Asosiy tarkibga o'tish

9.8-bo'lim: Polar koordinatali maydon

Ushbu bo'limda biz qutbli egri chiziqlar bilan o'ralgan maydonlarni ko'rib chiqamiz. Shuni ham yodda tutingki, biz odatda bu muammolarda bo'lgani kabi, "ostida" o'rniga "to'g'ridan-to'g'ri" dedik. Bu muammolar qutb koordinatalarida biroz boshqacha ishlaydi. Bu bo'limda biz topadigan maydon qanday ko'rinishga ega ekanligining eskizi.

This is a sketch that looks a lot like a slice of pie that is in the 1st quadrant. The “point” of the piece of pie is in the origin and is between the two angles from $\theta = \alpha$ to $\theta = \beta$ and the “crust” of the piece of pie is given by $r=f\left( \theta \right)$.

Biz yuqoridagi eskizdagi soyali joyni qidiramiz. Bu maydonni topish formulasi:

Polar egri chiziq bilan o'ralgan maydon

\[A = \int_{{\,\alpha }}^{{\,\beta }}{{\frac{1}{2}{r^2}\,d\theta }}\]

E'tibor bering, biz integralda \(f\left( \theta \right)\) o'rniga \(r\) dan foydalanamiz, shuning uchun integralni bajarayotganda ishonch hosil qiling va shunga mos ravishda almashtiring.

Keling, bir misolni ko'rib chiqaylik.

1-misol \(r = 2 + 4\cos \theta \) ning ichki halqasining maydonini aniqlang.
Yechim

Biz bu funktsiyani birinchi marta ko'rishni boshlaganimizda grafigini tuzdik qutb koordinatalari. Ushbu muammoni hal qilish uchun biz \(\theta \) qiymatlarini ham bilishimiz kerak bo'ladi, bu erda egri chiziq boshlang'ichdan o'tadi. Bularni tenglamani nolga tenglashtirib, yechish orqali olishimiz mumkin.

\[\begin{align*}0 & = 2 + 4\cos \theta \\ \cos \theta & = - \frac{1}{2}\hspace{0.5in} \Rightarrow \hspace{0.5in}\theta = \frac{{2\pi }}{3},\frac{{4\pi }}{3}\end{align*}\]

Mana bu egri chiziqning ichki halqasi soyalangan eskizi.

This is the graph of $r=2+4\cos \theta $. The heart shaped portion of this graph goes through the points (in Cartesian coordinates to make it a little easier to visualize the graph) (6,0), (0,2), (0,2) and the origin. The inner loop is in the 1st and 4th quadrant and goes up to the point (2,0). Also included are two lines moving out of the origin. The first is in the 2nd quadrant and forms an angle of $\frac{2\pi}{3}$ with the positive x-axis. The second is in the 3rd quadrant and forms an angle of $\frac{4\pi}{3}$ with the positive x-axis.

Nima uchun biz egri chiziqning kelib chiqishidan o'tadigan \(\theta \) qiymatlarini bilishimiz kerakligini tushuna olasizmi? Bu nuqtalar ichki halqaning qayerda boshlanishi va tugashini belgilaydi va shuning uchun ham formuladagi integratsiya chegaralari hisoblanadi.

Shunday qilib, maydon, keyin,

\[\begin{align*}A & = \int_{{\,\frac{{2\pi }}{3}}}^{{\,\frac{{4\pi }}{3}}}{{\frac{1}{2}{{\left( {2 + 4\cos \theta } \right)}^2}\,d\theta }}\\ & = \int_{{\,\frac{{2\pi }}{3}}}^{{\,\frac{{4\pi }}{3}}}{{\frac{1}{2}\left( {4 + 16\cos \theta + 16{{\cos }^2}\theta } \right)\,d\theta }}\\ & = \int_{{\,\frac{{2\pi }}{3}}}^{{\,\frac{{4\pi }}{3}}}{{2 + 8\cos \theta + 4\left( {1 + \cos \left( {2\theta } \right)} \right)\,d\theta }}\\ & = \int_{{\,\frac{{2\pi }}{3}}}^{{\,\frac{{4\pi }}{3}}}{{6 + 8\cos \theta + 4\cos \left( {2\theta } \right)\,d\theta }}\\ & = \left. {\left( {6\theta + 8\sin \theta + 2\sin \left( {2\theta } \right)} \right)} \right|_{\frac{{2\pi }}{3}}^{\frac{{4\pi }}{3}}\\ & = 4\pi - 6\sqrt 3 = 2.174\end{align*}\]

Siz ushbu integralda bajarilgan ishlarga amal qildingiz, shunday emasmi? Ushbu bo'limda siz trig-funksiyalarning bir nechta integrallariga duch kelasiz, shuning uchun agar kerak bo'lsa, sizga qaytishingiz kerak. Trig funksiyalarni o'z ichiga olgan integrallar bo'limlari va tez ko'rib chiqish.

Shunday qilib, biz bitta egri chiziq bilan o'ralgan maydonlarni aniqlaymiz, ammo ikkita egri chiziq orasidagi maydonni topmoqchi bo'lgan quyidagi eskiz kabi vaziyatlar haqida nima deyish mumkin.

This is a sketch that looks a lot like an upper portion of a slice of pie that is in the 1st quadrant. The region is between the two angles from $\theta = \alpha$ to $\theta = \beta$. The “outer” edge of the region is given by $r_{o}=g\left( \theta \right)$. The “inner” edge of the region is given by $r_{i}=f\left( \theta \right)$.

Bunday holda, ikkalasi tomonidan o'ralgan maydonni topish uchun yuqoridagi formuladan foydalanishimiz mumkin va keyin haqiqiy maydon ikkalasi orasidagi farqdir. Buning formulasi,

Ikki qutb egri chizig'i orasidagi maydon

\[A = \int_{{\,\alpha }}^{{\,\beta }}{{\frac{1}{2}\left( {r_o^2 - r_i^2} \right)\,d\theta }}\]

Keling, buning bir misolini ko'rib chiqaylik.

2-misol \(r = 3 + 2\sin \theta \) ichida va \(r = 2\) tashqarisida joylashgan maydonni aniqlang.
Yechim

Mana biz kutayotgan mintaqaning eskizi.

This is the graph of $r=3+2\sin \theta $. This is a vaguely heart shaped portion of this graph goes through the points (in Cartesian coordinates to make it a little easier to visualize the graph) (3,0), (0,5), (-3,0) and (0,-2). The “crease” in the heart is at (0,-2). Also included on the graph is the circle give by r=2. The crease of the first graph is inside the circle and the upper portion of the circle is inside the heart. The circle and heart intersect in the 3rd and 4th quadrant. The area from the heart down to the circle that is in the 1st and 2nd quadrant is shaded.

Ushbu maydonni aniqlash uchun biz ikkita egri chiziq kesishadigan \(\theta \) qiymatlarini bilishimiz kerak. Biz bu nuqtalarni ikkita tenglamani o'rnatish va yechish orqali aniqlashimiz mumkin.

\[\begin{align*}3 + 2\sin \theta & = 2\\ \sin \theta & = - \frac{1}{2}\hspace{0.5in} \Rightarrow \hspace{0.5in}\theta = \frac{{7\pi }}{6},\frac{{11\pi }}{6}\end{align*}\]

Mana bu burchaklar qo'shilgan shaklning eskizi.

This is the same graph as above except two lines have been added to the graph. The first line comes out of the origin and goes through the intersection in the 4th quadrant makes angles of $-\frac{\pi}{6}$ or $\frac{11\pi}{6}$ (depending on direction of rotation) with the positive x-axis. The second line comes out of the origin and goes through the intersection in the 3rd quadrant makes angles of $\frac{7\pi}{6}$ with the positive x-axis.

Shuni ham yodda tutingki, biz \(\frac{{11\pi }}{6}\) burchagining yana bir ko'rinishi \( - \frac{\pi }{6}\) ekanligini tan oldik. Bu muammo uchun muhim ahamiyatga ega. Yuqoridagi formuladan foydalanish uchun biz kichikroq burchakdan kattaroq burchakka o'tayotganda maydonni o'rab olishimiz kerak. Shunday qilib, agar biz \(\frac{{7\pi }}{6}\)-dan \(\frac{{11\pi }}{6}\)-ga foydalansak, biz soyali maydonni o'rab olmaymiz, aksincha, uchta mintaqaning pastki qismini qamrab olamiz. Biroq, agar biz \( - \frac{\pi }{6}\) dan \(\frac{{7\pi }}{6}\) burchaklaridan foydalansak, biz kutayotgan maydonni o'rab olamiz.

Shunday qilib, maydon, keyin,

\[\begin{align*}A & = \int_{{\, - \frac{\pi }{6}}}^{{\,\frac{{7\pi }}{6}}}{{\frac{1}{2}\left( {{{\left( {3 + 2\sin \theta } \right)}^2} - {{\left( 2 \right)}^2}} \right)\,d\theta }}\\ & = \int_{{\, - \frac{\pi }{6}}}^{{\,\frac{{7\pi }}{6}}}{{\frac{1}{2}\left( {5 + 12\sin \theta + 4{{\sin }^2}\theta } \right)\,d\theta }}\\ & = \int_{{\, - \frac{\pi }{6}}}^{{\,\frac{{7\pi }}{6}}}{{\frac{1}{2}\left( {7 + 12\sin \theta - 2\cos \left( {2\theta } \right)} \right)\,d\theta }}\\ & = \left. {\frac{1}{2}\left( {7\theta - 12\cos \theta - \sin \left( {2\theta } \right)} \right)} \right|_{ - \frac{\pi }{6}}^{\frac{{7\pi }}{6}}\\ & = \frac{{11\sqrt 3 }}{2} + \frac{{14\pi }}{3} = 24.187\end{align*}\]

Keling, oldingi misolni biroz o'zgartirish bilan ishlaylik.

3-misol \(r = 3 + 2\sin \theta \) tashqarisidagi va \(r = 2\) ichidagi hududning maydonini aniqlang.
Yechim

Bu safar biz quyidagi mintaqani qidirmoqdamiz.

This is the graph of $r=3+2\sin \theta $. This is a vaguely heart shaped portion of this graph goes through the points (in Cartesian coordinates to make it a little easier to visualize the graph) (3,0), (0,5), (-3,0) and (0,-2). The “crease” in the heart is at (0,-2). Also included on the graph is the circle give by r=2. The crease of the first graph is inside the circle and the upper portion of the circle is inside the heart. The circle and heart intersect in the 3rd and 4th quadrant. Two lines have been added to the graph. The first line comes out of the origin and goes through the intersection in the 4th quadrant makes angles of $-\frac{\pi}{6}$ or $\frac{11\pi}{6}$ (depending on direction of rotation) with the positive x-axis. The second line comes out of the origin and goes through the intersection in the 3rd quadrant makes angles of $\frac{7\pi}{6}$ with the positive x-axis. The area from the heart down to the circle that is in the 3rd and 4th quadrant is shaded.

Shunday qilib, bu biz \(\frac{{7\pi }}{6}\) chegaralaridan \(\frac{{11\pi }}{6}\) gacha bo'lgan hududni olamiz. Bu hudud uchun hudud,

\[\begin{align*}A & = \int_{{\,\frac{{7\pi }}{6}}}^{{\,\frac{{11\pi }}{6}}}{{\frac{1}{2}\left( {{{\left( 2 \right)}^2} - {{\left( {3 + 2\sin \theta } \right)}^2}} \right)\,d\theta }}\\ & = \int_{{\,\frac{{7\pi }}{6}}}^{{\,\frac{{11\pi }}{6}}}{{\frac{1}{2}\left( { - 5 - 12\sin \theta - 4{{\sin }^2}\theta } \right)\,d\theta }}\\ & = \int_{{\,\frac{{7\pi }}{6}}}^{{\,\frac{{11\pi }}{6}}}{{\frac{1}{2}\left( { - 7 - 12\sin \theta + 2\cos \left( {2\theta } \right)} \right)\,d\theta }}\\ & = \left. {\frac{1}{2}\left( { - 7\theta + 12\cos \theta + \sin \left( {2\theta } \right)} \right)} \right|_{\frac{{7\pi }}{6}}^{\frac{{11\pi }}{6}}\\ & = \frac{{11\sqrt 3 }}{2} - \frac{{7\pi }}{3} = 2.196\end{align*}\]

E'tibor bering, bu soha uchun "tashqi" va "ichki" funktsiyalar qarama-qarshi edi!

Keling, ushbu misolning yakuniy modifikatsiyasini qilaylik.

4-misol \(r = 3 + 2\sin \theta \) va \(r = 2\) ichidagi maydonni aniqlang.
Yechim

Mana bu misol uchun eskiz.

This is the graph of $r=3+2\sin \theta $. This is a vaguely heart shaped portion of this graph goes through the points (in Cartesian coordinates to make it a little easier to visualize the graph) (3,0), (0,5), (-3,0) and (0,-2). The “crease” in the heart is at (0,-2). Also included on the graph is the circle give by r=2. The crease of the first graph is inside the circle and the upper portion of the circle is inside the heart. The circle and heart intersect in the 3rd and 4th quadrant. Two lines have been added to the graph. The first line comes out of the origin and goes through the intersection in the 4th quadrant makes angles of $-\frac{\pi}{6}$ or $\frac{11\pi}{6}$ (depending on direction of rotation) with the positive x-axis. The second line comes out of the origin and goes through the intersection in the 3rd quadrant makes angles of $\frac{7\pi}{6}$ with the positive x-axis. The area that is inside both graphs has been shaded.

Biz bu muammoni oldingi ikkitasini qilgandek hal qila olmaymiz. Biridan ikkinchisiga ko'payganimizda, bu hududni o'rab olishga imkon beradigan cheklovlar to'plami yo'q. Esda tutingki, biz \(\theta \) ni oshirganimizda, biz kutayotgan maydon o'ralgan bo'lishi kerak. Biroq, biz ishlashimiz mumkin bo'lgan \(\theta \) uchun ikkita diapazon bu mintaqani emas, balki oldingi ikkita misoldagi hududni qamrab oladi.

Biroq, bu holda, bu katta muammo emas. Ushbu muammoda hududni olishning ikki yo'li mavjud. Biz ikkalasini ham ko'rib chiqamiz.

Yechim 1
Bunday holda, aylana ikki qismga bo'linganiga e'tibor bering va biz yuqori qismdan keyin turibmiz. Shuningdek, biz 3-misolda pastki qismning maydonini topganimizga e'tibor bering. Shuning uchun, maydon,

\[\begin{align*}{\mbox{Area}} & = {\mbox{Area of Circle}} - {\mbox{Area from Example 3}}\\ & = \pi {\left( 2 \right)^2} - 2.196\\ & = 10.370\end{align*}\]

Yechim 2
Bu holda biz deyarli bir xil ishni qilamiz, bundan tashqari bu safar biz 2-misolda ko'rib chiqilayotgan qismdan ko'ra limakonning boshqa qismi deb hisoblaymiz. Shuningdek, bu holatda limakonning maydonini hisoblashimiz kerak bo'ladi.

Shunday qilib, ushbu yondashuvdan foydalanadigan maydon

\[\begin{align*}{\mbox{Area}} & = {\mbox{Area of Limacon}} - {\mbox{Area from Example 2}}\\ & = \int_{{\,0}}^{{\,2\pi }}{{\frac{1}{2}{{\left( {3 + 2\sin \theta } \right)}^2}\,d\theta }} - 24.187\\ & = \int_{{\,0}}^{{\,2\pi }}{{\frac{1}{2}\left( {9 + 12\sin \theta + 4{{\sin }^2}\theta } \right)\,d\theta }} - 24.187\\ & = \int_{{\,0}}^{{\,2\pi }}{{\frac{1}{2}\left( {11 + 12\sin \theta - 2\cos \left( {2\theta } \right)} \right)\,d\theta }} - 24.187\\ & = \left. {\frac{1}{2}\left( {11\theta - 12\cos \left( \theta \right) - \sin \left( {2\theta } \right)} \right)} \right|_0^{2\pi } - 24.187\\ & = 11\pi - 24.187\\ & = 10.370\end{align*}\]

Biroz uzoqroq yondashuv, lekin ba'zida biz bu uzoqroq yondashuvni qo'llashga majbur bo'lamiz.

Ushbu so'nggi misol ko'rsatganidek, biz qutb koordinatalaridagi barcha maydonlarni integraldan to'g'ridan-to'g'ri ololmaydi.