15.1-bo'lim : Ikki tomonlama integrallar
1. \(f\left( {x,y} \right) = {x^2} + y\) ostidagi va \(xy\) tekisligida \( - 1 \le x \le 3\), \(0 \le y \le 4\) tomonidan berilgan to'g'ri to'rtburchak soha bo'yichagi hajmni baholash uchun O'rta nuqta qoidasidan foydalaning. \(x\) yo‘nalishida 4 ta bo‘limdan va \(y\) yo‘nalishida 2 ta bo‘limdan foydalaning.
Yechim
Xo'sh, avval biz bu erda ko'rib chiqilayotgan to'rtburchakning qisqacha eskizini olamiz.
Ochiq kulrang chiziqlar har bir yo'nalish uchun bo'linmalarni ko'rsatadi. Har bir "blok" ning markazidagi nuqtalar har bir blokning o'rta nuqtalari.
Pastki qatordagi har bir nuqtaning koordinatalari:
\[\left( { - \frac{1}{2},1} \right)\hspace{0.25in}\left( {\frac{1}{2},1} \right)\hspace{0.25in}\left( {\frac{3}{2},1} \right)\hspace{0.25in}\left( {\frac{5}{2},1} \right)\]va yuqori qatordagi har bir nuqtaning koordinatalari,
\[\left( { - \frac{1}{2},3} \right)\hspace{0.25in}\left( {\frac{1}{2},3} \right)\hspace{0.25in}\left( {\frac{3}{2},3} \right)\hspace{0.25in}\left( {\frac{5}{2},3} \right)\]2-qadam
Biz bilamizki, biz kutayotgan hajm oddiy,
\[V = \iint\limits_{R}{{f\left( {x,y} \right)\,dA}}\]va biz ushbu alohida holat uchun O'rta nuqta qoidasi quyidagi ikkilamchi yig'indi ekanligini bilamiz.
\[\iint\limits_{R}{{f\left( {x,y} \right)\,dA}} \approx \sum\limits_{i = 1}^4 {\sum\limits_{j = 1}^2 {f\left( {{{\overline{x}}_i},{{\overline{y}}_j}} \right)\,\Delta {\kern 1pt} A} } \hspace{0.25in}\hspace{0.25in}f\left( {x,y} \right) = {x^2} + y\]Esda tutingki, to'rtta \(x\) bo'linmasi bor va shuning uchun \(i\) yig'indisi 4 ga tushadi. Ikkita \(y\) bo'linmasi mavjud va \(j\) yig'indisi 2 ga boradi.
Formuladagi \(\left( {{{\overline{x}}_i},{{\overline{y}}_j}} \right)\) oddiygina har bir blokning o'rta nuqtalari (i.e. yuqoridagi 1-bosqichda sanab o'tilgan nuqtalar) va \(\Delta {\kern 1pt} A\) har bir blokning maydoni va \(\Delta {\kern 1pt} A = \left( 1 \right)\left( 2 \right) = 2\) ham shunday.
Shunday qilib, ushbu muammoning hajmi taxminan,
\[V \approx \sum\limits_{i = 1}^4 {\sum\limits_{j = 1}^2 {2f\left( {{{\overline{x}}_i},{{\overline{y}}_j}} \right)\,} } \hspace{0.25in}\hspace{0.25in}f\left( {x,y} \right) = {x^2} + y\]E'tibor bering, biz \(\Delta {\kern 1pt} A\) ni formulaga kiritdik.
3-qadam
Ushbu qo'sh yig'indini hisoblashning eng yaxshi usuli, ehtimol \(i\) ning har bir qiymati uchun "ichki" yig'indini hisoblashdir.
Ichki yig'indilarning har biri \(i\) ning belgilangan qiymati uchun amalga oshiriladi, chunki \(i\) \(i = 1\) dan \(i = 4\) gacha ishlaydi. Shuning uchun, to'rtta ichki yig'indi 1-bosqichdagi yuqoridagi eskizdagi har bir ustundagi bloklar uchun o'rta nuqtalardagi funktsiya qiymatlari yordamida hisoblanadi.
Bu bizga quyidagi jamlamalar kerakligini anglatadi.
\[\begin{align*}& i = 1\,\,\,\,:\,\,\,\,\,\sum\limits_{j = 1}^2 {2f\left( {{{\overline{x}}_1},{{\overline{y}}_j}} \right)\, = } \sum\limits_{j = 1}^2 {2f\left( { - \frac{1}{2},{{\overline{y}}_j}} \right)\,} = 2\left[ {f\left( { - \frac{1}{2},1} \right) + f\left( { - \frac{1}{2},3} \right)} \right] = 9\\ & i = 2\,\,\,\,:\,\,\,\,\,\sum\limits_{j = 1}^2 {2f\left( {{{\overline{x}}_2},{{\overline{y}}_j}} \right)\,} = \sum\limits_{j = 1}^2 {2f\left( {\frac{1}{2},{{\overline{y}}_j}} \right)\,} = 2\left[ {f\left( {\frac{1}{2},1} \right) + f\left( {\frac{1}{2},3} \right)} \right] = 9\\ & i = 3\,\,\,\,:\,\,\,\,\,\sum\limits_{j = 1}^2 {2f\left( {{{\overline{x}}_3},{{\overline{y}}_j}} \right) = \,} \sum\limits_{j = 1}^2 {2f\left( {\frac{3}{2},{{\overline{y}}_j}} \right)\,} = 2\left[ {f\left( {\frac{3}{2},1} \right) + f\left( {\frac{3}{2},3} \right)} \right] = 17\\ & i = 4\,\,\,\,:\,\,\,\,\,\sum\limits_{j = 1}^2 {2f\left( {{{\overline{x}}_4},{{\overline{y}}_j}} \right) = \,} \sum\limits_{j = 1}^2 {2f\left( {\frac{5}{2},{{\overline{y}}_j}} \right)\,} = 2\left[ {f\left( {\frac{5}{2},1} \right) + f\left( {\frac{5}{2},3} \right)} \right] = 33\end{align*}\]4-qadam
Endi biz "tashqi" yig'indini hisoblashimiz mumkin. Bu biz 3-bosqichda hisoblagan barcha ichki yig'indilarning yig'indisidir.
Keyin hajm taxminan,
\[V \approx \sum\limits_{i = 1}^4 {\sum\limits_{j = 1}^2 {2f\left( {{{\overline{x}}_i},{{\overline{y}}_j}} \right)\,} } = 9 + 9 + 17 + 33 = \require{bbox} \bbox[2pt,border:1px solid black]{{68}}\]Malumot uchun biz oxir-oqibat aniq hajm \(\frac{{208}}{3} = 69.333\overline 3 \) ekanligini tekshira olamiz va shuning uchun bu holatda taxminiylik juda yaqin.