17.4-bo'lim : Vektor maydonlarining sirt integrallari
5. \( \displaystyle \iint\limits_{S}{{\vec F\centerdot \,d\vec S}}\) ni baholang, bunda \(\vec F = y\,\vec i + 2x\,\vec j + \left( {z - 8} \right)\,\vec k\) va \(S\) - musbat orientatsiya bilan \(4x + 2y + z = 8\), \(z = 0\), \(y = 0\) va \(x = 0\) bilan chegaralangan qattiq jismning yuzasi. E'tibor bering, ushbu qattiq jismning barcha to'rtta yuzasi \(S\) tarkibiga kiritilgan.
Yechim
Keling, ushbu muammoni hal qilishda biz ishlayotgan sirtning qisqacha eskizi bilan boshlaylik.
Xo'sh, muammo bayonida ta'kidlanganidek, eskizdagi barcha to'rtta sirt (ikkitasi ko'rsatilmagan) \(S\) ning bir qismidir, shuning uchun ularning har birini quyidagicha aniqlaymiz.
\({S_1}:{\mbox{ Plane given by }}4x + 2y + z = 8{\mbox{ }}\left( {i.e{\mbox{ the top of the solid}}} \right)\)
\({S_2}:{\mbox{ Plane given by }}y = 0{\mbox{ }}\left( {i.e{\mbox{ the triangle on right side of the solid}}} \right)\)
\({S_3}:{\mbox{ Plane given by }}x = 0{\mbox{ }}\left( {i.e{\mbox{ the triangle at back of the solid - not shown in sketch}}} \right)\)
\({S_4}:{\mbox{ Plane given by }}z = 0{\mbox{ }}\left( {i.e{\mbox{ the triangle on bottom of the solid - not shown in sketch}}} \right)\)
Yuqoridagi ta'riflarda ta'kidlanganidek, dastlabki ikkita sirt eskizda ko'rsatilgan, ammo oxirgi ikkitasi qattiq jismning yo'nalishi tufayli ko'rsatilmagan. Quyida koordinatalar tekisliklariga mos keladigan uchta sirtning har birining eskizlari keltirilgan.
Har bir eskizda biz ularning har biri uchun o'zgaruvchilarga limitlar berdik, chunki har bir sirt bo'ylab sirt integralini bajarishni boshlaganimizda, oxir-oqibat bizga kerak bo'ladi.
Endi biz ushbu sirtlarning har biri uchun integralni ko'rib chiqishimiz va bajarishimiz kerak va biz ularni ushbu bo'limdagi birinchi bir nechta muammolarga qaraganda biroz tezroq o'tkazamiz.
2-qadam
\({S_1}\) bilan boshlaylik. Bu holda tekislikning tenglamasini quyidagicha yozishimiz mumkin:
\[f\left( {x,y,z} \right) = 4x + 2y + z - 8 = 0\]U holda sirt uchun birlik normal vektor,
\[\vec n = \frac{{\nabla f}}{{\left\| {\nabla f} \right\|}} = \frac{{\left\langle {4,2,1} \right\rangle }}{{\left\| {\nabla f} \right\|}}\]Biz gradientning kattaligini hisoblamadik, chunki biz integral bilan ishlashni boshlaganimizda u bekor qilinishini bilamiz.
Sirt ijobiy yo'nalishga ega va shuning uchun sirt bilan o'ralgan hududdan tashqariga qarab turishi kerak. Bu shuni anglatadiki, bu tekislikdagi normal vektorlar odatda yuqoriga qaragan bo'lishi kerak (i.e. ijobiy \(z\) komponentiga ega) bu normal vektor bajaradi.
Endi bizga quyidagi nuqta mahsuloti kerak bo'ladi va vektor maydonidagi \(z\) ga tekislik tenglamasini (albatta \(z\) uchun echilgan) ulashni unutmang.
\[\begin{align*}\vec F\left( {x,y,8 - 4x - 2y} \right)\centerdot \vec n & = \left\langle {y,2x, - 4x - 2y} \right\rangle \centerdot \frac{{\left\langle {4,2,1} \right\rangle }}{{\left\| {\nabla f} \right\|}}\\ & = \frac{1}{{\left\| {\nabla f} \right\|}}\left( {2y} \right)\end{align*}\]U holda integral,
\[\begin{align*}\iint\limits_{{{S_1}}}{{\vec F\centerdot \,d\vec S}} & = \iint\limits_{{{S_1}}}{{\frac{1}{{\left\| {\nabla f} \right\|}}\left( {2y} \right)\,dS}}\\ & = \iint\limits_{D}{{\frac{1}{{\left\| {\nabla f} \right\|}}\left( {2y} \right)\left\| {\nabla f} \right\|\,dA}}\\ & = \iint\limits_{D}{{2y\,dA}}\end{align*}\]Bu holda \(D\) shunchaki \({S_4}\) va shuning uchun biz endi integralni tugatishimiz mumkin.
\[\begin{align*}\iint\limits_{{{S_1}}}{{\vec F\centerdot \,d\vec S}} & = \iint\limits_{D}{{2y\,dA}}\\ & = \int_{0}^{2}{{\int_{0}^{{4 - 2x}}{{2y\,dy}}\,dx}}\\ & = \int_{0}^{2}{{\left. {{y^2}} \right|_0^{4 - 2x}\,dx}}\\ & = \int_{0}^{2}{{{{\left( {4 - 2x} \right)}^2}\,dx}}\\ & = \left. { - \frac{1}{6}{{\left( {4 - 2x} \right)}^3}} \right|_0^2 = \underline {\frac{{32}}{3}} \end{align*}\]3-qadam
Keyinchalik, biz \({S_2}\)-ga g'amxo'rlik qilamiz. Bu holda sirt uchun tenglama oddiygina \(y = 0\) va \(D\) 1-bosqichdagi \({S_2}\) eskizida berilgan.
Bu holda \({S_2}\) shunchaki \(xz\) tekisligining bir qismi bo'lib, biz ijobiy yo'nalishga egamiz, shuning uchun normal vektor sirt bilan o'ralgan hududdan uzoqroqqa yo'naltirilishi kerak. Bu shuni anglatadiki, bu holda oddiy vektor oddiygina \(\vec n = - \vec j = \left\langle {0, - 1,0} \right\rangle \).
Ushbu sirt uchun nuqta mahsuloti:
\[\vec F\left( {x,0,z} \right)\centerdot \vec n = \left\langle {0,2x,z - 8} \right\rangle \centerdot \frac{{\left\langle {0, - 1,0} \right\rangle }}{1} = - 2x\]\(y = 0\)-ni vektor maydoniga ulashni unutmang va gradientning kattaligi,
\[\left\| {\nabla f} \right\| = \sqrt {{{\left( 0 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 0 \right)}^2}} = 1\]U holda integral,
\[\begin{align*}\iint\limits_{{{S_2}}}{{\vec F\centerdot \,d\vec S}} & = \iint\limits_{{{S_2}}}{{ - 2x\,dS}}\\ & = \iint\limits_{D}{{ - 2x\left( 1 \right)\,dA}}\\ & = \int_{0}^{2}{{\int_{0}^{{8 - 4x}}{{ - 2x\,dz}}\,dx}}\\ & = \int_{0}^{2}{{\left. { - 2xz} \right|_0^{8 - 4x}\,dx}}\\ & = \int_{0}^{2}{{8{x^2} - 16x\,dx}}\\ & = \left. {\left( {\frac{8}{3}{x^3} - 8{x^2}} \right)} \right|_0^2 = \underline { - \frac{{32}}{3}} \end{align*}\]4-qadam
Endi biz \({S_3}\) bilan shug'ullanishimiz mumkin. Bu holda sirt uchun tenglama oddiygina \(x = 0\) va \(D\) 1-bosqichdagi \({S_3}\) eskizida berilgan.
Bu holda \({S_3}\) oddiygina \(yz\) tekisligining bir qismidir va biz ijobiy yo'nalishga egamiz, shuning uchun normal vektor sirt bilan o'ralgan hududdan uzoqroqqa yo'naltirilishi kerak. Bu shuni anglatadiki, bu holda oddiy vektor oddiygina \(\vec n = - \vec i = \left\langle { - 1,0,0} \right\rangle \).
Ushbu sirt uchun nuqta mahsuloti:
\[\vec F\left( {0,y,z} \right)\centerdot \vec n = \left\langle {y,0,z - 8} \right\rangle \centerdot \frac{{\left\langle { - 1,0,0} \right\rangle }}{1} = - y\]\(x = 0\)-ni vektor maydoniga ulashni unutmang va gradientning kattaligi,
\[\left\| {\nabla f} \right\| = \sqrt {{{\left( 1 \right)}^2} + {{\left( 0 \right)}^2} + {{\left( 0 \right)}^2}} = 1\]U holda integral,
\[\begin{align*}\iint\limits_{{{S_3}}}{{\vec F\centerdot \,d\vec S}} & = \iint\limits_{{{S_3}}}{{ - y\,dS}}\\ & = \iint\limits_{D}{{ - y\left( 1 \right)\,dA}}\\ & = \int_{0}^{4}{{\int_{0}^{{8 - 2y}}{{ - y\,dz}}\,dy}}\\ & = \int_{0}^{4}{{\left. { - yz} \right|_0^{8 - 2y}\,dy}}\\ & = \int_{0}^{4}{{2{y^2} - 8y\,dy}}\\ & = \left. {\left( {\frac{2}{3}{y^3} - 4{y^2}} \right)} \right|_0^4 = \underline { - \frac{{64}}{3}} \end{align*}\]5-qadam
Va nihoyat, \({S_4}\)-ga g'amxo'rlik qilaylik. Bu holda sirt uchun tenglama oddiygina \(z = 0\) va \(D\) 1-bosqichdagi \({S_4}\) eskizida berilgan.
Bu holda \({S_4}\) shunchaki \(xy\) tekisligining bir qismi bo'lib, biz ijobiy yo'nalishga egamiz, shuning uchun normal vektor sirt bilan o'ralgan hududdan uzoqroqqa yo'naltirilishi kerak. Bu shuni anglatadiki, bu holda normal vektor oddiygina \(\vec n = - \vec k = \left\langle {0,0, - 1} \right\rangle \).
Ushbu sirt uchun nuqta mahsuloti:
\[\vec F\left( {x,y,0} \right)\centerdot \vec n = \left\langle {y,2x, - 8} \right\rangle \centerdot \frac{{\left\langle {0,0, - 1} \right\rangle }}{1} = 8\]\(z = 0\)-ni vektor maydoniga ulashni unutmang va gradientning kattaligi,
\[\left\| {\nabla f} \right\| = \sqrt {{{\left( 0 \right)}^2} + {{\left( 0 \right)}^2} + {{\left( 1 \right)}^2}} = 1\]U holda integral,
\[\begin{align*}\iint\limits_{{{S_4}}}{{\vec F\centerdot \,d\vec S}} & = \iint\limits_{{{S_4}}}{{8\,dS}}\\ & = \iint\limits_{D}{{8\,dA}}\\ & = 8\iint\limits_{D}{{dA}}\\ & = 8\left( {{\mbox{Area of }}D} \right)\\ & = 8\left( {\frac{1}{2}} \right)\left( 2 \right)\left( 4 \right) = \underline {32} \end{align*}\]Bunday holda, biz ikki tomonlama integralni hisoblashimiz shart emasligiga e'tibor bering, chunki \(D\) shunchaki to'g'ri burchakli uchburchak edi va biz uning maydonini osongina hisoblashimiz mumkin.
6-qadam
Endi to'liq sirt bo'yicha integralning qiymatini olish uchun biz qilishimiz kerak bo'lgan narsa - yuqoridagi to'rtta sirt ustidagi har bir integralning qiymatlarini jamlash. Shunday qilib,
\[\iint\limits_{S}{{\vec F\centerdot \,d\vec S}} = \left( {\frac{{32}}{3}} \right) + \left( { - \frac{{32}}{3}} \right) + \left( { - \frac{{64}}{3}} \right) + \left( {32} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{32}}{3}}}\]Biz har bir alohida integral qiymatlar qayerdan kelganini ko'rsatish uchun qavslar qo'yamiz. Umuman olganda, bular, albatta, kerak emas.