Keling, endi integralni o'rnatamiz. Bu holda sirt shaklda bo'ladi,

\[y = g\left( {x,z} \right) = 3{x^2} + 3{z^2}\]

shuning uchun sirt integrali uchun quyidagi formuladan foydalanamiz.

\[\iint\limits_{S}{{f\left( {x,y,z} \right)\,dS}} = \iint\limits_{D}{{f\left( {x,g\left( {x,z} \right),z} \right)\sqrt {{{\left( {\frac{{\partial g}}{{\partial x}}} \right)}^2} + 1 + {{\left( {\frac{{\partial g}}{{\partial z}}} \right)}^2}} \,dA}}\]

U holda integral,

\[\begin{align*}\iint\limits_{S}{{40y\,dS}} & = \iint\limits_{D}{{40\left( {3{x^2} + 3{z^2}} \right)\sqrt {{{\left( {6x} \right)}^2} + 1 + {{\left( {6z} \right)}^2}} \,dA}}\\ & = \iint\limits_{D}{{120\left( {{x^2} + {z^2}} \right)\sqrt {36\left( {{x^2} + {z^2}} \right) + 1} \,dA}}\end{align*}\]

Sirt tenglamasini integranddagi \(y\) ga ulashni unutmang va esda tutingki, \(D\) biz 1-bosqichda topilgan diskdir.

Endi bu muammo uchun biz integralni bajarish uchun qutb koordinatalaridan foydalanishni xohlashimiz aniq bo'lishi kerak. Biz quyidagi qutb koordinatalaridan foydalanamiz.

\[x = r\cos \theta \hspace{0.25in}\hspace{0.25in}z = r\sin \theta \hspace{0.25in}\hspace{0.25in}{x^2} + {z^2} = {r^2}\]

Bundan tashqari, \(D\) \({x^2} + {z^2} \le 2\) diski bo'lgani uchun integral uchun chegaralar shunday bo'ladi,

\[\begin{array}{c}0 \le \theta \le 2\pi \\ 0 \le r \le \sqrt 2 \end{array}\]

Integralni qutbli koordinatalarga aylantirish,

\[\begin{align*}\iint\limits_{S}{{40y\,dS}} & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 2 }}{{120{r^2}\sqrt {36{r^2} + 1} \,\left( r \right)dr}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{0}^{{\sqrt 2 }}{{120{r^3}\sqrt {36{r^2} + 1} \,dr}}\,d\theta }}\end{align*}\]

\(dA\) ni qutb koordinatalariga o'tkazishda qo'shimcha \(r\) ni olishni unutmang.

Endi biz qilishimiz kerak bo'lgan narsa - ikki karrali integralni baholash va agar siz ilgari bunday integralni ko'rmagan bo'lsangiz, bu biroz qiyin bo'lishi mumkin.

Integralni bajarish uchun quyidagi almashtirishdan foydalanamiz.

\[u = 36{r^2} + 1\hspace{0.25in} \to \hspace{0.25in}du = 72r\,dr\hspace{0.25in} \to \hspace{0.25in}\frac{1}{{72}}du = r\,dr\]

Muammo shundaki, bu birinchi qarashda ishlamaydi, chunki differentsial faqat ildiz oldidagi uchta \(r\) dan birini yo'q qiladi. Biroq, biz \({r^2}\) ni almashtirishni ham hal qila olamiz,

\[{r^2} = \frac{1}{{36}}\left( {u - 1} \right)\]

va endi biz qolgan ikkita \(r\) ni \(u\) ga aylantira olamiz.

Shunday qilib, almashtirishdan foydalanib, integral,

\[\begin{align*}\iint\limits_{S}{{40y\,dS}}& = \int_{0}^{{2\pi }}{{\int_{1}^{{73}}{{120\left( {\frac{1}{{72}}} \right)\left( {\frac{1}{{36}}} \right)\left( {u - 1} \right){u^{\frac{1}{2}}}\,du}}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\int_{1}^{{73}}{{\frac{5}{{108}}\left( {{u^{\frac{3}{2}}} - {u^{\frac{1}{2}}}} \right)\,du}}\,d\theta }}\end{align*}\]

Esda tutingki, biz “yangi” \(u\) chegaralarini olish uchun “eski” \(r\) chegaralarini almashtirishga ulab, asl integraldagi \(r\) chegaralarini \(u\) chegaralariga aylantirdik.

Endi biz integralni baholashni osongina tugatishimiz mumkin.

\[\begin{align*}\iint\limits_{S}{{40y\,dS}} & = \int_{0}^{{2\pi }}{{\left. {\frac{5}{{108}}\left( {\frac{2}{5}{u^{\frac{5}{2}}} - \frac{2}{3}{u^{\frac{3}{2}}}} \right)} \right|_1^{73}\,d\theta }}\\ & = \int_{0}^{{2\pi }}{{\frac{5}{{108}}\left[ {\frac{2}{5}\left( {{{73}^{\frac{5}{2}}}} \right) - \frac{2}{3}\left( {{{73}^{\frac{3}{2}}}} \right) - \left( { - \frac{4}{{15}}} \right)} \right]\,d\theta }}\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{5\pi }}{{54}}\left[ {\frac{2}{5}\left( {{{73}^{\frac{5}{2}}}} \right) - \frac{2}{3}\left( {{{73}^{\frac{3}{2}}}} \right) + \frac{4}{{15}}} \right]\, = 5176.8958}}\end{align*}\]

Noto'g'ri javob bilan chalkash integral, lekin bu ba'zida sodir bo'ladi, shuning uchun bu sodir bo'lganda biz bundan juda hayajonlanmasligimiz kerak.