StokesTheorem - 2-masala yechimi | MathModel.uz

17.5-bo'lim : Stoks teoremasi

2. $\displaystyle \iint\limits_{S}{{{\mathop{\rm curl}\nolimits} \vec F\centerdot d\vec S}}$ ni baholash uchun Stokes teoremasidan foydalaning, bunda $\vec F = \left( {{z^2} - 1} \right)\,\vec i + \left( {z + x{y^3}} \right)\,\vec j + 6\,\vec k$ va $S$ $x = - 2$ oldidagi $x = 6 - 4{y^2} - 4{z^2}$ qismidir.

Yechim

Keling, ushbu muammoni hal qilishda biz ishlayotgan sirtning qisqacha eskizi bilan boshlaylik.

$A three dimensional sketch of the surface drawn with the traditional axes. It is the portion of the elliptic paraboloid $x = 6 - 4{y^2} - 4{z^2}$ in front of $x = - 2$, which lies on its side pointing along the x-axis and opens towards the viewer. Heavy curved arrows around the open circular end show the direction of travel along the boundary curve $C$.$

$A three dimensional sketch of the same paraboloid, this time drawn with a set of “box” axes to make the surface easier to visualize. Heavy curved arrows around the open end again show the direction of travel along the boundary curve $C$.$

Biz an'anaviy o'qlari bo'lgan eskizni va sirtni ingl.

Sirtning yo'nalishi manfiy $x$ o'qi tomon yo'naltirilganligi sababli barcha normal vektorlar sirt bilan o'ralgan hududga ishora qiladi. Shunday qilib, agar biz sirtning chekkasi bo'ylab yursak, i.e. egri $C$, ko'rsatilgan yo'nalishda boshimiz sirt bilan o'ralgan hududga ishora qiladi (i.e. normal vektorlar bilan bir xil yo'nalishda) keyin bizning chap qo'limiz mintaqa ustida bo'ladi. Shuning uchun, eskizda ko'rsatilgan yo'nalish $C$ ning ijobiy yo'nalishidir.

Agar siz egri chiziq yo'nalishini tasavvur qilishda muammoga duch kelsangiz, shunchaki chashka yoki piyola oling va uni orqasiga qog'oz varaqasi bilan yon tomonga qo'ying. Vizualizatsiyaga yordam berish uchun qog'oz varag'iga chashka / piyola oldida o'tirgan tekislikni aks ettiruvchi o'qlar to'plamini chizing. Keyin kichkina tayoqchani kesib oling va uning "old" tomoniga yuz qo'ying va chap qo'lni yorqin rangga bo'yab, uni tezda ko'rishingiz mumkin. Endi, chashka / piyola / nima bo'lishidan qat'iy nazar, chetiga tayoq shaklini qo'ying, uning boshi normal vektorlar yo'nalishi bo'yicha (bizning holatda paraboloid / chashka / piyola ichiga) chap qo'l bilan sirt ustida. Tayoq figurasidagi "yuz" qaragan yo'nalish ijobiy yo'nalishni olish uchun sirt bo'ylab yurishingiz kerak bo'lgan yo'nalishdir. $C$.

2-qadam

Biz Stokes teoremasidan quyidagi yo'nalishda foydalanamiz.

\[\iint\limits_{S}{{{\mathop{\rm curl}\nolimits} \vec F\centerdot d\vec S}} = \int\limits_{C}{{\vec F\centerdot d\vec r}}\]

Muammo bayonotida vektor maydoni bizga berilgan, shuning uchun biz bu haqda tashvishlanishga hojat yo'q. Biz $C$ bilan shug'ullanishimiz kerak.

Bu holda $C$ - muammo bayonotidagi ikkita tenglamani teng o'rnatish orqali biz oladigan egri chiziq. Shunday qilib,

\[ - 2 = 6 - 4{y^2} - 4{z^2}\hspace{0.25in} \to \hspace{0.25in}4{y^2} + 4{z^2} = 8\hspace{0.25in} \Rightarrow \hspace{0.25in}{y^2} + {z^2} = 2\]

Agar biz paraboloid oldida bo'lsak va to'g'ridan-to'g'ri $x$ o'qi bo'ylab qarasak, $C$ ning quyidagi eskizini ko'ramiz.

$A sketch of the curve $C$ as seen from in front of the paraboloid looking straight down the x-axis, so $y$ is the horizontal axis and $z$ is the vertical axis. It is the circle ${y^2} + {z^2} = 2$ of radius $\sqrt 2 $ centered at the origin and the arrows show the direction it is traveled.$

$C$ ning mumkin bo'lgan parametrlaridan biri quyidagicha berilgan:

\[\vec r\left( t \right) = \left\langle { - 2,\sqrt 2 \sin t,\sqrt 2 \cos t} \right\rangle \hspace{0.25in}\hspace{0.25in}0 \le t \le 2\pi \]

Parametrlashning $x$ komponenti -2 ga teng, chunki $C$ $x = - 2$ da joylashgan.

3-qadam

Biz oxir-oqibat chiziqli integralni bajarishimiz kerakligini bilganimiz sababli, bizga quyidagi nuqta mahsuloti kerakligini bilamiz.

\[\begin{align*}\vec F\left( {\vec r\left( t \right)} \right)\centerdot \vec r'\left( t \right) & = \left\langle {2{{\cos }^2}t - 1,\sqrt 2 \cos t - 4\sqrt 2 {{\sin }^3}t,6} \right\rangle \centerdot \left\langle {0,\sqrt 2 \cos t, - \sqrt 2 \sin t} \right\rangle \\ & = \sqrt 2 \cos t\left( {\sqrt 2 \cos t - 4\sqrt 2 {{\sin }^3}t} \right) - 6\sqrt 2 \sin t\\ & = 2{\cos ^2}t - 8\cos t{\sin ^3}t - 6\sqrt 2 \sin t\\ & = \left( {1 + \cos \left( {2t} \right)} \right) - 8\cos t{\sin ^3}t - 6\sqrt 2 \sin t\end{align*}\]

$C$ parametrlarini vektor maydoniga ulashni unutmang!

Biz integratsiyani ko'zlagan holda birinchi muddatda biroz soddalashtirdik.

4-qadam

Mayli, keling, Stoks teoremasi yordamida integralni baholaylik.

\[\begin{align*}\iint\limits_{S}{{{\mathop{\rm curl}\nolimits} \vec F\centerdot d\vec S}} & = \int\limits_{C}{{\vec F\centerdot d\vec r}}\\ & = \int_{0}^{{2\pi }}{{\left( {1 + \cos \left( {2t} \right)} \right) - 8\cos t{{\sin }^3}t - 6\sqrt 2 \sin t\,dt}}\\ & = \left. {\left( {t + \frac{1}{2}\sin \left( {2t} \right) - 2{{\sin }^4}t + 6\sqrt 2 \cos t\,dt} \right)} \right|_0^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{2\pi }}\end{align*}\]