លក្ខណៈនៃអាំងតេក្រាលឌុប
[កែប្រែ]
(សំនុំទទេ) នោះគេបាន
![{\displaystyle \iint _{D}f(x,y)dxdy=\iint _{D_{1}}f(x,y)dxdy+\iint _{D_{2}}f(x,y)dxdy}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4fc42f2c22730a807bda4d6100394b01a8a094c)
- ប្រសិនបើ
នៅលើដែនកំនត់ D គេបាន
![{\displaystyle \iint _{D}f(x,y)dxdy\leq \iint _{D}g(x,y)dxdy}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9735491b59bd73a6a6e3b190c9f853b08e7803c)
ដោយយោងតាមការប្តូរ
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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b0be17d766881046bdc093b5ba2b6cd0a775290)
នោះគេបាន
ឧទាហរណ៍៖ គណនាអាំងតេក្រាលនៃ
ដំណោះស្រាយ៖
តាង
គេបាន
![{\displaystyle x={\frac {1}{5}}(3u+2v),\qquad y={\frac {1}{5}}(-u+v)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c30a94f8a0eaddb061452b3a8ff5824d0d85ccf2)
- 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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/970bc404446ca1299db5b43639fb32a4e96a9fb0)
![{\displaystyle M:0\leq u\leq 1,\qquad 0\leq v\leq 1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba4271b161a2a55a588965213eeaede35b1dd111)
គេបាន
ឧទាហរណ៍១៖ គណនាអាំងតេក្រាលឌុបនៃ
ដែល
ដំណោះស្រាយ
តាង
និង
គេបាន
![{\displaystyle \Rightarrow r^{2}-2r\cos \theta \leq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dbd602f09ba8957cfc9d5dec08cea428bf968073)
![{\displaystyle \Rightarrow 0\leq r\leq 2\cos \theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/445b4a31409e8351a594e3ab494013822e17e6e7)
![{\displaystyle \Rightarrow \cos \theta \geq 0,\,\,r\sin \theta \geq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9dda3db026ea84e13f4520528ee39fba9f319415)
![{\displaystyle \Rightarrow 0\leq \theta \leq {\frac {\pi }{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab784405cb7e5ae0f6c7e18836262c38a2d3f626)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a7542988ed4865d12a7d07d8b568912159ac376)
អាំងតេក្រាលប្តូរអថេរដ៍សំខាន់
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ឧទាហរណ៍ទី២៖ ចំពោះ ចូរគណនាអាំងតេក្រាលឌុបនៃ
គន្លឹះសំខាន់ៗ
តាង នោះគេបាន តាង គេបាន
![{\displaystyle J={\begin{vmatrix}x_{u}&x_{v}\\y_{u}&y_{v}\end{vmatrix}}={\begin{vmatrix}a&0\\0&b\end{vmatrix}}=ab}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32603bc7e608daccc86f82d500da11efe76141b4)
ហេតុនេះ
![{\displaystyle \iint _{D}xydxdy=\iint _{D_{1}}(au)(bv)\cdot abdudv=\iint _{D_{1}}a^{2}b^{2}uvdudv}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fcb6b1534ad2f0b7dc00e976ace59deb70425c4)
ដោយយោងតាមវិធីប្តូរអថេរក្នុងកូអរដោនេប៉ូលែរគេបាន
![{\displaystyle u=r\cos \theta ,\,\,v=r\sin \theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/49625a451dc48a0a04a40c41748067f4a518d0e3)
![{\displaystyle x=ar\cos \theta ,\,\,y=br\sin \theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/bb64c6c115fd37eb12b46406a192f7cd2f86c737)
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ដំណោះស្រាយ
តាង គេបានដែនកំនត់
![{\displaystyle M:0\leq r\leq 1,\,\,0\leq \theta {\frac {\pi }{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4d057bb20652ad120c906192e9e628e3d6e63e4c)
ដេទែមីណង់យ៉ាកូបី
![{\displaystyle {\begin{aligned}J&={\begin{vmatrix}x_{r}&x_{\theta }\\y_{r}&y_{\theta }\end{vmatrix}}={\begin{vmatrix}a\cos \theta &-ar\sin \theta \\b\sin \theta &br\cos \theta \end{vmatrix}}\\&=abr(\cos ^{2}\theta +\sin ^{2}\theta )=abr\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0c4cd6b89b398d7b3cc1c5c93a48c2c475a12dd9)
![{\displaystyle {\begin{aligned}\therefore \iint _{D}xydxdy&=\iint _{M}(ar\cos \theta )(br\sin \theta )abrdrd\theta \\&=a^{2}b^{2}\int _{0}^{\frac {\pi }{2}}\cos \theta \sin \theta d\theta \int _{0}^{1}r^{3}dr\\&=a^{2}b^{2}{\begin{bmatrix}{\frac {1}{2}}\sin ^{2}\theta \end{bmatrix}}_{0}^{1}={\frac {1}{8}}a^{2}b^{2}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8505499fdf85239194da5c772a429977313ebd08)
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