17.6-bo'lim : Divergensiya teoremasi
3. \( \displaystyle \iint\limits_{S}{{\vec F\centerdot d\vec S}}\) ni baholash uchun Divergensiya teoremasidan foydalaning, bunda \(\vec F = 2xz\vec i + \left( {1 - 4x{y^2}} \right)\,\vec j + \left( {2z - {z^2}} \right)\,\vec k\) va \(S\) \(z = 6 - 2{x^2} - 2{y^2}\) va \(z = 0\) tekisligi bilan chegaralangan qattiq jismning yuzasidir. E'tibor bering, ushbu qattiq jismning ikkala yuzasi \(S\) tarkibiga kiradi.
Yechim
Keling, ushbu muammoni hal qilishda biz ishlayotgan sirtning qisqacha eskizi bilan boshlaylik.
Biz an'anaviy o'qlari bo'lgan eskizni va sirtni ingl. Elliptik paraboloidning pastki "qopqog'i" ham sirtga kiritilgan, ammo ko'rsatilmagan.
Shuni ham yodda tutingki, biz yuqorida ko'rsatilgan ikkala sirtni ham o'z ichiga olganmiz, chunki sirt mintaqani o'rab oladi (yoki agar siz ushbu terminologiyadan foydalanmoqchi bo'lsangiz, chegara egri chizig'i).
2-qadam
Divergensiya teoremasidan quyidagi yo'nalishda foydalanamiz.
\[\iint\limits_{S}{{\vec F\centerdot d\vec S}} = \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F\,dV}}\]Bu erda \(E\) 1-qadamdagi eskizlarda ko'rsatilgan qattiq narsadir.
Uch karrali integralni takrorlangan integralga aylantirish uchun bizga kerak bo'ladigan \(D\) mintaqasi muammo bayonidagi ikkita sirtning kesishishidir. Bu,
\[0 = 6 - 2{x^2} - 2{y^2}\hspace{0.25in} \to \hspace{0.25in} {x^2} + {y^2} = 3\]Shunday qilib, \(D\) bu disk va shuning uchun biz oxir-oqibat bu integral uchun silindrsimon koordinatalarni qilamiz. Mana \(E\) mintaqasi uchun silindrsimon chegaralar.
\[\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le \sqrt 3 \\ 0 \le z \le 6 - 2{x^2} - 2{y^2} = 6 - 2{r^2}\end{array}\]\(z\) chegaralarini silindrsimon koordinatalarga aylantirishni ham unutmang!
Bizga vektor maydonining divergentsiyasi ham kerak bo'ladi, shuning uchun bu erda.
\[{\mathop{\rm div}\nolimits} \vec F = \frac{\partial }{{\partial x}}\left( {2xz} \right) + \frac{\partial }{{\partial y}}\left( {1 - 4x{y^2}} \right) + \frac{\partial }{{\partial z}}\left( {2z - {z^2}} \right) = 2 - 8xy\]3-qadam
Endi integralga Divergensiya teoremasini qo'llaymiz va uni silindrsimon koordinatalarga aylantiramiz.
\[\begin{align*}\iint\limits_{S}{{\vec F\centerdot d\vec S}} & = \iiint\limits_{E}{{{\mathop{\rm div}\nolimits} \vec F\,dV}}\\ & = \iiint\limits_{E}{{2 - 8xy\,dV}}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{\int_{0}^{{6 - 2{r^2}}}{{\left( {2 - 8{r^2}\cos \theta \sin \theta } \right)r\,dz}}\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{\int_{0}^{{6 - 2{r^2}}}{{2r - 8{r^3}\cos \theta \sin \theta \,dz}}\,dr}}\,d\theta }}\end{align*}\]\(dV\) ni silindrsimon koordinatalarga o'tkazishda \(r\) ni olishni unutmang.
4-qadam
Keyin integralni baholashda qilishimiz kerak.
\[\begin{align*}\iint\limits_{S}{{\vec F\centerdot d\vec S}} & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{\int_{0}^{{6 - 2{r^2}}}{{2r - 8{r^3}\cos \theta \sin \theta \,dz}}\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{\left. {\left( {2r - 8{r^3}\cos \theta \sin \theta } \right)z} \right|_0^{6 - 2{r^2}}\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{\left( {2r - 8{r^3}\cos \theta \sin \theta } \right)\left( {6 - 2{r^2}} \right)\,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 3 }}{{12r - 4{r^3} - \left( {48{r^3} - 16{r^5}} \right)\cos \theta \sin \theta \,dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\left. {\left( {6{r^2} - {r^4} - \left( {12{r^4} - \frac{8}{3}{r^6}} \right)\cos \theta \sin \theta } \right)} \right|_0^{\sqrt 3 }\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{9 - 36\cos \theta \sin \theta \,d\theta }}\\ & = \int_{0}^{{2\pi }}{{9 - 18\sin \left( {2\theta } \right)\,d\theta }}\\ & = \left. {\left( {9\theta + 9\cos \left( {2\theta } \right)} \right)} \right|_0^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{18\pi }}\end{align*}\]