17.3-bo'lim : Yuzaki integrallar
3. \( \displaystyle \iint\limits_{S}{{2y\,dS}}\) ni baholang, bu erda \(S\) \({y^2} + {z^2} = 4\) ning \(x = 0\) va \(x = 3 - z\) o'rtasidagi qismidir.
Yechim
Keling, ushbu muammoni hal qilishda biz ishlayotgan sirtning qisqacha eskizi bilan boshlaylik.
E'tibor bering, bu muammodagi sirt faqat silindrning o'zi. Tsilindrning "qopqoqlari" eskizdagi qizil "qopqoq" ga qaramasdan, bu sirtning bir qismi emas. Bu eskizda silindrning old chetini aniq qilish uchun eskizga kiritilgan. Agar muammo aniq bo'lishi uchun qandaydir tarzda belgilab qo'ygan bo'lsa, biz "shlyapalar" ni kiritgan bo'lardik.
2-qadam
Endi bizning sirtimiz silindr bo'lgani uchun biz uni parametrlashimiz va sirt integrali uchun quyidagi formuladan foydalanishimiz kerak.
\[\iint\limits_{S}{{f\left( {x,y,z} \right)\,dS}} = \iint\limits_{D}{{f\left( {\vec r\left( {u,v} \right)} \right)\left\| {{{\vec r}_u} \times {{\vec r}_v}} \right\|\,dA}}\]Bu erda parametrlashtirishni amalga oshirishda kerak bo'lganda \(u\) va \(v\) tanlanadi.
Oldingi bo'limda tsilindrni qanday parametrlash kerakligini ko'rdik, shuning uchun parametrlash uchun batafsil ma'lumot bermaymiz. Parametrlash - bu,
\[\vec r\left( {x,\theta } \right) = \left\langle {x,2\sin \theta ,2\cos \theta } \right\rangle \hspace{0.25in}\,\,\,\,\,0 \le \theta \le 2\pi ,\,\,\,\,\,0 \le x \le 3 - z = 3 - 2\cos \theta \]Biz \(\theta \) ning to‘liq diapazonidan foydalanamiz, chunki biz uning butun atrofida aylanishiga ruxsat beramiz. x‑o'qi. \(x\) chegaralari silindrni "bog'laydigan" ikkita tekislikdan kelib chiqadi va biz parametrlash yordamida yuqori chegarani aylantirishimiz kerak.
Keyinchalik, biz o'zaro mahsulotni hisoblashimiz kerak.
\[{\vec r_x} = \left\langle {1,0,0} \right\rangle \hspace{0.25in}{\vec r_\theta } = \left\langle {0,2\cos \theta , - 2\sin \theta } \right\rangle \] \[{\vec r_x} \times {\vec r_\theta } = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\1&0&0\\0&{2\cos \theta }&{ - 2\sin \theta }\end{array}} \right| = 2\sin \theta \vec j + 2\cos \theta \vec k\]O'zaro mahsulotning kattaligi,
\[\left\| {{{\vec r}_x} \times {{\vec r}_\theta }} \right\| = \sqrt {4{{\sin }^2}\theta + 4{{\cos }^2}\theta } = 2\]U holda integral,
\[\iint\limits_{S}{{2y\,dS}} = \iint\limits_{D}{{2\left( {2\sin \theta } \right)\left( 2 \right)\,dA}} = \iint\limits_{D}{{8\sin \theta \,dA}}\]Yuzaki parametrlashtirishning \(y\) komponentini integrandaga ulashni unutmang va \(D\) bu faqat \(x\) va \(\theta \) uchun biz yuqorida parametrlashda qayd etgan chegaralardir.
3-qadam
Endi biz qilishimiz kerak bo'lgan yagona narsa - ikki karrali integralni baholash va bu hozir juda qiyin bo'lmasligi kerak.
U holda integral,
\[\begin{align*}\iint\limits_{S}{{2y\,dS}} & = \int_{0}^{{2\pi }}{{\int_{0}^{{3 - 2\cos \theta }}{{8\sin \theta \,dx}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\left. {\left( {8x\sin \theta } \right)} \right|_0^{3 - 2\cos \theta }\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{8\left( {3 - 2\cos \theta } \right)\sin \theta \,d\theta }}\\ & = \int_{0}^{{2\pi }}{{24\sin \theta - 16\sin \theta \cos \theta \,d\theta }}\\ & = \int_{0}^{{2\pi }}{{24\sin \theta - 8\sin \left( {2\theta } \right)\,d\theta }}\\ & = \left. {\left( { - 24\cos \theta + 4\cos \left( {2\theta } \right)\,} \right)} \right|_0^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{0}\end{align*}\]