ដោយយោងតាមការប្តូរ { x = φ ( u , v ) y = ψ ( u , v ) {\displaystyle {\begin{cases}x=\varphi (u,v)\\y=\psi (u,v)\end{cases}}} ដែនកំនត់ K នៃប្លង់ uv ឆ្លុះគ្នានឹងដែនកំនត់ D នៃប្លង់ xy និងដេទែមីណង់យ៉ាកូបី
នោះគេបាន ∬ D f ( x , y ) d x d y = ∬ K f ( φ ( u , v ) , ψ ( u , v ) ) J d u d v {\displaystyle \iint _{D}f(x,y)dxdy=\iint _{K}f(\varphi (u,v),\psi (u,v))\,J\,dudv}
ឧទាហរណ៍៖ គណនាអាំងតេក្រាលនៃ ∬ D ( x + y ) d x d y , D : 0 ≤ x − 2 y ≤ 1 , 0 ≤ x + 3 y ≤ 1 {\displaystyle \iint _{D}(x+y)dxdy,\qquad D:0\leq x-2y\leq 1,\qquad 0\leq x+3y\leq 1}
ដំណោះស្រាយ៖
តាង x − 2 y = u , x + 3 y = v {\displaystyle x-2y=u,\qquad x+3y=v} គេបាន
គេបាន
∬ D ( x + y ) d x d y = ∬ M 1 5 ( 2 u + 3 v ) ⋅ 1 5 d u d v = 1 25 ∫ 0 1 ( ∫ 0 1 ( 2 u + 3 v ) d v ) d u = 1 25 ∫ 0 1 [ 2 u v + 3 2 v 2 ] v = 0 v = 1 d u = 1 25 ∫ 0 1 ( 2 u + 3 2 ) d u = 1 25 [ u 2 + 3 2 u ] 0 1 = 1 25 ⋅ 5 2 = 1 10 {\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}}}
តាង { x = r cos θ y = r sin θ {\displaystyle {\begin{cases}x=r\cos \theta \\y=r\sin \theta \end{cases}}\,} នោះគេបាន J = | cos θ − r sin θ sin θ r cos θ | = r {\displaystyle J={\begin{vmatrix}\cos \theta &-r\sin \theta \\\sin \theta &r\cos \theta \end{vmatrix}}=r\,\,\,} និង ∬ D f ( x , y ) d x d y = ∬ f ( r cos θ , r sin θ ) r d r d θ {\displaystyle \iint _{D}f(x,y)dxdy=\iint f(r\cos \theta ,r\sin \theta )r\,drd\theta }
ឧទាហរណ៍១៖ គណនាអាំងតេក្រាលឌុបនៃ ∬ D x y d x d y {\displaystyle \iint _{D}xydxdy\,} ដែល D = { ( x , y ) | ( x − 1 ) 2 + y 2 ≤ 1 y ≥ 0 } {\displaystyle D={\begin{Bmatrix}(x,y)|(x-1)^{2}+y^{2}\leq 1\,\,\,y\geq 0\end{Bmatrix}}}
ដំណោះស្រាយ
តាង x = r cos θ , y = r sin θ {\displaystyle x=r\cos \theta ,\,\,y=r\sin \theta } និង J = r {\displaystyle J=r\,} គេបាន
M = { ( r , θ ) | 0 ≤ r ≤ 2 cos θ , 0 ≤ θ ≤ π 2 } {\displaystyle M={\begin{Bmatrix}(r,\theta )|0\leq r\leq 2\cos \theta ,\,\,0\leq \theta \leq {\frac {\pi }{2}}\end{Bmatrix}}}
ហេតុនេះ
ដែនកំនត់ K នៃប្លង់ uv ឆ្លុះគ្នានឹងដែនកំនត់ D នៃប្លង់ xy និងដេទែមីណង់យ៉ាកូបី
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7911e26972eeaf6601189b4e828bc3c4473ee381)
ដែល 
និង 
គេបាន
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a7542988ed4865d12a7d07d8b568912159ac376)
![{\displaystyle \iint _{D}[f(x,y)\pm g(x,y)]dxdy=\iint _{D}f(x,y)dxdy\pm \iint _{D}g(x,y)dxdy}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8f74e5a88e23503be8097743ae215847001e068)
