អាំងតេក្រាលឌុប

លក្ខណៈនៃអាំងតេក្រាលឌុប

1. ${\displaystyle \iint _{D}[f(x,y)\pm g(x,y)]dxdy=\iint _{D}f(x,y)dxdy\pm \iint _{D}g(x,y)dxdy}$
   
   
2. ${\displaystyle \iint _{D}kf(x,y)dxdy=k\iint _{D}f(x,y)dxdy}$
   
   
3. ${\displaystyle D=D_{1}\cup D_{2},\,\,D_{1}\cap D_{2}=\varnothing \,\,\,}$ (សំនុំទទេ) នោះគេបាន
      ${\displaystyle \iint _{D}f(x,y)dxdy=\iint _{D_{1}}f(x,y)dxdy+\iint _{D_{2}}f(x,y)dxdy}$
4. ប្រសិនបើ ${\displaystyle f(x,y)\leq g(x,y)\,\,\,}$ នៅលើដែនកំនត់ D គេបាន
       ${\displaystyle \iint _{D}f(x,y)dxdy\leq \iint _{D}g(x,y)dxdy}$

វិធីសាស្រ្តប្តូរអថេរ

វិធីសាស្រ្តទូទៅ

ដោយយោងតាមការប្តូរ ${\displaystyle {\begin{cases}x=\varphi (u,v)\\y=\psi (u,v)\end{cases}}}$    ដែនកំនត់ K នៃប្លង់ uv ឆ្លុះគ្នានឹងដែនកំនត់ D នៃប្លង់ xy និងដេទែមីណង់យ៉ាកូបី

${\displaystyle J={\frac {\delta (x,y)}{\delta (u,v)}}={\begin{vmatrix}x_{u}&x_{v}\\y_{u}&y_{v}\end{vmatrix}}=x_{u}y_{v}-x_{v}y_{u}>0}$

នោះគេបាន ${\displaystyle \iint _{D}f(x,y)dxdy=\iint _{K}f(\varphi (u,v),\psi (u,v))\,J\,dudv}$

ឧទាហរណ៍៖ គណនាអាំងតេក្រាលនៃ ${\displaystyle \iint _{D}(x+y)dxdy,\qquad D:0\leq x-2y\leq 1,\qquad 0\leq x+3y\leq 1}$

ដំណោះស្រាយ៖

តាង ${\displaystyle x-2y=u,\qquad x+3y=v}$ គេបាន

${\displaystyle x={\frac {1}{5}}(3u+2v),\qquad y={\frac {1}{5}}(-u+v)}$

ដេទែមីណង់យ៉ាកូបី ${\displaystyle J={\begin{vmatrix}x_{u}&x_{v}\\y_{u}&y_{v}\end{vmatrix}}={\begin{vmatrix}{\frac {3}{5}}&{\frac {2}{5}}\\-{\frac {1}{5}}&{\frac {1}{5}}\end{vmatrix}}={\frac {1}{5}}}$

${\displaystyle M:0\leq u\leq 1,\qquad 0\leq v\leq 1}$

គេបាន

${\displaystyle {\begin{aligned}\iint _{D}(x+y)dxdy&=\iint _{M}{\frac {1}{5}}(2u+3v)\cdot {\frac {1}{5}}dudv\\&={\frac {1}{25}}\int _{0}^{1}(\int _{0}^{1}(2u+3v)dv)du\\&={\frac {1}{25}}\int _{0}^{1}[2uv+{\frac {3}{2}}v^{2}]_{v=0}^{v=1}du\\&={\frac {1}{25}}\int _{0}^{1}(2u+{\frac {3}{2}})du={\frac {1}{25}}[u^{2}+{\frac {3}{2}}u]_{0}^{1}\\&={\frac {1}{25}}\cdot {\frac {5}{2}}={\frac {1}{10}}\end{aligned}}}$

ក្នុងកូអរសោនេប៉ូលែ

តាង ${\displaystyle {\begin{cases}x=r\cos \theta \\y=r\sin \theta \end{cases}}\,}$     នោះគេបាន ${\displaystyle J={\begin{vmatrix}\cos \theta &-r\sin \theta \\\sin \theta &r\cos \theta \end{vmatrix}}=r\,\,\,}$ និង
      ${\displaystyle \iint _{D}f(x,y)dxdy=\iint f(r\cos \theta ,r\sin \theta )r\,drd\theta }$

ឧទាហរណ៍១៖ គណនាអាំងតេក្រាលឌុបនៃ ${\displaystyle \iint _{D}xydxdy\,}$ ដែល
      ${\displaystyle D={\begin{Bmatrix}(x,y)|(x-1)^{2}+y^{2}\leq 1\,\,\,y\geq 0\end{Bmatrix}}}$

ដំណោះស្រាយ

តាង ${\displaystyle x=r\cos \theta ,\,\,y=r\sin \theta }$ និង ${\displaystyle J=r\,}$ គេបាន

${\displaystyle D:(r\cos \theta -1)^{2}+(r\sin \theta )^{2}\leq 1}$

${\displaystyle \Rightarrow r^{2}-2r\cos \theta \leq 0}$

${\displaystyle \Rightarrow 0\leq r\leq 2\cos \theta }$

${\displaystyle \Rightarrow \cos \theta \geq 0,\,\,r\sin \theta \geq 0}$

${\displaystyle \Rightarrow 0\leq \theta \leq {\frac {\pi }{2}}}$

${\displaystyle M={\begin{Bmatrix}(r,\theta )|0\leq r\leq 2\cos \theta ,\,\,0\leq \theta \leq {\frac {\pi }{2}}\end{Bmatrix}}}$

ហេតុនេះ

${\displaystyle {\begin{aligned}\iint _{D}xydxdy&=\iint _{M}(r\cos \theta )(r\sin \theta )rdrd\theta \\&=\int _{0}^{\frac {\pi }{2}}\cos \theta \sin \theta (\int _{0}^{2\cos \theta }r^{3}dr)\,d\theta \\&=\int _{0}^{\frac {\pi }{2}}\cos \theta \sin \theta [{\frac {1}{4}}r^{4}]{_{0}^{2\cos \theta }}d\theta \\&=\int _{0}^{\frac {\pi }{2}}\cos \theta \sin \theta \cdot 4\cos ^{4}\theta d\theta \\&=4\int _{0}^{\frac {\pi }{2}}\cos ^{5}\theta \sin \theta d\theta \\&=4{\begin{bmatrix}-{\frac {1}{6}}\cos ^{6}\theta \end{bmatrix}}_{0}^{\frac {\pi }{2}}\\&=4\cdot {\frac {1}{6}}={\frac {2}{3}}\end{aligned}}}$