អាំងតេក្រាលឌុប

លក្ខណៈនៃអាំងតេក្រាលឌុប

1. $\underset{D}{\iint }[f(x,y)\pm g(x,y)]dxdy=\underset{D}{\iint }f(x,y)dxdy\pm \underset{D}{\iint }g(x,y)dxdy$
   
   
2. $\underset{D}{\iint }kf(x,y)dxdy=k\underset{D}{\iint }f(x,y)dxdy$
   
   
3. $D=D_1\cup D_2,D_1\cap D_2=\varnothing$ (សំនុំទទេ) នោះគេបាន
   ទំព័រគំរូ:Spaces$\underset{D}{\iint }f(x,y)dxdy=\underset{D_1}{\iint }f(x,y)dxdy+\underset{D_2}{\iint }f(x,y)dxdy$
4. ប្រសិនបើ $f(x,y)\le g(x,y)$ នៅលើដែនកំនត់ D គេបាន
   ទំព័រគំរូ:Spaces $\underset{D}{\iint }f(x,y)dxdy\le \underset{D}{\iint }g(x,y)dxdy$

វិធីសាស្រ្តប្តូរអថេរ

វិធីសាស្រ្តទូទៅ

ដោយយោងតាមការប្តូរ $\{\begin{array}{c}x=\phi (u,v)\\ y=\psi (u,v)\end{array}$ ទំព័រគំរូ:Spacesដែនកំនត់ K នៃប្លង់ uv ឆ្លុះគ្នានឹងដែនកំនត់ D នៃប្លង់ xy និងដេទែមីណង់យ៉ាកូបី

$J=\frac{\delta (x,y)}{\delta (u,v)}=\left|\begin{array}{cc}{x}_u& x_v\\ y_u& y_v\end{array}\right|=x_u{y}_v-x_v{y}_u>0$

នោះគេបាន $\underset{D}{\iint }f(x,y)dxdy=\underset{K}{\iint }f(\phi (u,v),\psi (u,v))Jdudv$

ឧទាហរណ៍៖ គណនាអាំងតេក្រាលនៃ $\underset{D}{\iint }(x+y)dxdy,D:0\le x-2y\le 1,0\le x+3y\le 1$

ដំណោះស្រាយ៖

តាង $x-2y=u,x+3y=v$ គេបាន

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

ដេទែមីណង់យ៉ាកូបី $J=\left|\begin{array}{cc}{x}_u& x_v\\ y_u& y_v\end{array}\right|=\left|\begin{array}{cc}\frac{3}{5}& \frac{2}{5}\\ -\frac{1}{5}& \frac{1}{5}\end{array}\right|=\frac{1}{5}$

$M:0\le u\le 1,0\le v\le 1$

គេបាន

$\begin{array}{cc}\underset{D}{\iint }(x+y)dxdy& =\underset{M}{\iint }\frac{1}{5}(2u+3v)\cdot \frac{1}{5}dudv\\ & =\frac{1}{25}\underset{0}{\overset{1}{\int }}(\underset{0}{\overset{1}{\int }}(2u+3v)dv)du\\ & =\frac{1}{25}\underset{0}{\overset{1}{\int }}[2uv+\frac{3}{2}{v}^2{]}_{v=0}^{v=1}du\\ & =\frac{1}{25}\underset{0}{\overset{1}{\int }}(2u+\frac{3}{2})du=\frac{1}{25}[u^2+\frac{3}{2}u{]}_0^1\\ & =\frac{1}{25}\cdot \frac{5}{2}=\frac{1}{10}\end{array}$

ក្នុងកូអរសោនេប៉ូលែ

តាង $\{\begin{array}{c}x=r\cos \theta \\ y=r\sin \theta \end{array}$ ទំព័រគំរូ:Spaces នោះគេបាន $J=\left|\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right|=r$ និង
ទំព័រគំរូ:Spaces $\underset{D}{\iint }f(x,y)dxdy=\iint f(r\cos \theta ,r\sin \theta )rdrd\theta$

ឧទាហរណ៍១៖ គណនាអាំងតេក្រាលឌុបនៃ $\underset{D}{\iint }xydxdy$ ដែល
ទំព័រគំរូ:Spaces $D=\left\{\begin{array}{c}(x,y)|(x-1{)}^2+y^2\le 1y\ge 0\end{array}\right\}$

ដំណោះស្រាយ

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

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

$\Rightarrow r^2-2r\cos \theta \le 0$

$\Rightarrow 0\le r\le 2\cos \theta$

$\Rightarrow \cos \theta \ge 0,r\sin \theta \ge 0$

$\Rightarrow 0\le \theta \le \frac{\pi }{2}$

$M=\left\{\begin{array}{c}(r,\theta )|0\le r\le 2\cos \theta ,0\le \theta \le \frac{\pi }{2}\end{array}\right\}$

ហេតុនេះ

$\begin{array}{cc}\underset{D}{\iint }xydxdy& =\underset{M}{\iint }(r\cos \theta )(r\sin \theta )rdrd\theta \\ & =\underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \theta \sin \theta (\underset{0}{\overset{2\cos \theta }{\int }}{r}^{3}dr)d\theta \\ & =\underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \theta \sin \theta [\frac{1}{4}{r}^4]{}_0^{2\cos \theta }d\theta \\ & =\underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \theta \sin \theta \cdot 4{\cos }^4\theta d\theta \\ & =4\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^5\theta \sin \theta d\theta \\ & =4{\left[\begin{array}{c}-\frac{1}{6}{\cos }^6\theta \end{array}\right]}_0^{\frac{\pi }{2}}\\ & =4\cdot \frac{1}{6}=\frac{2}{3}\end{array}$