15.4-bo'lim : Qutb koordinatalarida ikki karrali integrallar
1. \(\displaystyle \iint\limits_{D}{{{y^2} + 3x\,dA}}\) ni baholang, bu erda \(D\) 3-banddagi mintaqadir.rd \({x^2} + {y^2} = 1\) va \({x^2} + {y^2} = 9\) o'rtasidagi kvadrant.
Yechim
Quyida \(D\) hududining qisqacha eskizi keltirilgan.
Qutb koordinatalari qo'llaniladigan ikki karrali integrallar uchun \(D\) eskizi ko'pincha umumiy mintaqadagi kabi foydali emas.
Ammo, agar boshqa hech narsa bo'lmasa, bu muammoni hal qilish uchun qutb koordinatalari kerakligini aniq ko'rsatib beradi. Bu mintaqani Dekart koordinatalari nuqtai nazaridan tavsiflash mumkin, ammo muammoni hal qilish uchun ikkita integral kerak bo'ladi va chegaralarning aksariyati ko'pincha (har doim emas, lekin tez-tez) tartibsiz integral ishlarga olib keladigan ildizlarni o'z ichiga oladi.
Eskiz shuni ko'rsatadiki, mintaqa hech bo'lmaganda qisman dumaloq va bu har doim qutb koordinatalari hech bo'lmaganda o'ylash uchun yomon narsa emasligini ko'rsatishi kerak. Bunday holda, yuqorida muhokama qilingan Dekart chegaralari tufayli qutb koordinatalari bu integralni bajarishning deyarli yagona oson usuli hisoblanadi.
Shuni ham yodda tutingki, bizda qutb chegaralarini aniqlaydigan eskiz juda oddiy bo'lishi kerak.
2-qadam
Mayli, \(D\) uzukning faqat bir qismi, shuning uchun chegaralarni o'rnatish unchalik qiyin bo'lmasligi kerak. Mana ular,
\[\begin{array}{c}\pi \le \theta \le \frac{3}{2}\pi \\ \,1 \le r \le 3\end{array}\]3-qadam
U holda qutb koordinatalari bo'yicha integral,
\[\begin{align*}\iint\limits_{D}{{{y^2} + 3x\,dA}} & = \int_{\pi }^{{\frac{3}{2}\pi }}{{\int_{1}^{3}{{\left( {{{\left( {r\sin \theta } \right)}^2} + 3r\cos \theta } \right)r\,dr}}\,d\theta }}\\ & = \int_{\pi }^{{\frac{3}{2}\pi }}{{\int_{1}^{3}{{{r^3}{{\sin }^2}\theta + 3{r^2}\cos \theta \,dr}}\,d\theta }}\end{align*}\]Integralni aylantirishda \(x\) va \(y\) ni qutb koordinatalariga aylantirishni unutmang. Bundan tashqari, \(dA = r\,dr\,d\theta \) ni unutmang va shuning uchun biz integranda qo'shimcha \(r\) ni olamiz. Qo'shimcha \(r\) ni unutish bu turdagi muammolar bilan eng ko'p uchraydigan xatolardan biridir.
4-qadam
Mana \(r\) integratsiyasi.
\[\begin{align*}\iint\limits_{D}{{{y^2} + 3x\,dA}} & = \int_{\pi }^{{\frac{3}{2}\pi }}{{\int_{1}^{3}{{{r^3}{{\sin }^2}\theta + 3{r^2}\cos \theta \,dr}}\,d\theta }}\\ & = \int_{\pi }^{{\frac{3}{2}\pi }}{{\left. {\left( {\frac{1}{4}{r^4}{{\sin }^2}\theta + {r^3}\cos \theta } \right)} \right|_1^3\,d\theta }}\\ & = \int_{\pi }^{{\frac{3}{2}\pi }}{{20{{\sin }^2}\theta + 26\cos \theta \,d\theta }}\end{align*}\]5-qadam
Va nihoyat, \(\theta \) integratsiyasi.
\[\begin{align*}\iint\limits_{D}{{{y^2} + 3x\,dA}} & = \int_{\pi }^{{\frac{3}{2}\pi }}{{20{{\sin }^2}\theta + 26\cos \theta \,d\theta }}\\ & = \int_{\pi }^{{\frac{3}{2}\pi }}{{10\left( {1 - \cos \left( {2\theta } \right)} \right) + 26\cos \theta \,d\theta }}\\ & = \left. {\left( {10\theta - 5\sin \left( {2\theta } \right) + 26\sin \theta } \right)} \right|_\pi ^{\frac{3}{2}\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{5\pi - 26}}\end{align*}\]Siz qutbli integrallarda juda ko'p \({\cos ^2}\theta \), \({\sin ^2}\theta \) va \(\sin \theta \cos \theta \) atamalarini ko'rasiz, shuning uchun bu atamalarni qanday integrallashni bilganingizga ishonch hosil qiling! Bu holda biz \({\sin ^2}\theta \) ni birlashtira oladigan narsaga kamaytirish uchun yarim burchak formulasidan foydalandik.