Shunday qilib, yuqorida aytib o'tilganidek, \(r\) chegaralari (umid qilamanki) juda aniq. \(\theta \) chegaralarini topish uchun ikkita tenglamani teng o'rnatish kifoya. Shunday qilib,

\[\begin{align*}4 + 2\sin \theta &= 3 - \sin \theta \\ 3\sin \theta & = - 1\\ \sin \theta & = - \frac{1}{3}\hspace{0.5in} \Rightarrow \hspace{0.5in}\theta = {\sin ^{ - 1}}\left( { - \frac{1}{3}} \right) = - 0.3398\end{align*}\]

Bu to'rtinchi kvadrantdagi burchak, i.e. \(\alpha = - 0.3398\). Ikkinchi burchakni topish uchun ikkita qora chiziqning \(x\) o'qi bilan yaratadigan burchaklari bir xil ekanligini ta'kidlashimiz mumkin (albatta, belgiga e'tibor bermaslik). Shunday qilib, ikkinchi burchak oddiygina \(\beta = \pi + 0.3398 = 3.4814\) bo'ladi.

Keyin chegaralar,

\[\begin{array}{c} - 0.3398 \le \theta \le 3.4814\\ 3 - \sin \theta \le r \le 4 + 2\sin \theta \end{array}\]

Biz bilamizki, quyidagi ikki karrali integral har qanday \(D\) mintaqasining maydonini beradi.

\[A = \iint\limits_{D}{{dA}}\]

Bizning muammomiz uchun \(D\) mintaqasi muammo bayonotida berilgan mintaqadir va shuning uchun qutb koordinatalari bo'yicha integral u holda,

\[A = \iint\limits_{D}{{dA}} = \int_{{ - 0.3398}}^{{3.4814}}{{\int_{{3 - \sin \theta }}^{{4 + 2\sin \theta }}{{\,r\,dr}}\,d\theta }}\]

Integralni konvertatsiya qilishda \(dA = r\,dr\,d\theta \) 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 va bu kabi maydon integrallari uchun yanada to'g'riroq ko'rinadi.

Mana \(r\) integratsiyasi.

\[\begin{align*}A & = \int_{{ - 0.3398}}^{{3.4814}}{{\int_{{3 - \sin \theta }}^{{4 + 2\sin \theta }}{{\,r\,dr}}\,d\theta }}\\ & = \int_{{ - 0.3398}}^{{3.4814}}{{\left. {\frac{1}{2}{r^2}} \right|_{3 - \sin \theta }^{4 + 2\sin \theta }\,d\theta }}\\ & = \int_{{ - 0.3398}}^{{3.4814}}{{\frac{1}{2}\left[ {{{\left( {4 + 2\sin \theta } \right)}^2} - {{\left( {3 - \sin \theta } \right)}^2}} \right]\,d\theta }}\end{align*}\]

E'tibor bering, agar siz ushbu turdagi hududdagi muammolarni eslab qolsangiz orqaga Calculus II materialida bu integral juda tanish ko'rinishi kerak, chunki u biz ushbu materialda ishlatgan formulaga amal qiladi.

Va nihoyat, \(\theta \) integratsiyasi.

\[\begin{align*}A & = \int_{{ - 0.3398}}^{{3.4814}}{{\frac{1}{2}\left[ {{{\left( {4 + 2\sin \theta } \right)}^2} - {{\left( {3 - \sin \theta } \right)}^2}} \right]\,d\theta }}\\ & = \frac{1}{2}\int_{{ - 0.3398}}^{{3.4814}}{{16 + 16\sin \theta + 4{{\sin }^2}\theta - \left( {9 - 6\sin \theta + {{\sin }^2}\theta } \right)\,d\theta }}\\ & = \frac{1}{2}\int_{{ - 0.3398}}^{{3.4814}}{{7 + 22\sin \theta + 3{{\sin }^2}\theta \,d\theta }}\\ & = \frac{1}{2}\int_{{ - 0.3398}}^{{3.4814}}{{7 + 22\sin \theta + \frac{3}{2}\left( {1 - \cos \left( {2\theta } \right)} \right)\,d\theta }}\\ & = \frac{1}{2}\int_{{ - 0.3398}}^{{3.4814}}{{\frac{{17}}{2} + 22\sin \theta - \frac{3}{2}\cos \left( {2\theta } \right)\,d\theta }}\\ & = \frac{1}{2}\left. {\left( {\frac{{17}}{2}\theta - 22\cos \theta - \frac{3}{4}\sin \left( {2\theta } \right)} \right)} \right|_{ - 0.3398}^{3.4814} = \require{bbox} \bbox[2pt,border:1px solid black]{{36.5108}}\end{align*}\]

Ushbu muammoni hal qilish uchun juda ko'p soddalashtirish kerak edi. Ko'pincha shunday bo'ladi. Bularning hech biri ayniqsa qiyin emas edi, shunchaki zerikarli va agar e'tibor bermasangiz, xato qilish oson. Bu erda soddalashtirishning "eng qiyin" qismi \({\sin ^2}\theta \) ni biz birlashtira oladigan narsaga kamaytirish uchun yarim burchak formulasidan foydalanish edi.