \iint\limits_{{{D_2}}}{{6{x^2} - 40y\,dA}}\\ & = \int_{{\,0}}^{{\,1}}{{\int_{{\, - 2x + 3}}^{{\,3}}{{6{x^2} - 40y\,dy}}\,dx}} + \int_{{\,1}}^{{\,5}}{{\int_{{\,\frac{1}{2}x + \frac{1}{2}}}^{{\,3}}{{6{x^2} - 40y\,dy}}\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{\left. {\left( {6{x^2}y - 20{y^2}} \right)} \right|_{ - 2x + 3}^3\,dx}} + \int_{{\,1}}^{{\,5}}{{\left. {\left( {6{x^2}y - 20{y^2}} \right)} \right|_{\frac{1}{2}x + \frac{1}{2}}^3\,dx}}\\ & = \int_{{\,0}}^{{\,1}}{{12{x^3} - 180 + 20{{\left( {3 - 2x} \right)}^2}\,dx}} + \int_{{\,1}}^{{\,5}}{{ - 3{x^3} + 15{x^2} - 180 + 20{{\left( {\frac{1}{2}x + \frac{1}{2}} \right)}^2}\,dx}}\\ & = \left. {\left( {3{x^4} - 180x - \frac{{10}}{3}{{\left( {3 - 2x} \right)}^3}} \right)} \right|_0^1 + \left. {\left( { - \frac{3}{4}{x^4} + 5{x^3} - 180x + \frac{{40}}{3}{{\left( {\frac{1}{2}x + \frac{1}{2}} \right)}^3}} \right)} \right|_1^5\\ & = - \frac{{935}}{3}\end{align*}\]

Bu juda ko'p ish edi. E'tibor bering, biz birinchi almashtirishni amalga oshirganimizdan keyin hammasini ko'paytirmadik. Ikki kvadratik atamani asosiy Calc I almashtirish bilan osongina birlashtirish mumkin va shuning uchun biz ularni ko'paytirish bilan bezovta qilmadik. Biz buni ba'zi bir integrallarni biroz osonroq qilish uchun qilamiz.

Yechim 2
Bu yechim juda kam ish bo'ladi, chunki biz faqat bitta integral qilamiz.

\[\begin{align*}\iint\limits_{D}{{6{x^2} - 40y\,dA}} &= \int_{1}^{3}{{\int_{{ - \frac{1}{2}y + \frac{3}{2}}}^{{2y - 1}}{{6{x^2} - 40y\,dx}}\,dy}}\\ & = \int_{1}^{3}{{\left. {\left( {2{x^3} - 40xy} \right)} \right|_{ - \frac{1}{2}y + \frac{3}{2}}^{2y - 1}\,dy}}\\ & = \int_{1}^{3}{{100y - 100{y^2} + 2{{\left( {2y - 1} \right)}^3} - 2{{\left( { - \frac{1}{2}y + \frac{3}{2}} \right)}^3}\,dy}}\\ & = \left. {\left( {50{y^2} - \frac{{100}}{3}{y^3} + \frac{1}{4}{{\left( {2y - 1} \right)}^4} + {{\left( { - \frac{1}{2}y + \frac{3}{2}} \right)}^4}} \right)} \right|_1^3\\ & = - \frac{{935}}{3}\end{align*}\]

Shunday qilib, raqamlar biroz chalkash edi, ammo bundan tashqari, xuddi shunday natija uchun ish juda kam edi. Shuni ham yodda tutingki, biz yana ikkita atamani kublarga ajratmadik, chunki ularni Calc I almashtirishdan foydalanish osonroq.

Oldingi misolning oxirgi qismi bizga ko'rsatganidek, biz bu integrallarni ikkala tartibda ham integrallashimiz mumkin (i.e. $x$, keyin $y$ yoki $y$, keyin esa $x$ keladi, garchi ko'pincha bitta buyurtma boshqasidan osonroq bo'ladi. Darhaqiqat, shunday paytlar bo'ladiki, integralni bir tartibda bajarish ham mumkin bo'lmasa, boshqa tartibda integralni bajarish mumkin bo'ladi.

Bundan tashqari, Calculus I almashtirish haqida unutmang. Talabalar ko'pincha shoshilib, integral baholashni amalga oshirgandan so'ng hamma narsani ko'paytiradilar va natijada hamma narsani ko'paytirish bilan bog'liq qiyinchiliklardan qochadi. Calculus I almashtirishlari har doim ham ko'rinmaydi, lekin ular bajarilganda ular deyarli har doim muammoning qolgan qismidagi ishni soddalashtiradi.

Keling, bunday integrallarning bir nechta misollarini ko'rib chiqaylik.

2-misol Avval integrallash tartibini teskari qilib, quyidagi integrallarni baholang.

$\displaystyle \int_{{\,0}}^{{\,3}}{{\int_{{\,{x^2}}}^{{\,9}}{{{x^3}{{\bf{e}}^{{y^3}}}\,dy}}\,dx}}$
$\displaystyle \int_{{\,0}}^{{\,8}}{{\int_{{\,\sqrt[3]{y}}}^{{\,2}}{{\sqrt {{x^4} + 1} \,dx}}\,dy}}$

a $\displaystyle \int_{{\,0}}^{{\,3}}{{\int_{{\,{x^2}}}^{{\,9}}{{{x^3}{{\bf{e}}^{{y^3}}}\,dy}}\,dx}}$ Yechim

Birinchidan, agar biz $y$ ga nisbatan integrallashga harakat qilsak, integralni bajara olmaymiz, chunki $y$ integratsiyasini amalga oshirish uchun eksponensial oldida ${y^2}$ kerak bo'ladi. Agar biz integratsiya tartibini o'zgartirsak, biz qila oladigan integralni olamiz deb umid qilamiz.

Endi biz integratsiya tartibini o'zgartiramiz, desak, bu biz avval $x$, keyin esa $y$ bilan integratsiya qilishni xohlayotganimizni anglatadi. Shuni ham yodda tutingki, biz asl chegaralarni saqlab, integrallarni almashtira olmaymiz va u bilan bajara olmaymiz. Bu bizning asl muammomizni hal qilmaydi va $x$ ga nisbatan integratsiya qilish uchun bizda $x$ integrallari chegarasida bo'lishi mumkin emas. Agar biz javobimiz kerak bo'lgandek doimiy bo'lmasligini e'tiborsiz qoldirsak ham.

Shunday qilib, keling, integratsiya tartibini qanday o'zgartirishimizni ko'rib chiqaylik. Integratsiya tartibini teskari o'zgartirishning eng yaxshi usuli birinchi navbatda integratsiyaning dastlabki chegaralari bilan berilgan hududning eskizini chizishdir. Integraldan biz ushbu mintaqani belgilaydigan tengsizliklar ekanligini ko'ramiz.

\[\begin{array}{c}0 \le x \le 3\\ {x^2} \le y \le 9\end{array}\]

Ushbu tengsizliklar bizga $y = {x^2}$ pastki chegarada va $y = 9$ yuqori chegarada $x = 0$ va $x = 3$ o'rtasida joylashgan mintaqani xohlashimizni aytadi. Mana o'sha hududning eskizi.

$This is the 2D graph on the domain 0<x<9 of y=9 and $y=x^{2}$. In this domain the graph of y=9 is always larger than the graph of $y=x^{2}$. It is also noted that the lower curve could also be written as $x=\sqrt{y}$. The area between the two functions has been shaded in.$

Biz $x$ bilan integratsiya qilmoqchi bo'lganimiz sababli, avval $x$ chegaralarini aniqlashimiz kerak (ehtimol $y$ nuqtai nazaridan) va keyin $y$ chegaralarini olishimiz kerak. Mana ular bu mintaqa uchun.

\[\begin{array}{c}0 \le x \le \sqrt y \\ 0 \le y \le 9\end{array}\]

Ushbu mintaqada chizilgan har qanday gorizontal chiziq $x = 0$ da boshlanadi va $x = \sqrt y $ da tugaydi va shuning uchun bu $x$ ning chegaralari va mintaqalar uchun $y$ diapazoni 0 dan 9 gacha.

Tartib teskari bo'lgan integral endi,

\[\int_{{\,0}}^{{\,3}}{{\int_{{\,{x^2}}}^{{\,9}}{{{x^3}{{\bf{e}}^{{y^3}}}\,dy}}\,dx}} = \int_{{\,0}}^{{\,9}}{{\int_{{\,0}}^{{\,\sqrt y }}{{{x^3}{{\bf{e}}^{{y^3}}}\,dx}}\,dy}}\]

va biz ushbu tartib bilan birinchi integratsiyani amalga oshirishimiz mumkinligiga e'tibor bering. Umid qilamizki, bu biz qila oladigan ikkinchi integralni beradi. Mana bu integral uchun ish.

\[\begin{align*}\int_{{\,0}}^{{\,3}}{{\int_{{\,{x^2}}}^{{\,9}}{{{x^3}{{\bf{e}}^{{y^3}}}\,dy}}\,dx}} & = \int_{{\,0}}^{{\,9}}{{\int_{{\,0}}^{{\,\sqrt y }}{{{x^3}{{\bf{e}}^{{y^3}}}\,dx}}\,dy}}\\ & = \int_{{\,0}}^{{\,9}}{{\left. {\frac{1}{4}{x^4}{{\bf{e}}^{{y^3}}}} \right|_0^{\sqrt y }\,dy}}\\ & = \int_{{\,0}}^{{\,9}}{{\frac{1}{4}{y^2}{{\bf{e}}^{{y^3}}}\,dy}}\\ & = \left. {\frac{1}{{12}}{{\bf{e}}^{{y^3}}}} \right|_0^9\\ & = \frac{1}{{12}}\left( {{{\bf{e}}^{729}} - 1} \right)\end{align*}\]

Shunday qilib, biz kutganimizdek, integratsiya tartibini almashtirganimizdan so'ng, biz integralni bajara oldik.

b $\displaystyle \int_{{\,0}}^{{\,8}}{{\int_{{\,\sqrt[3]{y}}}^{{\,2}}{{\sqrt {{x^4} + 1} \,dx}}\,dy}}$ Yechim

Birinchi integralda bo'lgani kabi, biz bu integralni birinchi navbatda $x$ ga nisbatan integrallash orqali amalga oshira olmaymiz, shuning uchun biz integratsiya tartibini o'zgartirib, biz integrallashimiz mumkin bo'lgan narsani olamiz deb umid qilamiz. Mana bu integraldan oladigan o'zgaruvchilar uchun chegaralar.

\[\begin{array}{c}\sqrt[3]{y} \le x \le 2\\ 0 \le y \le 8\end{array}\]

va bu hududning eskizi.

$This is the 2D graph on the domain 0<x<2 of $y=x^{3}$. The right edge is marked as the line x=2. It is also noted that the curve could also be written as $x=\sqrt[3]{y}$. The area between the function and the x-axis has been shaded in.$

Shunday qilib, agar biz integratsiya tartibini o'zgartirsak, biz quyidagi chegaralarni olamiz.

\[\begin{array}{c}0 \le x \le 2\\ 0 \le y \le {x^3}\end{array}\]

U holda integral,

\[\begin{align*}\int_{{\,0}}^{{\,8}}{{\int_{{\,\sqrt[3]{y}}}^{{\,2}}{{\sqrt {{x^4} + 1} \,dx}}\,dy}} & = \int_{{\,0}}^{{\,2}}{{\int_{{\,0}}^{{\,{x^3}}}{{\sqrt {{x^4} + 1} \,dy}}\,dx}}\\ & = \int_{{\,0}}^{{\,2}}{{\left. {y\sqrt {{x^4} + 1} } \right|_0^{{x^3}}\,dx}}\\ & = \int_{{\,0}}^{{\,2}}{{{x^3}\sqrt {{x^4} + 1} \,dx}} = \frac{1}{6}\left( {{{17}^{\frac{3}{2}}} - 1} \right)\end{align*}\]

Ushbu bo'limning yakuniy mavzusi qo'sh integralning ikkita geometrik talqini. Birinchi talqin - bu birinchisida qo'sh integral g'oyasini ishlab chiqishda foydalangan fikrning kengaytmasi bo'limi ushbu bobdan. Biz buni $xy$ tekisligida $z = f\left( {x,y} \right)$ funksiyasi yuzasidan va $R$ to‘rtburchaklar ustida joylashgan qattiq jismning hajmini ko‘rib chiqdik. Ushbu g'oyani yanada umumiy hududlarga kengaytirish mumkin.

$z = f\left( {x,y} \right)$ tomonidan berilgan sirt ostida va $xy$ tekisligida $D$ mintaqasidan yuqorida joylashgan qattiq jismning hajmi quyidagicha ifodalanadi:

\[V = \iint\limits_{D}{{f\left( {x,y} \right)\,dA}}\]

3-misol $z = 16xy + 200$ tomonidan berilgan sirt ostida joylashgan va $y = {x^2}$ va $y = 8 - {x^2}$ bilan chegaralangan $xy$-tekisligida mintaqadan yuqorida joylashgan qattiq jismning hajmini toping.

Yechim

Mana sirt grafigi va biz mintaqani sirt ostidagi $xy$ tekisligida ko'rsatishga harakat qildik.

Fazoviy koordinatalar sistemasi va egri chiziq grafigi

$This is a boxed 3D coordinate system. The z-axis is left vertical edge of the box, the x-axis is the bottom left edge of the box and the y-axis is the bottom front edge of the box. This is the graph of a surface that appears to be almost a disk that has been slightly twisted. The left edge appears to be mostly flat (i.e. parallel to the xy-plane) and the right edge appears to have been twisted in such a way that the front edge raises up and the back end lowers down. Below the surface in the xy-plane is a graph of the two functions $y=x^{2}$ and $y=8-x^{2}$ on the range -2<x<2 and the area between them is shaded.$

Mana $xy$ tekisligidagi mintaqaning o'z-o'zidan eskizi.

$This is the 2D graph on the domain -2<x<2 of $y=x^{2}$ and $y=8-x^{2}$. In this domain the graph of $y=8-x^{2}$ is always larger than the graph of $y=x^{2}$. The area between the two functions has been shaded in.$

Ikki chegaralovchi tenglamani tenglashtirib, ular $x = 2$ va $x = - 2$ da kesishishini ko'rishimiz mumkin. Shunday qilib, $xy$ tekisligida $D$ mintaqasini aniqlaydigan tengsizliklar:

\[\begin{array}{c} - 2 \le x \le 2\\ {x^2} \le y \le 8 - {x^2}\end{array}\]

Keyin hajm quyidagicha beriladi:

\[\begin{align*}V &= \iint\limits_{D}{{16xy + 200\,dA}}\\ & = \int_{{ - 2}}^{{\,2}}{{\int_{{{x^2}}}^{{8 - {x^2}}}{{16xy + 200\,dy}}\,dx}}\\ & = \int_{{ - 2}}^{{\,2}}{{\left. {\left( {8x{y^2} + 200y} \right)} \right|_{{x^2}}^{8 - {x^2}}\,dx}}\\ & = \int_{{ - 2}}^{{\,2}}{{ - 128{x^3} - 400{x^2} + 512x + 1600\,dx}}\\ & = \left. {\left( { - 32{x^4} - \frac{{400}}{3}{x^3} + 256{x^2} + 1600x} \right)} \right|_{ - 2}^2 = \frac{{12800}}{3}\end{align*}\]

4-misol $4x + 2y + z = 10$, $y = 3x$, $z = 0$, $x = 0$ tekisliklari bilan o‘ralgan qattiq jismning hajmini toping.

Yechim

Bu misol avvalgisidan bir oz farq qiladi. Bu erda $D$ mintaqasi aniq ko'rsatilmagan, shuning uchun biz uni topishimiz kerak. Birinchidan, e'tibor bering, oxirgi ikkita samolyot haqiqatan ham biz ularga etib kelganimizda $xy$ va $yz$ samolyotlaridan o'tmasligimizni aytmoqda.

Birinchi tekislik, $4x + 2y + z = 10$, tovushning yuqori qismidir va shuning uchun biz haqiqatan ham ostidagi hajmni qidirmoqdamiz,

\[z = 10 - 4x - 2y\]

va $xy$ tekisligida $D$ mintaqasidan yuqorida. Ikkinchi tekislik, $y = 3x$ (ha, bu tekislik), quyida ko'rsatilganidek, hajmning tomonlaridan birini beradi.

This is a boxed 3D coordinate system. The z-axis is right vertical edge of the box, the x-axis is the top left edge of the box and the y-axis is the bottom front edge of the box. This the graph of two triangular planes in the 1st octant that form a solid. The top triangle has vertices (0,0,10), (0,5,0) and (1,3,0) and it is noted that its equation is 4x+2y+z=10. The front triangle has vertices (0,0,0), (0,0,10) and (1,3,0) and it is noted that its equation is y=3x. The back and bottom triangle that completely form the solid are not shown do to the orientation of the solid.

$D$ mintaqasi $xy$ tekisligidagi mintaqa bo'ladi (i.e. $z = 0$) $y = 3x$, $x = 0$ va $z + 4x + 2y = 10$ $xy$ tekisligi bilan kesishgan chiziq bilan chegaralanadi. $z + 4x + 2y = 10$ $xy$ tekisligi bilan qayerda kesishishini $z = 0$-ni unga ulash orqali aniqlashimiz mumkin.

\[0 + 4x + 2y = 10\hspace{0.25in} \Rightarrow \hspace{0.25in}2x + y = 5\hspace{0.25in} \Rightarrow \hspace{0.25in}y = - 2x + 5\]

Mana, $D$ hududining eskizi.

$D$ mintaqasi haqiqatan ham bu qattiq jism $xy$ tekisligida o'tiradigan joy va bu erda mintaqani belgilaydigan tengsizliklar.

\[\begin{array}{c}0 \le x \le 1\\ 3x \le y \le - 2x + 5\end{array}\]

Mana bu qattiqning hajmi.

\[\begin{align*}V & = \iint\limits_{D}{{10 - 4x - 2y\,dA}}\\ & = \int_{0}^{{\,1}}{{\int_{{3x}}^{{ - 2x + 5}}{{10 - 4x - 2y\,dy}}\,dx}}\\ & = \int_{0}^{{\,1}}{{\left. {\left( {10y - 4xy - {y^2}} \right)} \right|_{3x}^{ - 2x + 5}\,dx}}\\ & = \int_{0}^{{\,1}}{{25{x^2} - 50x + 25\,dx}}\\ & = \left. {\left( {\frac{{25}}{3}{x^3} - 25{x^2} + 25x} \right)} \right|_0^1 = \frac{{25}}{3}\end{align*}\]

E'tibor bering, umuman olganda,

\[V = \iint\limits_{D}{{f\left( {x,y} \right)\,dA}}\]

$xy$ tekisligida $z = f\left( {x,y} \right)$ grafigi va $D$ mintaqasi orasidagi aniq hajmni beradi. $xy$ tekisligidan pastda joylashgan hududlar manfiy hajmga, $xy$ tekisligidan yuqori bo'lgan hududlar esa ijobiy hajmga ega.

Biz shunga o'xshash fikrni Hisoblash Ida ko'rdik, bu erda

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

$y = f\left( x \right)$ tomonidan berilgan egri chiziq va $\left[ {a,b} \right]$ oralig'idagi $x$ o'qi orasidagi aniq maydonni beradi.

Ikkilamchi integralning ikkinchi geometrik talqini quyidagicha.

Ikki tomonlama integral sifatida mintaqaning maydoni

\[{\mbox{Area of }}D = \iint\limits_{D}{{dA}}\]

Umuman olganda, nima uchun bu haqiqat ekanligini tushunish oson. Aytaylik, biz quyida ko'rsatilgan hududning maydonini topmoqchimiz.

Hisoblashdan men bilamizki, bu maydonni integral bilan topish mumkin,

\[A = \int_{{\,a}}^{{\,b}}{{{g_2}\left( x \right) - {g_1}\left( x \right)\,dx}}\]

Yoki qo'sh integral nuqtai nazaridan bizda,

\[\begin{align*}{\mbox{Area of }}D & = \iint\limits_{D}{{dA}}\\ & = \int_{{\,a}}^{{\,b}}{{\int_{{\,{g_1}\left( x \right)}}^{{\,{g_2}\left( x \right)}}{{dy}}\,dx}}\\ & = \int_{{\,a}}^{{\,b}}{{\left. y \right|_{{g_1}\left( x \right)}^{{g_2}\left( x \right)}\,dx}} = \int_{{\,a}}^{{\,b}}{{{g_2}\left( x \right) - {g_1}\left( x \right)\,dx}}\end{align*}\]

Bu biz Calculus I-da bo'lgan bir xil formuladir.