雅可比矩陣和變數替換在多元微積分中被證明非常有用，當我們想要改變變數時。它們非常有用，因為如果我們想要積分一個函式，例如
∬ R e x + y x − y d A {\displaystyle \iint _{R}e^{\frac {x+y}{x-y}}\,dA} R {\displaystyle R} ( 1 , 0 ) , ( 2 , 0 ) , ( 0 , − 2 ) , ( 0 , − 1 ) {\displaystyle (1,0),(2,0),(0,-2),(0,-1)}
如果我們能將 x + y {\displaystyle x+y} u {\displaystyle u} x − y {\displaystyle x-y} v {\displaystyle v} e u v {\displaystyle e^{\frac {u}{v}}}
讓我們從介紹變數變換的過程開始。假設我們有一個函式 f ( x , y ) {\displaystyle f(x,y)}
∬ R f ( x , y ) d A {\displaystyle \iint _{R}f(x,y)\,dA}
其中 R {\displaystyle R} x y {\displaystyle xy} d A {\displaystyle dA}
然而， R {\displaystyle R} x , y {\displaystyle x,y} R {\displaystyle R}
如果可以將變數更改為更方便的變數，那就容易多了。假設還有兩個變數 u , v {\displaystyle u,v} x , y {\displaystyle x,y}
x = x ( u , v ) and y = y ( u , v ) . {\displaystyle x=x(u,v)\quad {\text{and}}\quad y=y(u,v).}
原始積分可以改寫為
∬ S f ( x ( u , v ) , y ( u , v ) ) | ∂ ( x , y ) ∂ ( u , v ) | d u d v {\displaystyle \iint _{S}f(x(u,v),y(u,v))\left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|\,du\,dv}
其中 S {\displaystyle S} u v {\displaystyle uv} x y {\displaystyle xy} R {\displaystyle R} | ∂ ( x , y ) ∂ ( u , v ) | {\displaystyle \left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|} | ∂ ( x , y ) ∂ ( u , v ) | {\displaystyle \left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|} 下一節 中的用途和意義。
事實上，我們在 R 3 {\displaystyle \mathbb {R} ^{3}}
第一個例子是在積分中使用極座標，而第二個例子是在積分中使用球座標。在積分中使用極座標是一種變數變化，因為我們實際上將變數 x , y , z {\displaystyle x,y,z} r , θ , z {\displaystyle r,\theta ,z}
x = r cos θ y = r sin θ z = z {\displaystyle x=r\cos \theta \quad y=r\sin \theta \quad z=z}
因此，被積函式 f ( x , y ) {\displaystyle f(x,y)} f ( x ( r , θ , z ) , y ( r , θ , z ) , z ( r , θ , z ) ) {\displaystyle f(x(r,\theta ,z),y(r,\theta ,z),z(r,\theta ,z))}
∭ R f ( x , y , z ) d V = ∭ S f ( r cos θ , r sin θ , z ) d z r d r d θ {\displaystyle \iiint _{R}f(x,y,z)dV=\iiint _{S}f(r\cos \theta ,r\sin \theta ,z)dz\ r\ dr\ d\theta }
第二個例子，球座標積分，提供了一個類似的解釋。原始變數 x , y , z {\displaystyle x,y,z} ρ , θ , ϕ {\displaystyle \rho ,\theta ,\phi }
{ x = ρ sin ϕ cos θ y = ρ sin ϕ sin θ z = ρ cos ϕ {\displaystyle {\begin{cases}x=\rho \sin \phi \cos \theta \\y=\rho \sin \phi \sin \theta \\z=\rho \cos \phi \end{cases}}}
這些關係可以得到
∭ R f ( x , y , z ) d V = ∭ S f ( ρ sin ϕ cos θ , ρ sin ϕ sin θ , ρ cos ϕ ) ρ 2 sin ϕ d ρ d θ d ϕ {\displaystyle \iiint _{R}f(x,y,z)dV=\iiint _{S}f(\rho \sin \phi \cos \theta ,\rho \sin \phi \sin \theta ,\rho \cos \phi )\rho ^{2}\sin \phi \ d\rho \ d\theta \ d\phi }
我們理解了從笛卡爾座標系到極座標系和球座標系的變換。然而，這兩種變換隻是變數變換的具體例子。我們應該將範圍擴充套件到所有型別的變換。我們將不再討論特定的變換，比如 x = r cos θ , y = r sin θ and z = z {\displaystyle x=r\cos \theta ,y=r\sin \theta {\text{ and }}z=z}
我們考慮一個由變換 T {\displaystyle T} u v {\displaystyle uv} x y {\displaystyle xy}
T ( u , v ) = ( x , y ) {\displaystyle T(u,v)=(x,y)} ( x , y ) {\displaystyle (x,y)} ( u , v ) {\displaystyle (u,v)}
在這個變換中， x , y {\displaystyle x,y} u , v {\displaystyle u,v}
x = g ( u , v ) y = h ( u , v ) {\displaystyle x=g(u,v)\quad y=h(u,v)}
我們通常假設 T {\displaystyle T} C 1 {\displaystyle C^{1}} g , h {\displaystyle g,h}
如果 T ( u 1 , v 1 ) = ( x 1 , y 1 ) {\displaystyle T(u_{1},v_{1})=(x_{1},y_{1})} ( x 1 , y 1 ) {\displaystyle (x_{1},y_{1})} ( u 1 , v 1 ) {\displaystyle (u_{1},v_{1})} 像
如果沒有任何兩個點具有相同的像，就像函式一樣， T {\displaystyle T} 一一對應
T {\displaystyle T} S {\displaystyle S} R {\displaystyle R} R {\displaystyle R} S {\displaystyle S} T ( S ) = R {\displaystyle T(S)=R}
如果 T {\displaystyle T} 一一對應 逆變換 T − 1 {\displaystyle T^{-1}} x y {\displaystyle xy} u v {\displaystyle uv} T − 1 ( R ) = S {\displaystyle T^{-1}(R)=S}
回顧一下，我們已經建立了變換 T ( S ) = R {\displaystyle T(S)=R} S {\displaystyle S} u v {\displaystyle uv} R {\displaystyle R} x y {\displaystyle xy} S {\displaystyle S} T {\displaystyle T} R {\displaystyle R}
x = u 2 − v 2 y = 2 u v {\displaystyle x=u^{2}-v^{2}\quad y=2uv}
求 S {\displaystyle S} S = { ( u , v ) : 0 ≤ u ≤ 1 , 0 ≤ v ≤ 1 } {\displaystyle S=\{(u,v):0\leq u\leq 1,\ 0\leq v\leq 1\}}
在這種情況下，我們需要知道區域 S {\displaystyle S}
u = 0 u = 1 v = 0 v = 1 {\displaystyle u=0\quad u=1\quad v=0\quad v=1}
如果我們能使用 x , y {\displaystyle x,y} u , v {\displaystyle u,v} S {\displaystyle S}
When u = 0 ( 0 ≤ v ≤ 1 ) { x = − v 2 y = 0 (substitution) Thus, y = 0 ( − 1 ≤ x ≤ 0 ) {\displaystyle {\begin{aligned}{\text{When }}\quad &u=0\quad (0\leq v\leq 1)\\&{\begin{cases}x=-v^{2}\\y=0\\\end{cases}}\quad {\text{(substitution)}}\\{\text{Thus, }}\quad &y=0\quad (-1\leq x\leq 0)\\\end{aligned}}} When u = 1 ( 0 ≤ v ≤ 1 ) { x = 1 − v 2 y = 2 v Thus, x = 1 − y 2 4 ( 0 ≤ x ≤ 1 ) {\displaystyle {\begin{aligned}{\text{When }}\quad &u=1\quad (0\leq v\leq 1)\\&{\begin{cases}x=1-v^{2}\\y=2v\\\end{cases}}\\{\text{Thus, }}\quad &x=1-{\frac {y^{2}}{4}}\quad (0\leq x\leq 1)\\\end{aligned}}} When v = 0 ( 0 ≤ 0 ≤ 1 ) { x = u 2 y = 0 Thus, y = 0 ( 0 ≤ x ≤ 1 ) {\displaystyle {\begin{aligned}{\text{When }}\quad &v=0\quad (0\leq 0\leq 1)\\&{\begin{cases}x=u^{2}\\y=0\\\end{cases}}\\{\text{Thus, }}\quad &y=0\quad (0\leq x\leq 1)\\\end{aligned}}} When v = 1 ( 0 ≤ u ≤ 1 ) { x = u 2 − 1 y = 2 u Thus, x = y 2 4 − 1 ( − 1 ≤ x ≤ 0 ) {\displaystyle {\begin{aligned}{\text{When }}\quad &v=1\quad (0\leq u\leq 1)\\&{\begin{cases}x=u^{2}-1\\y=2u\\\end{cases}}\\{\text{Thus, }}\quad &x={\frac {y^{2}}{4}}-1\quad (-1\leq x\leq 0)\\\end{aligned}}}
因此， S {\displaystyle S} R = { ( x , y ) : 0 ≤ y ≤ 2 , y 2 4 − 1 ≤ x ≤ 1 − y 2 4 } {\displaystyle R=\{(x,y):0\leq y\leq 2,\ {\frac {y^{2}}{4}}-1\leq x\leq 1-{\frac {y^{2}}{4}}\}}
我們可以使用相同的方法計算 S {\displaystyle S} R {\displaystyle R}
雅可比矩陣是本章最重要的概念之一。它“補償”了當我們改變變數時面積的變化，從而使在改變變數後，積分的結果不會發生變化。回想一下，在上節的開頭，我們保留了對 | ∂ ( x , y ) ∂ ( u , v ) | {\displaystyle \left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|} ∬ S f ( x ( u , v ) , y ( u , v ) ) | ∂ ( x , y ) ∂ ( u , v ) | d u d v {\displaystyle \iint _{S}f(x(u,v),y(u,v))\left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|\ du\ dv}
回想一下，當我們討論 u {\displaystyle u}
∫ a b f ( x ) d x = ∫ c d f ( x ( u ) ) d x d u d u where c = x ( a ) , d = x ( b ) {\displaystyle \int _{a}^{b}f(x)\ dx=\int _{c}^{d}f(x(u))\ {\frac {dx}{du}}\ du\quad {\text{where }}c=x(a),d=x(b)}
例如，
∫ sin ( ln ( x ) ) x d x {\displaystyle \int {\frac {\sin(\ln(x))}{x}}\ dx}
Let u = ln ( x ) Thus, d u d x = 1 x ⇒ d u = 1 x d x {\displaystyle {\begin{aligned}{\text{Let }}\quad &u=\ln(x)\\{\text{Thus, }}\quad &{\frac {du}{dx}}={\frac {1}{x}}\\\Rightarrow \ &du={\frac {1}{x}}dx\\\end{aligned}}}
∫ sin ( ln ( x ) ) x d x = ∫ sin ( ln ( x ) ) ( 1 x d x ) rearrangement = ∫ sin ( u ) d u let u = ln ( x ) ⟹ d u = 1 x d x = − cos ( u ) + C integration = − cos ( ln ( x ) ) + C resubstitution {\displaystyle {\begin{aligned}\int {\frac {\sin(\ln(x))}{x}}\ dx&=\int \sin(\ln(x))\ {\Big (}{\frac {1}{x}}\ dx{\Big )}&\quad {\text{rearrangement}}\\&=\int \sin(u)\ du&\quad {\text{let }}u=\ln(x)\implies du={\frac {1}{x}}\ dx\\&=-\cos(u)+C&\quad {\text{integration}}\\&=-\cos(\ln(x))+C&\quad {\text{resubstitution}}\\\end{aligned}}}
如果我們在積分中新增端點，結果將是
∫ e e 2 sin ( ln ( x ) ) x d x = ∫ e e 2 sin ( ln ( x ) ) ( 1 x d x ) rearrangement = ∫ 1 2 sin ( u ) d u remember u = ln ( x ) and d u = 1 x d x = [ − cos ( u ) ] 1 2 integration = cos ( 1 ) − cos ( 2 ) {\displaystyle {\begin{aligned}\int _{e}^{e^{2}}{\frac {\sin(\ln(x))}{x}}\ dx&=\int _{e}^{e^{2}}\sin(\ln(x))\ {\Big (}{\frac {1}{x}}\ dx{\Big )}&\quad {\text{rearrangement}}\\&=\int _{1}^{2}\sin(u)\ du&\quad {\text{remember }}u=\ln(x){\text{ and }}du={\frac {1}{x}}\ dx\\&={\Big [}-\cos(u){\Big ]}_{1}^{2}&\quad {\text{integration}}\\&=\cos(1)-\cos(2)\\\end{aligned}}}
如果我們仔細觀察解中的“重排”和“記住”部分，我們會發現我們有效地將我們的變數從 x {\displaystyle x} u {\displaystyle u}
∫ a b f ( x ) d x = ∫ x = a x = b f ( x ( u ) ) d ( x ( u ) ) = ∫ u = x ( a ) u = x ( b ) f ( x ( u ) ) d x d u d u {\displaystyle \int _{a}^{b}f(x)dx=\int _{x=a}^{x=b}f(x(u))\ d(x(u))=\int _{u=x(a)}^{u=x(b)}f(x(u))\ {\frac {dx}{du}}\ du}
術語 d x d u {\displaystyle {\frac {dx}{du}}} f ( x ) {\displaystyle f(x)} f ( x ( u ) ) {\displaystyle f(x(u))} 改變了積分割槽域 “拉伸” “壓縮” d u d x {\displaystyle {\frac {du}{dx}}} d x d u {\displaystyle {\frac {dx}{du}}} d x d u ( d u d x ) = 1 {\displaystyle {\frac {dx}{du}}\left({\frac {du}{dx}}\right)=1}
x , y {\displaystyle x,y} u , v {\displaystyle u,v} 改變了積分割槽域
所以，繼續我們的思路，應該也存在一個推匯出的項來抵消區域的變化。換句話說
∬ R f ( x , y ) d A 1 = ∬ S f ( x ( u , v ) , y ( u , v ) ) d A 1 d A 2 ⏟ informal term d A 2 {\displaystyle \iint _{R}f(x,y)\,dA_{1}=\iint _{S}f(x(u,v),y(u,v))\underbrace {\frac {dA_{1}}{dA_{2}}} _{\text{informal term}}\,dA_{2}}
請注意，這裡使用的符號僅用於 直觀目的
在這種情況下，當我們將函式從 f ( x , y ) {\displaystyle f(x,y)} f ( x ( u ) , y ( u ) ) {\displaystyle f(x(u),y(u))} “拉伸” “壓縮” d A 2 d A 1 {\displaystyle {\frac {dA_{2}}{dA_{1}}}} d A 1 d A 2 {\displaystyle {\frac {dA_{1}}{dA_{2}}}} d A 1 d A 2 {\displaystyle {\frac {dA_{1}}{dA_{2}}}} u , v {\displaystyle u,v} u , v {\displaystyle u,v} x , y {\displaystyle x,y} u , v {\displaystyle u,v} x {\displaystyle x} y {\displaystyle y}
現在，我們來推導雅可比矩陣。在上面的回顧中，我們已經非正式地 d A 1 d A 2 {\displaystyle {\frac {dA_{1}}{dA_{2}}}} d A 1 {\displaystyle dA_{1}} x y {\displaystyle xy} R {\displaystyle R} d A 2 {\displaystyle dA_{2}} u v {\displaystyle uv} S {\displaystyle S} x , y {\displaystyle x,y} u , v {\displaystyle u,v} u , v {\displaystyle u,v} u v {\displaystyle uv} d A 1 {\displaystyle dA_{1}} d A 2 {\displaystyle dA_{2}}
讓我們先從 d A 2 {\displaystyle dA_{2}} S 0 {\displaystyle S_{0}} S {\displaystyle S} u v {\displaystyle uv} ( u 0 , v 0 ) {\displaystyle (u_{0},v_{0})} Δ u , Δ v {\displaystyle \Delta u,\Delta v} S 0 {\displaystyle S_{0}}
Δ A 2 = Δ u Δ v {\displaystyle \Delta A_{2}=\Delta u\Delta v}
S 0 {\displaystyle S_{0}} R 0 {\displaystyle R_{0}} x y {\displaystyle xy} T ( S 0 ) = R 0 {\displaystyle T(S_{0})=R_{0}} ( x 0 , y 0 ) = T ( u 0 , v 0 ) {\displaystyle (x_{0},y_{0})=T(u_{0},v_{0})} r {\displaystyle \mathbf {r} } R 0 {\displaystyle R_{0}} ( u , v ) {\displaystyle (u,v)} r {\displaystyle \mathbf {r} } R 0 {\displaystyle R_{0}}
r ( u , v ) = x ( u , v ) i + y ( u , v ) j where u 0 ≤ u ≤ u 0 + Δ u , v 0 ≤ v ≤ v 0 + Δ v {\displaystyle \mathbf {r} (u,v)=x(u,v)\mathbf {i} +y(u,v)\mathbf {j} \quad {\text{where }}u_{0}\leq u\leq u_{0}+\Delta u,\ v_{0}\leq v\leq v_{0}+\Delta v}
區域 R 0 {\displaystyle R_{0}} u , v {\displaystyle u,v} r ( u , v ) {\displaystyle \mathbf {r} (u,v)} d A 1 {\displaystyle dA_{1}}
變換 R 0 = T ( S 0 ) {\displaystyle R_{0}=T(S_{0})} R 0 {\displaystyle R_{0}} a {\displaystyle \mathbf {a} } b {\displaystyle \mathbf {b} }
Δ A 1 = | a × b | {\displaystyle \Delta A_{1}=|\mathbf {a} \times \mathbf {b} |}
在這個平行四邊形中，兩個向量 a {\displaystyle \mathbf {a} } b {\displaystyle \mathbf {b} } u , v {\displaystyle u,v}
a = r ( u 0 + Δ u , v 0 ) − r ( u 0 , v 0 ) and b = r ( u 0 , v 0 + Δ v ) − r ( u 0 , v 0 ) {\displaystyle \mathbf {a} =\mathbf {r} (u_{0}+\Delta u,v_{0})-\mathbf {r} (u_{0},v_{0})\quad {\text{ and }}\quad \mathbf {b} =\mathbf {r} (u_{0},v_{0}+\Delta v)-\mathbf {r} (u_{0},v_{0})}
這看起來非常類似於偏導數的定義
r u = lim Δ u → 0 r ( u 0 + Δ u , v 0 ) − r ( u 0 , v 0 ) Δ u and r v = lim Δ v → 0 r ( u 0 , v 0 + Δ v ) − r ( u 0 , v 0 ) Δ v {\displaystyle \mathbf {r} _{u}=\lim _{\Delta u\rightarrow 0}{\frac {\mathbf {r} (u_{0}+\Delta u,v_{0})-\mathbf {r} (u_{0},v_{0})}{\Delta u}}\quad {\text{ and }}\quad \mathbf {r} _{v}=\lim _{\Delta v\rightarrow 0}{\frac {\mathbf {r} (u_{0},v_{0}+\Delta v)-\mathbf {r} (u_{0},v_{0})}{\Delta v}}}
因此，我們可以近似地認為
a = r ( u 0 + Δ u , v 0 ) − r ( u 0 , v 0 ) ≈ Δ u r u and b = r ( u 0 , v 0 + Δ v ) − r ( u 0 , v 0 ) ≈ Δ v r v {\displaystyle {\begin{aligned}&\mathbf {a} =\mathbf {r} (u_{0}+\Delta u,v_{0})-\mathbf {r} (u_{0},v_{0})\approx \Delta u\mathbf {r} _{u}\quad {\text{ and }}\quad \\&\mathbf {b} =\mathbf {r} (u_{0},v_{0}+\Delta v)-\mathbf {r} (u_{0},v_{0})\approx \Delta v\mathbf {r} _{v}\\\end{aligned}}}
現在，我們計算 r u , r v {\displaystyle \mathbf {r} _{u},\mathbf {r} _{v}} r ( u , v ) = x ( u , v ) i + y ( u , v ) j {\displaystyle \mathbf {r} (u,v)=x(u,v)\mathbf {i} +y(u,v)\mathbf {j} }
r u = x u ( u , v ) i + y u ( u , v ) j = ∂ x ∂ u i + ∂ y ∂ u j and r v = x v ( u , v ) i + y v ( u , v ) j = ∂ x ∂ v i + ∂ y ∂ v j {\displaystyle {\begin{aligned}&\mathbf {r} _{u}=x_{u}(u,v)\ \mathbf {i} +y_{u}(u,v)\ \mathbf {j} ={\frac {\partial x}{\partial u}}\ \mathbf {i} +{\frac {\partial y}{\partial u}}\ \mathbf {j} \quad {\text{ and }}\\&\mathbf {r} _{v}=x_{v}(u,v)\ \mathbf {i} +y_{v}(u,v)\ \mathbf {j} ={\frac {\partial x}{\partial v}}\ \mathbf {i} +{\frac {\partial y}{\partial v}}\ \mathbf {j} \\\end{aligned}}}
我們可以計算 Δ A 1 = | | a × b | | {\displaystyle \Delta A_{1}=||\mathbf {a} \times \mathbf {b} ||} 7.1 章中的叉積。請注意，|| 內部的豎線用於計算大小（或範數），而 || 外部的豎線用於取絕對值。
Δ A 1 = | | a × b | | = | | ( Δ u r u ) × ( Δ v r v ) | | approximation = | | r u × r v | | Δ u Δ v = | | det ( i j k ∂ x ∂ u ∂ y ∂ u 0 ∂ x ∂ v ∂ y ∂ v 0 ) | | Δ u Δ v cross product = | | det ( ∂ x ∂ u ∂ x ∂ v ∂ y ∂ u ∂ y ∂ v ) k | | Δ u Δ v evaluation = | det ( ∂ x ∂ u ∂ x ∂ v ∂ y ∂ u ∂ y ∂ v ) | k | ⏟ 1 | Δ u Δ v {\displaystyle {\begin{aligned}\Delta A_{1}&=||\mathbf {a} \times \mathbf {b} ||\\&=||(\Delta u\ \mathbf {r} _{u})\times (\Delta v\ \mathbf {r} _{v})||&\quad {\text{approximation}}\\&=||\mathbf {r} _{u}\times \mathbf {r} _{v}||\Delta u\Delta v\\&={\begin{vmatrix}{\begin{vmatrix}\det {\begin{pmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\{\frac {\partial x}{\partial u}}&{\frac {\partial y}{\partial u}}&0\\{\frac {\partial x}{\partial v}}&{\frac {\partial y}{\partial v}}&0\\\end{pmatrix}}\\\end{vmatrix}}\end{vmatrix}}\Delta u\Delta v&\quad {\text{cross product}}\\&={\begin{vmatrix}{\begin{vmatrix}\det {\begin{pmatrix}{\frac {\partial x}{\partial u}}&{\frac {\partial x}{\partial v}}\\{\frac {\partial y}{\partial u}}&{\frac {\partial y}{\partial v}}\\\end{pmatrix}}\mathbf {k} \end{vmatrix}}\end{vmatrix}}\Delta u\Delta v&\quad {\text{evaluation}}\\&={\begin{vmatrix}\det {\begin{pmatrix}{\frac {\partial x}{\partial u}}&{\frac {\partial x}{\partial v}}\\{\frac {\partial y}{\partial u}}&{\frac {\partial y}{\partial v}}\\\end{pmatrix}}\underbrace {|\mathbf {k} |} _{1}\end{vmatrix}}\Delta u\Delta v\\\end{aligned}}}
然後，我們可以代入我們新推導的項。
Δ A 1 Δ A 2 = | det ( ∂ x ∂ u ∂ x ∂ v ∂ y ∂ u ∂ y ∂ v ) | Δ u Δ v Δ u Δ v = | det ( ∂ x ∂ u ∂ x ∂ v ∂ y ∂ u ∂ y ∂ v ) | = | ∂ x ∂ u ∂ y ∂ v − ∂ x ∂ v ∂ y ∂ u | {\displaystyle {\frac {\Delta A_{1}}{\Delta A_{2}}}={\frac {{\begin{vmatrix}\det {\begin{pmatrix}{\frac {\partial x}{\partial u}}&{\frac {\partial x}{\partial v}}\\{\frac {\partial y}{\partial u}}&{\frac {\partial y}{\partial v}}\\\end{pmatrix}}\end{vmatrix}}\Delta u\Delta v}{\Delta u\Delta v}}={\begin{vmatrix}\det {\begin{pmatrix}{\frac {\partial x}{\partial u}}&{\frac {\partial x}{\partial v}}\\{\frac {\partial y}{\partial u}}&{\frac {\partial y}{\partial v}}\\\end{pmatrix}}\end{vmatrix}}=\left|{\frac {\partial x}{\partial u}}{\frac {\partial y}{\partial v}}-{\frac {\partial x}{\partial v}}{\frac {\partial y}{\partial u}}\right|}
最後，我們推匯出雅可比行列式的絕對值
然後，我們將在積分變數替換中使用雅可比行列式。新增絕對值是為了防止出現負面積。
∬ R f ( x , y ) d A ≈ ∑ i = 1 m ∑ j = 1 n f ( x i , y j ) Δ A ≈ ∑ i = 1 m ∑ j = 1 n f ( x ( u i , v j ) , y ( u i , v j ) ) Δ A 2 Since Δ A 2 ≈ | ∂ ( x , y ) ∂ ( u , v ) | Δ u Δ v ∑ i = 1 m ∑ j = 1 n f ( x ( u i , v j ) , y ( u i , v j ) ) Δ A 2 ≈ ∑ i = 1 m ∑ j = 1 n f ( x ( u i , v j ) , y ( u i , v j ) ) | ∂ ( x , y ) ∂ ( u , v ) | Δ u Δ v ≈ ∬ S f ( x ( u , v ) , y ( u , v ) ) | ∂ ( x , y ) ∂ ( u , v ) | d u d v {\displaystyle {\begin{aligned}\iint _{R}f(x,y)\,dA&\approx \sum _{i=1}^{m}\sum _{j=1}^{n}f(x_{i},y_{j})\Delta A\\&\approx \sum _{i=1}^{m}\sum _{j=1}^{n}f(x(u_{i},v_{j}),y(u_{i},v_{j}))\Delta A_{2}\\{\text{Since }}&\Delta A_{2}\approx \left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|\Delta u\Delta v\\\sum _{i=1}^{m}\sum _{j=1}^{n}f(x(u_{i},v_{j}),y(u_{i},v_{j}))\Delta A_{2}&\approx \sum _{i=1}^{m}\sum _{j=1}^{n}f(x(u_{i},v_{j}),y(u_{i},v_{j}))\ \left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|\Delta u\Delta v\\&\approx \iint _{S}f(x(u,v),y(u,v))\ \left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|\ du\ dv\\\end{aligned}}}
以下是二重積分變數替換定理，我們已經 直觀地解釋 並非 證明 等式 證明
定理. T {\displaystyle T} C 1 {\displaystyle C^{1}} u v {\displaystyle uv} S {\displaystyle S} x y {\displaystyle xy} R {\displaystyle R} x = x ( u , v ) {\displaystyle x=x(u,v)} y = y ( u , v ) {\displaystyle y=y(u,v)} f {\displaystyle f} R {\displaystyle R} ∬ R f ( x , y ) d x d y = ∬ S f ( x ( u , v ) , y ( u , v ) ) | ∂ ( x , y ) ∂ ( u , v ) | d u d v {\displaystyle \iint _{R}f(x,y)\,dx\,dy=\iint _{S}f(x(u,v),y(u,v))\left|{\frac {\partial (x,y)}{\partial (u,v)}}\right|\,du\,dv}
備註. 如果我們改變一些符號，我們可以得到 ∬ S f ( u ( x , y ) , v ( x , y ) ) | ∂ ( u , v ) ∂ ( x , y ) | d x d y = ∬ R f ( u , v ) d u d v , {\displaystyle \iint _{S}f(u(x,y),v(x,y))\left|{\frac {\partial (u,v)}{\partial (x,y)}}\right|\,dx\,dy=\iint _{R}f(u,v)\,du\,dv,} T {\displaystyle T} x y {\displaystyle xy} S {\displaystyle S} u v {\displaystyle uv} R {\displaystyle R}
證明。
Let u = x + y {\displaystyle u=x+y} v = y x {\displaystyle v={\frac {y}{x}}} D ′ {\displaystyle D'} { u = x + y v = y x ⟹ { y = u − x v = y x ⟹ v = u − x x ⟹ x = u v + 1 ⟹ y = u − u v + 1 = u v v + 1 . {\displaystyle {\begin{cases}u=x+y\\v={\frac {y}{x}}\end{cases}}\implies {\begin{cases}y=u-x\\v={\frac {y}{x}}\end{cases}}\implies v={\frac {u-x}{x}}\implies x={\frac {u}{v+1}}\implies y=u-{\frac {u}{v+1}}={\frac {uv}{v+1}}.} ∂ ( x , y ) ∂ ( u , v ) = | ∂ x ∂ u ∂ x ∂ v ∂ y ∂ u ∂ y ∂ v | = | 1 v + 1 − u ( v + 1 ) 2 v v + 1 ( v + 1 ) u − u v ( v + 1 ) 2 | = 1 v + 1 ⋅ u ( v + 1 ) 2 − − u ( v + 1 ) 2 ⋅ v v + 1 = u + u v ( v + 1 ) 3 = u ( v + 1 ) 2 . {\displaystyle {\frac {\partial (x,y)}{\partial (u,v)}}={\begin{vmatrix}{\frac {\partial x}{\partial u}}&{\frac {\partial x}{\partial v}}\\{\frac {\partial y}{\partial u}}&{\frac {\partial y}{\partial v}}\end{vmatrix}}={\begin{vmatrix}{\frac {1}{v+1}}&{\frac {-u}{(v+1)^{2}}}\\{\frac {v}{v+1}}&{\frac {(v+1)u-uv}{(v+1)^{2}}}\end{vmatrix}}={\frac {1}{v+1}}\cdot {\frac {u}{(v+1)^{2}}}-{\frac {-u}{(v+1)^{2}}}\cdot {\frac {v}{v+1}}={\frac {u+uv}{(v+1)^{3}}}={\frac {u}{(v+1)^{2}}}.} x + y {\displaystyle x+y} y x {\displaystyle {\frac {y}{x}}} D {\displaystyle D} 2 ≤ x + y ≤ 3 {\displaystyle 2\leq x+y\leq 3} 4 ≤ y x ≤ 5 {\displaystyle 4\leq {\frac {y}{x}}\leq 5} u {\displaystyle u} v {\displaystyle v} D ′ {\displaystyle D'} 2 ≤ u ≤ 3 {\displaystyle 2\leq u\leq 3} 4 ≤ v ≤ 5 {\displaystyle 4\leq v\leq 5} ∬ D y ( x + y ) x d x = ∫ 4 5 ∫ 2 3 ( u v ⋅ u ( v + 1 ) 2 ⏟ > 0 because of the bounds ) d u d v = ∫ 4 5 v ( v + 1 ) 2 ( 3 3 3 − 2 3 3 ) d v = 19 3 ∫ 4 + 1 5 + 1 w − 1 w 2 d w let w = v + 1 ⟹ d w = d v = 19 3 [ ln | w | − 1 w ] 5 6 = 19 3 ( ln 6 − ln 5 + 1 30 ) = 19 3 ( ln ( 6 5 ) + 1 30 ) by properties of logarithm {\displaystyle {\begin{aligned}\iint _{D}{\frac {y(x+y)}{x}}dx&=\int _{4}^{5}\int _{2}^{3}\left(uv\cdot \underbrace {\frac {u}{(v+1)^{2}}} _{>0{\text{ because of the bounds}}}\right)\,du\,dv\\&=\int _{4}^{5}{\frac {v}{(v+1)^{2}}}\left({\frac {3^{3}}{3}}-{\frac {2^{3}}{3}}\right)\,dv\\&={\frac {19}{3}}\int _{4+1}^{5+1}{\frac {w-1}{w^{2}}}\,dw\qquad {\text{let }}w=v+1\implies dw=dv\\&={\frac {19}{3}}\left[\ln |w|-{\frac {1}{w}}\right]_{5}^{6}\\&={\frac {19}{3}}\left(\ln 6-\ln 5+{\frac {1}{30}}\right)\\&={\frac {19}{3}}\left(\ln \left({\frac {6}{5}}\right)+{\frac {1}{30}}\right)\qquad {\text{by properties of logarithm}}\end{aligned}}}
如果我們繼續思考，我們也可以找到三個變數的雅可比行列式。假設有一個函式 f ( x , y , z ) {\displaystyle f(x,y,z)} x , y , z {\displaystyle x,y,z} u , v , w {\displaystyle u,v,w}
x = x ( u , v , w ) , y = y ( u , v , w ) , and z = z ( u , v , w ) {\displaystyle x=x(u,v,w),\quad y=y(u,v,w),\quad {\text{and}}\quad z=z(u,v,w)}
R {\displaystyle R} x y z {\displaystyle xyz} S {\displaystyle S} u v w {\displaystyle uvw} T ( S ) = R {\displaystyle T(S)=R}
為了計算三個變數的雅可比行列式，我們進行類似的過程。變換過程將是：將 u v w {\displaystyle uvw} Δ u , Δ v , Δ w {\displaystyle \Delta u,\Delta v,\Delta w} x y z {\displaystyle xyz} Δ V 2 = Δ u Δ v Δ w {\displaystyle \Delta V_{2}=\Delta u\Delta v\Delta w}
r ( u , v , w ) = x ( u , v , w ) i + y ( u , v , w ) j + z ( u , v , w ) k {\displaystyle \mathbf {r} (u,v,w)=x(u,v,w)\ \mathbf {i} +y(u,v,w)\ \mathbf {j} +z(u,v,w)\ \mathbf {k} }
平行六面體的三個邊可以用位置向量描述為
a = r ( u + Δ u , v , w ) − r ( u , v , w ) , b = r ( u , v + Δ v , w ) − r ( u , v , w ) , and c = r ( u , v , w + Δ w ) − r ( u , v , w ) . {\displaystyle {\begin{aligned}&\mathbf {a} =\mathbf {r} (u+\Delta u,v,w)-\mathbf {r} (u,v,w),\\&\mathbf {b} =\mathbf {r} (u,v+\Delta v,w)-\mathbf {r} (u,v,w),\quad {\text{and}}\\&\mathbf {c} =\mathbf {r} (u,v,w+\Delta w)-\mathbf {r} (u,v,w).\\\end{aligned}}}
由於 r {\displaystyle \mathbf {r} }
r u = lim Δ u → 0 r ( u + Δ u , v , w ) − r ( u , v , w ) Δ u , r v = lim Δ v → 0 r ( u , v + Δ v , w ) − r ( u , v , w ) Δ v , and r w = lim Δ w → 0 r ( u , v , w + Δ w ) − r ( u , v , w ) Δ w . {\displaystyle {\begin{aligned}&\mathbf {r} _{u}=\lim _{\Delta u\rightarrow 0}{\frac {\mathbf {r} (u+\Delta u,v,w)-\mathbf {r} (u,v,w)}{\Delta u}},\\&\mathbf {r} _{v}=\lim _{\Delta v\rightarrow 0}{\frac {\mathbf {r} (u,v+\Delta v,w)-\mathbf {r} (u,v,w)}{\Delta v}},\quad {\text{and}}\\&\mathbf {r} _{w}=\lim _{\Delta w\rightarrow 0}{\frac {\mathbf {r} (u,v,w+\Delta w)-\mathbf {r} (u,v,w)}{\Delta w}}.\\\end{aligned}}}
三個向量 a , b , c {\displaystyle \mathbf {a} ,\mathbf {b} ,\mathbf {c} }
a ≈ Δ u r u , b ≈ Δ v r v , and c ≈ Δ w r w {\displaystyle \mathbf {a} \approx \Delta u\ \mathbf {r} _{u},\quad \mathbf {b} \approx \Delta v\ \mathbf {r} _{v},\quad {\text{and}}\quad \mathbf {c} \approx \Delta w\ \mathbf {r} _{w}}
由於位置向量 r {\displaystyle \mathbf {r} } r ( u , v , w ) = x ( u , v , w ) i + y ( u , v , w ) j + z ( u , v , w ) k {\displaystyle \mathbf {r} (u,v,w)=x(u,v,w)\ \mathbf {i} +y(u,v,w)\ \mathbf {j} +z(u,v,w)\ \mathbf {k} } r {\displaystyle \mathbf {r} }
r u = ∂ x ∂ u i + ∂ y ∂ u j + ∂ z ∂ u k , r v = ∂ x ∂ v i + ∂ y ∂ v j + ∂ z ∂ v k , and r v = ∂ x ∂ w i + ∂ y ∂ w j + ∂ z ∂ w k {\displaystyle {\begin{aligned}&\mathbf {r} _{u}={\frac {\partial x}{\partial u}}\ \mathbf {i} +{\frac {\partial y}{\partial u}}\ \mathbf {j} +{\frac {\partial z}{\partial u}}\ \mathbf {k} ,\\&\mathbf {r} _{v}={\frac {\partial x}{\partial v}}\ \mathbf {i} +{\frac {\partial y}{\partial v}}\ \mathbf {j} +{\frac {\partial z}{\partial v}}\ \mathbf {k} ,\quad {\text{and}}\\&\mathbf {r} _{v}={\frac {\partial x}{\partial w}}\ \mathbf {i} +{\frac {\partial y}{\partial w}}\ \mathbf {j} +{\frac {\partial z}{\partial w}}\ \mathbf {k} \\\end{aligned}}}
回想一下，由向量 a , b , c {\displaystyle \mathbf {a} ,\mathbf {b} ,\mathbf {c} }
V = | ( a × b ) ⋅ c | {\displaystyle V=|(\mathbf {a} \times \mathbf {b} )\ \cdot \ \mathbf {c} |}
我們只需要將向量替換為我們得到的向量。
Δ V 1 = | ( a × b ) ⋅ c | = | ( Δ u r u ) × ( Δ v r v ) ⋅ Δ w r w | = | r u × r v ⋅ r w | Δ u Δ v Δ w = | | i j k ∂ x ∂ u ∂ y ∂ u ∂ z ∂ u ∂ x ∂ v ∂ y ∂ v ∂ z ∂ v | ⋅ ( ∂ x ∂ w ∂ y ∂ w ∂ z ∂ w ) | Δ u Δ v Δ w cross product = | ( ∂ y ∂ u ∂ z ∂ v − ∂ z ∂ u ∂ y ∂ v ∂ z ∂ u ∂ x ∂ v − ∂ x ∂ u ∂ z ∂ v ∂ x ∂ u ∂ y ∂ v − ∂ y ∂ u ∂ x ∂ v ) ⋅ ( ∂ x ∂ w ∂ y ∂ w ∂ z ∂ w ) | Δ u Δ v Δ w = | ∂ x ∂ w ∂ y ∂ u ∂ z ∂ v − ∂ x ∂ w ∂ y ∂ v ∂ z ∂ u + ∂ x ∂ v ∂ y ∂ w ∂ z ∂ u − ∂ x ∂ u ∂ y ∂ w ∂ z ∂ v + ∂ x ∂ u ∂ y ∂ v ∂ z ∂ w − ∂ x ∂ v ∂ y ∂ u ∂ z ∂ w | Δ u Δ v Δ w dot product = | ∂ x ∂ u ( ∂ y ∂ v ∂ z ∂ w − ∂ y ∂ w ∂ z ∂ v ) + ∂ x ∂ v ( ∂ y ∂ w ∂ z ∂ u − ∂ y ∂ u ∂ z ∂ w ) + ∂ x ∂ w ( ∂ y ∂ u ∂ z ∂ v − ∂ y ∂ v ∂ z ∂ u ) | Δ u Δ v Δ w rearrangement = | | ∂ x ∂ u ∂ x ∂ v ∂ x ∂ w ∂ y ∂ u ∂ y ∂ v ∂ y ∂ w ∂ z ∂ u ∂ z ∂ v ∂ z ∂ w | | Δ u Δ v Δ w cross product {\displaystyle {\begin{aligned}\Delta V_{1}=|(\mathbf {a} \times \mathbf {b} )\ \cdot \ \mathbf {c} |&=|(\Delta u\ \mathbf {r} _{u})\times (\Delta v\ \mathbf {r} _{v})\ \cdot \ \Delta w\ \mathbf {r} _{w}|\\&=|\mathbf {r} _{u}\times \mathbf {r} _{v}\ \cdot \ \mathbf {r} _{w}|\Delta u\Delta v\Delta w\\&={\begin{vmatrix}{\begin{vmatrix}\mathbf {i} &\mathbf {j} &\mathbf {k} \\{\frac {\partial x}{\partial u}}&{\frac {\partial y}{\partial u}}&{\frac {\partial z}{\partial u}}\\{\frac {\partial x}{\partial v}}&{\frac {\partial y}{\partial v}}&{\frac {\partial z}{\partial v}}\\\end{vmatrix}}\ \cdot \ {\begin{pmatrix}{\frac {\partial x}{\partial w}}\\{\frac {\partial y}{\partial w}}\\{\frac {\partial z}{\partial w}}\\\end{pmatrix}}\end{vmatrix}}\Delta u\Delta v\Delta w&\quad {\text{cross product}}\\&={\begin{vmatrix}{\begin{pmatrix}{\frac {\partial y}{\partial u}}{\frac {\partial z}{\partial v}}-{\frac {\partial z}{\partial u}}{\frac {\partial y}{\partial v}}\\{\frac {\partial z}{\partial u}}{\frac {\partial x}{\partial v}}-{\frac {\partial x}{\partial u}}{\frac {\partial z}{\partial v}}\\{\frac {\partial x}{\partial u}}{\frac {\partial y}{\partial v}}-{\frac {\partial y}{\partial u}}{\frac {\partial x}{\partial v}}\\\end{pmatrix}}\ \cdot \ {\begin{pmatrix}{\frac {\partial x}{\partial w}}\\{\frac {\partial y}{\partial w}}\\{\frac {\partial z}{\partial w}}\\\end{pmatrix}}\end{vmatrix}}\Delta u\Delta v\Delta w\\&=\left|{\frac {\partial x}{\partial w}}{\frac {\partial y}{\partial u}}{\frac {\partial z}{\partial v}}-{\frac {\partial x}{\partial w}}{\frac {\partial y}{\partial v}}{\frac {\partial z}{\partial u}}+{\frac {\partial x}{\partial v}}{\frac {\partial y}{\partial w}}{\frac {\partial z}{\partial u}}-{\frac {\partial x}{\partial u}}{\frac {\partial y}{\partial w}}{\frac {\partial z}{\partial v}}+{\frac {\partial x}{\partial u}}{\frac {\partial y}{\partial v}}{\frac {\partial z}{\partial w}}-{\frac {\partial x}{\partial v}}{\frac {\partial y}{\partial u}}{\frac {\partial z}{\partial w}}\right|\Delta u\Delta v\Delta w&\quad {\text{dot product}}\\&=\left|{\frac {\partial x}{\partial u}}{\bigg (}{\frac {\partial y}{\partial v}}{\frac {\partial z}{\partial w}}-{\frac {\partial y}{\partial w}}{\frac {\partial z}{\partial v}}{\bigg )}+{\frac {\partial x}{\partial v}}{\bigg (}{\frac {\partial y}{\partial w}}{\frac {\partial z}{\partial u}}-{\frac {\partial y}{\partial u}}{\frac {\partial z}{\partial w}}{\bigg )}+{\frac {\partial x}{\partial w}}{\bigg (}{\frac {\partial y}{\partial u}}{\frac {\partial z}{\partial v}}-{\frac {\partial y}{\partial v}}{\frac {\partial z}{\partial u}}{\bigg )}\right|\Delta u\Delta v\Delta w&\quad {\text{rearrangement}}\\&={\begin{vmatrix}{\begin{vmatrix}{\frac {\partial x}{\partial u}}&{\frac {\partial x}{\partial v}}&{\frac {\partial x}{\partial w}}\\{\frac {\partial y}{\partial u}}&{\frac {\partial y}{\partial v}}&{\frac {\partial y}{\partial w}}\\{\frac {\partial z}{\partial u}}&{\frac {\partial z}{\partial v}}&{\frac {\partial z}{\partial w}}\\\end{vmatrix}}\end{vmatrix}}\Delta u\Delta v\Delta w&\quad {\text{cross product}}\\\end{aligned}}}
因此， Δ V 1 Δ V 2 = | | ∂ x ∂ u ∂ x ∂ v ∂ x ∂ w ∂ y ∂ u ∂ y ∂ v ∂ y ∂ w ∂ z ∂ u ∂ z ∂ v ∂ z ∂ w | | Δ u Δ v Δ w Δ u Δ v Δ w = | | ∂ x ∂ u ∂ x ∂ v ∂ x ∂ w ∂ y ∂ u ∂ y ∂ v ∂ y ∂ w ∂ z ∂ u ∂ z ∂ v ∂ z ∂ w | | {\displaystyle {\frac {\Delta V_{1}}{\Delta V_{2}}}={\frac {{\begin{vmatrix}{\begin{vmatrix}{\frac {\partial x}{\partial u}}&{\frac {\partial x}{\partial v}}&{\frac {\partial x}{\partial w}}\\{\frac {\partial y}{\partial u}}&{\frac {\partial y}{\partial v}}&{\frac {\partial y}{\partial w}}\\{\frac {\partial z}{\partial u}}&{\frac {\partial z}{\partial v}}&{\frac {\partial z}{\partial w}}\\\end{vmatrix}}\end{vmatrix}}\Delta u\Delta v\Delta w}{\Delta u\Delta v\Delta w}}={\begin{vmatrix}{\begin{vmatrix}{\frac {\partial x}{\partial u}}&{\frac {\partial x}{\partial v}}&{\frac {\partial x}{\partial w}}\\{\frac {\partial y}{\partial u}}&{\frac {\partial y}{\partial v}}&{\frac {\partial y}{\partial w}}\\{\frac {\partial z}{\partial u}}&{\frac {\partial z}{\partial v}}&{\frac {\partial z}{\partial w}}\\\end{vmatrix}}\end{vmatrix}}}
定義。 x = x ( u , v , w ) , y = y ( u , v , w ) {\displaystyle x=x(u,v,w),y=y(u,v,w)} z = z ( u , v , w ) {\displaystyle z=z(u,v,w)} T {\displaystyle T} ∂ ( x , y , z ) ∂ ( u , v , w ) = | ∂ x ∂ u ∂ x ∂ v ∂ x ∂ w ∂ y ∂ u ∂ y ∂ v ∂ y ∂ w ∂ z ∂ u ∂ z ∂ v ∂ z ∂ w | {\displaystyle {\frac {\partial (x,y,z)}{\partial (u,v,w)}}={\begin{vmatrix}{\frac {\partial x}{\partial u}}&{\frac {\partial x}{\partial v}}&{\frac {\partial x}{\partial w}}\\{\frac {\partial y}{\partial u}}&{\frac {\partial y}{\partial v}}&{\frac {\partial y}{\partial w}}\\{\frac {\partial z}{\partial u}}&{\frac {\partial z}{\partial v}}&{\frac {\partial z}{\partial w}}\\\end{vmatrix}}}
新增絕對值是為了防止出現負體積。
∭ R f ( x , y , z ) d V ≈ ∑ i = 1 m ∑ j = 1 n ∑ k = 1 p f ( x i , y j , z k ) Δ V ≈ ∑ i = 1 m ∑ j = 1 n ∑ k = 1 p f ( x ( u i , v j , w k ) , y ( u i , v j , w k ) , z ( u i , v j , w k ) ) Δ V 2 Since Δ V 2 ≈ | ∂ ( x , y , z ) ∂ ( u , v , w ) | Δ u Δ v Δ w ∑ i = 1 m ∑ j = 1 n ∑ k = 1 p f ( x ( u i , v j , w k ) , y ( u i , v j , w k ) , z ( u i , v j , w k ) ) Δ V 2 ≈ ∑ i = 1 m ∑ j = 1 n ∑ k = 1 p f ( x ( u i , v j , w k ) , y ( u i , v j , w k ) , z ( u i , v j , w k ) ) | ∂ ( x , y , z ) ∂ ( u , v , w ) | Δ u Δ v Δ w ≈ ∭ S f ( x ( u , v , w ) , y ( u , v , w ) , z ( u , v , w ) ) | ∂ ( x , y , z ) ∂ ( u , v , w ) | d u d v d w {\displaystyle {\begin{aligned}\iiint _{R}f(x,y,z)dV&\approx \sum _{i=1}^{m}\sum _{j=1}^{n}\sum _{k=1}^{p}f(x_{i},y_{j},z_{k})\Delta V\\&\approx \sum _{i=1}^{m}\sum _{j=1}^{n}\sum _{k=1}^{p}f(x(u_{i},v_{j},w_{k}),y(u_{i},v_{j},w_{k}),z(u_{i},v_{j},w_{k}))\Delta V_{2}\\{\text{Since }}&\Delta V_{2}\approx \left|{\frac {\partial (x,y,z)}{\partial (u,v,w)}}\right|\Delta u\Delta v\Delta w\\\sum _{i=1}^{m}\sum _{j=1}^{n}\sum _{k=1}^{p}f(x(u_{i},v_{j},w_{k}),y(u_{i},v_{j},w_{k}),z(u_{i},v_{j},w_{k}))\Delta V_{2}&\approx \sum _{i=1}^{m}\sum _{j=1}^{n}\sum _{k=1}^{p}f(x(u_{i},v_{j},w_{k}),y(u_{i},v_{j},w_{k}),z(u_{i},v_{j},w_{k}))\ \left|{\frac {\partial (x,y,z)}{\partial (u,v,w)}}\right|\Delta u\Delta v\Delta w\\&\approx \iiint _{S}f(x(u,v,w),y(u,v,w),z(u,v,w))\ \left|{\frac {\partial (x,y,z)}{\partial (u,v,w)}}\right|\ du\ dv\ dw\\\end{aligned}}}
然後，我們有以下定理，它類似於二重積分的定理。再次提醒，以上解釋並非
定理。 T {\displaystyle T} C 1 {\displaystyle C^{1}} u v w {\displaystyle uvw} S {\displaystyle S} x y z {\displaystyle xyz} R {\displaystyle R} x = x ( u , v , w ) , y = y ( u , v , w ) {\displaystyle x=x(u,v,w),y=y(u,v,w)} z = z ( u , v , w ) {\displaystyle z=z(u,v,w)} f {\displaystyle f} R {\displaystyle R} ∭ R f ( x , y , z ) d x d y d z = ∭ S f ( x ( u , v , w ) , y ( u , v , w ) , z ( u , v , w ) ) | ∂ ( x , y , z ) ∂ ( u , v , w ) | d u d v d w {\displaystyle \iiint _{R}f(x,y,z)\,dx\,dy\,dz=\iiint _{S}f(x(u,v,w),y(u,v,w),z(u,v,w))\left|{\frac {\partial (x,y,z)}{\partial (u,v,w)}}\right|\,du\,dv\,dw}
備註. d x d y d z {\displaystyle dx\,dy\,dz} d V {\displaystyle dV} 如果我們改變一些記號，我們可以得到 ∭ S f ( u ( x , y , z ) , v ( x , y , z ) , w ( x , y , z ) ) | ∂ ( u , v , w ) ∂ ( x , y , z ) | d x d y d z = ∭ R f ( u , v , w ) d u d v d w . {\displaystyle \iiint _{S}f(u(x,y,z),v(x,y,z),w(x,y,z))\left|{\frac {\partial (u,v,w)}{\partial (x,y,z)}}\right|\,dx\,dy\,dz=\iiint _{R}f(u,v,w)\,du\,dv\,dw.} T {\displaystyle T} x y z {\displaystyle xyz} S {\displaystyle S} u v w {\displaystyle uvw} R {\displaystyle R}
現在我們瞭解了雅可比矩陣的用途和推導，是時候應用這些新知識來解決一些例子了。前兩個例子包括將座標系從笛卡爾座標系變換到極座標系，以及將笛卡爾座標系變換到球座標系。
有時，我們可能將積分割槽域變換到其他座標系中的另一個區域。這可以簡化積分的計算，特別是在笛卡爾座標系中的區域與圓形相關時，例如球體、圓錐體、圓形等。
讓我們從將座標系從笛卡爾座標系變換到極座標系開始。
命題。 f ( x , y ) {\displaystyle f(x,y)} 笛卡爾座標 g ( r , θ ) = f ( r cos θ , r sin θ ) {\displaystyle g(r,\theta )=f(r\cos \theta ,r\sin \theta )} 極座標 相同 S {\displaystyle S} R {\displaystyle R} ∬ R f ( x , y ) d x d y = ∬ S g ( r , θ ) r d r d θ . {\displaystyle \iint _{R}f(x,y)\,dx\,dy=\iint _{S}g(r,\theta ){\color {green}{r}}\,dr\,d\theta .}
證明。 x = r cos θ and y = r sin θ . {\displaystyle x=r\cos \theta {\text{ and }}y=r\sin \theta .} ∂ ( x , y ) ∂ ( r , θ ) = | ∂ x ∂ r ∂ x ∂ θ ∂ y ∂ r ∂ y ∂ θ | = | cos θ − r sin θ sin θ r cos θ | = r ( cos 2 θ + sin 2 θ ) = r . {\displaystyle {\frac {\partial (x,y)}{\partial (r,\theta )}}={\begin{vmatrix}{\frac {\partial x}{\partial r}}&{\frac {\partial x}{\partial \theta }}\\{\frac {\partial y}{\partial r}}&{\frac {\partial y}{\partial \theta }}\end{vmatrix}}={\begin{vmatrix}\cos \theta &-r\sin \theta \\\sin \theta &r\cos \theta \end{vmatrix}}=r\left(\cos ^{2}\theta +\sin ^{2}\theta \right)=r.} ∬ R f ( x , y ) d x d y = ∬ S f ( r cos θ , r sin θ ) | ∂ ( x , y ) ∂ ( r , θ ) | d r d θ = ∬ S g ( r , θ ) r d r d θ . {\displaystyle \iint _{R}f(x,y)\,dx\,dy=\iint _{S}f(r\cos \theta ,r\sin \theta )\left|{\frac {\partial (x,y)}{\partial (r,\theta )}}\right|\,dr\,d\theta =\iint _{S}g(r,\theta )r\,dr\,d\theta .} ◻ {\displaystyle \Box }
命題. f ( x , y , z ) {\displaystyle f(x,y,z)} 笛卡爾座標 g ( r , θ , z ) = f ( r cos θ , r sin θ , z ) {\displaystyle g(r,\theta ,z)=f(r\cos \theta ,r\sin \theta ,z)} 柱座標 相同 S {\displaystyle S} R {\displaystyle R} ∭ R f ( x , y , z ) d x d y d z = ∭ S g ( r , θ , z ) r d r d θ d z . {\displaystyle \iiint _{R}f(x,y,z)\,dx\,dy\,dz=\iiint _{S}g(r,\theta ,z){\color {green}{r}}\,dr\,d\theta \,dz.}
命題. f ( x , y , z ) {\displaystyle f(x,y,z)} 笛卡爾座標 g ( ρ , ϕ , θ ) = f ( ρ sin ϕ cos θ , ρ sin ϕ sin θ , ρ cos ϕ ) {\displaystyle g(\rho ,\phi ,\theta )=f(\rho \sin \phi \cos \theta ,\rho \sin \phi \sin \theta ,\rho \cos \phi )} 球座標 相同 S {\displaystyle S} R {\displaystyle R} ∭ R f ( x , y , z ) d x d y d z = ∭ S g ( ρ , ϕ , θ ) ρ 2 sin ϕ d ρ d ϕ d θ . {\displaystyle \iiint _{R}f(x,y,z)\,dx\,dy\,dz=\iiint _{S}g(\rho ,\phi ,\theta ){\color {green}{\rho ^{2}\sin \phi }}\,d\rho \,d\phi \,d\theta .}
證明。
薩魯斯法則的示意圖。紅色箭頭對應正項，藍色箭頭對應負項。 If we change from Cartesian coordinates to spherical coordinates, we have the relationships x = ρ sin ϕ cos θ , y = ρ sin ϕ sin θ and z = ρ cos ϕ . {\displaystyle x=\rho \sin \phi \cos \theta ,\,y=\rho \sin \phi \sin \theta \;{\text{and}}\;z=\rho \cos \phi .} ∂ ( x , y , z ) ∂ ( ρ , ϕ , θ ) = | ∂ x ∂ ρ ∂ x ∂ ϕ ∂ x ∂ θ ∂ y ∂ ρ ∂ y ∂ ϕ ∂ y ∂ θ ∂ z ∂ ρ ∂ z ∂ ϕ ∂ z ∂ θ | = | sin ϕ cos θ ρ cos ϕ cos θ − ρ sin ϕ sin θ sin ϕ sin θ ρ cos ϕ sin θ ρ sin ϕ sin θ cos ϕ − ρ sin ϕ 0 | = 0 + ρ 2 sin ϕ cos 2 ϕ cos 2 θ + ρ 2 sin 3 ϕ sin 2 θ − ( − ρ 2 sin ϕ cos 2 ϕ sin 2 θ ) − ( − ρ 2 sin 3 ϕ cos 2 θ ) − 0 by Rule of Sarrus = ρ 2 sin ϕ cos 2 ϕ ( sin 2 θ + cos 2 θ ⏟ 1 ) + ρ 2 sin 3 ϕ ⏟ sin ϕ sin 2 ϕ ( sin 2 θ + cos 2 θ ⏟ 1 ) = ρ 2 sin ϕ ( cos 2 ϕ + sin 2 ϕ ⏟ 1 ) = ρ 2 sin ϕ . {\displaystyle {\begin{aligned}{\frac {\partial (x,y,z)}{\partial (\rho ,\phi ,\theta )}}&={\begin{vmatrix}{\frac {\partial x}{\partial \rho }}&{\frac {\partial x}{\partial \phi }}&{\frac {\partial x}{\partial \theta }}\\{\frac {\partial y}{\partial \rho }}&{\frac {\partial y}{\partial \phi }}&{\frac {\partial y}{\partial \theta }}\\{\frac {\partial z}{\partial \rho }}&{\frac {\partial z}{\partial \phi }}&{\frac {\partial z}{\partial \theta }}\\\end{vmatrix}}\\&={\begin{vmatrix}\sin \phi \cos \theta &\rho \cos \phi \cos \theta &-\rho \sin \phi \sin \theta \\\sin \phi \sin \theta &\rho \cos \phi \sin \theta &\rho \sin \phi \sin \theta \\\cos \phi &-\rho \sin \phi &0\end{vmatrix}}\\&=0{\color {purple}+\rho ^{2}\sin \phi \cos ^{2}\phi \cos ^{2}\theta }{\color {brown}+\rho ^{2}\sin ^{3}\phi \sin ^{2}\theta }{\color {purple}-(-\rho ^{2}\sin \phi \cos ^{2}\phi \sin ^{2}\theta )}{\color {brown}-(-\rho ^{2}\sin ^{3}\phi \cos ^{2}\theta )}-0\qquad {\text{by Rule of Sarrus}}\\&={\color {purple}\rho ^{2}\sin \phi \cos ^{2}\phi (\underbrace {\sin ^{2}\theta +\cos ^{2}\theta } _{1})}{\color {brown}+\rho ^{2}\underbrace {\sin ^{3}\phi } _{\sin \phi \sin ^{2}\phi }(\underbrace {\sin ^{2}\theta +\cos ^{2}\theta } _{1})}\\&=\rho ^{2}\sin \phi \left(\underbrace {{\color {purple}\cos ^{2}\phi }{\color {brown}+\sin ^{2}\phi }} _{1}\right)\\&=\rho ^{2}\sin \phi .\end{aligned}}} ∭ R f ( x , y , z ) d x d y d z = ∭ S f ( ρ sin ϕ cos θ , ρ sin ϕ sin θ , ρ cos ϕ ) | ∂ ( x , y , z ) ∂ ( ρ , ϕ , θ ) | d ρ d ϕ d θ = ∭ S g ( ρ , ϕ , θ ) ρ 2 ⏟ ≥ 0 sin ϕ ⏞ ≥ 0 since 0 ≤ ϕ ≤ π d ρ d ϕ d θ . {\displaystyle \iiint _{R}f(x,y,z)\,dx\,dy\,dz=\iiint _{S}f(\rho \sin \phi \cos \theta ,\rho \sin \phi \sin \theta ,\rho \cos \phi )\left|{\frac {\partial (x,y,z)}{\partial (\rho ,\phi ,\theta )}}\right|\,d\rho \,d\phi \,d\theta =\iiint _{S}g(\rho ,\phi ,\theta )\underbrace {\rho ^{2}} _{\geq 0}\overbrace {\sin \phi } ^{\geq 0\;{\text{since}}\;0\leq \phi \leq \pi }\,d\rho \,d\phi \,d\theta .} ◻ {\displaystyle \Box }
證明。
如果我們從笛卡爾座標系變換到柱座標系 x = r cos θ , y = r sin θ and z = z . {\displaystyle x=r\cos \theta ,\,y=r\sin \theta \;{\text{and}}\;{\color {green}{z=z}}.} ∂ ( x , y , z ) ∂ ( r , θ , z ) = | ∂ x ∂ r ∂ x ∂ θ ∂ x ∂ z ∂ y ∂ r ∂ y ∂ θ ∂ y ∂ z ∂ z ∂ r ∂ z ∂ θ ∂ z ∂ z | = | cos θ − r sin θ 0 sin θ r cos θ 0 0 0 1 | = 0 − 0 + | cos θ − r sin θ sin θ r cos θ | ⏟ by cofactor expansion along 3rd row = r . {\displaystyle {\frac {\partial (x,y,{\color {green}z})}{\partial (r,\theta ,{\color {green}z})}}={\begin{vmatrix}{\frac {\partial x}{\partial r}}&{\frac {\partial x}{\partial \theta }}&{\color {green}{\frac {\partial x}{\partial z}}}\\{\frac {\partial y}{\partial r}}&{\frac {\partial y}{\partial \theta }}&{\color {green}{\frac {\partial y}{\partial z}}}\\{\color {green}{\frac {\partial z}{\partial r}}}&{\color {green}{\frac {\partial z}{\partial \theta }}}&{\color {green}{\frac {\partial z}{\partial z}}}\end{vmatrix}}={\begin{vmatrix}\cos \theta &-r\sin \theta &{\color {green}{0}}\\\sin \theta &r\cos \theta &{\color {green}{0}}\\{\color {green}0}&{\color {green}0}&{\color {green}1}\end{vmatrix}}=\underbrace {0-0+{\begin{vmatrix}\cos \theta &-r\sin \theta \\\sin \theta &r\cos \theta \end{vmatrix}}} _{\text{by cofactor expansion along 3rd row}}=r.} 三重 ∭ R f ( x , y , z ) d x d y d z = ∭ S f ( r cos θ , r sin θ , z ) | ∂ ( x , y , z ) ∂ ( r , θ , z ) | d r d θ d z = ∭ S g ( r , θ , z ) r d r d θ d z . {\displaystyle \iiint _{R}f(x,y,{\color {green}z})\,dx\,dy\,{\color {green}dz}=\iiint _{S}f(r\cos \theta ,r\sin \theta ,{\color {green}z})\left|{\frac {\partial (x,y,{\color {green}z})}{\partial (r,\theta ,{\color {green}z})}}\right|\,dr\,d\theta \,{\color {green}dz}=\iiint _{S}g(r,\theta ,{\color {green}z})r\,dr\,d\theta \,{\color {green}dz}.} ◻ {\displaystyle \Box }
證明。
首先，將圓錐按提示放置。設圓錐在直角座標系和柱座標系中所包圍的區域分別為 C {\displaystyle C} C ′ {\displaystyle C'} ∭ C 1 d V = ∭ C ′ r d r d θ d z . {\displaystyle \iiint _{C}1\,dV=\iiint _{C'}r\,dr\,d\theta \,dz.} C ′ {\displaystyle C'} r , θ {\displaystyle r,\theta } z {\displaystyle z}
首先， θ {\displaystyle \theta } 0 ≤ θ ≤ 2 π {\displaystyle 0\leq \theta \leq 2\pi }
Then, given a fixed θ {\displaystyle \theta } r z {\displaystyle rz} r {\displaystyle r} z {\displaystyle z} r z {\displaystyle rz} x z {\displaystyle xz} θ = 0 {\displaystyle \theta =0} triangle ( 0 , 0 ) , ( 0 , h ) {\displaystyle (0,0),\,(0,h)} ( a , 0 ) {\displaystyle (a,0)} z ≤ − h a r + h ⟹ z h ≤ − r a + 1 ⟹ r a + z h ≤ 1 {\displaystyle {\color {green}z}\leq {\frac {-h}{a}}{\color {green}r}+h\implies {\frac {\color {green}z}{h}}\leq -{\frac {\color {green}r}{a}}+1\implies {\frac {\color {green}r}{a}}+{\frac {\color {green}z}{h}}\leq 1} θ {\displaystyle \theta } r a ≤ r a + z h ⏟ ≥ 0 ≤ 1 ⟹ 0 ≤ ⏟ by definition r ≤ a , {\displaystyle {\frac {r}{a}}\leq {\frac {r}{a}}+\underbrace {\frac {z}{h}} _{\geq 0}\leq 1\implies \underbrace {0\leq } _{\text{by definition}}r\leq a,} r {\displaystyle r} θ {\displaystyle \theta } r , θ {\displaystyle r,\theta } z h + r a ≤ 1 ⟹ 0 ≤ z ⏟ by cone ≤ h ( 1 − r a ) . {\displaystyle {\frac {z}{h}}+{\frac {r}{a}}\leq 1\implies \underbrace {0\leq z} _{\text{by cone}}\leq h\left(1-{\frac {r}{a}}\right).} z {\displaystyle z} θ {\displaystyle \theta } ∭ C ′ r d r d θ d z = ∫ 0 2 π ∫ 0 a ∫ 0 h ( 1 − r a ) r d z d r d θ by generalized Fubini's theorem for triple integration = ∫ 0 2 π ∫ 0 a r h ( 1 − r a ) d r d θ = ∫ 0 2 π [ h r 2 2 − h r 3 3 a ] r = 0 r = a d θ = h ∫ 0 2 π ( a 2 2 − a 3 2 3 a ⏟ a 2 / 6 ) d θ = ( 2 π ) a 2 6 = 1 3 π a 2 h {\displaystyle {\begin{aligned}\iiint _{C'}r\,dr\,d\theta \,dz&=\int _{0}^{2\pi }\int _{0}^{a}\int _{0}^{h\left(1-{\frac {r}{a}}\right)}r\,dz\,dr\,d\theta \qquad {\text{by generalized Fubini's theorem for triple integration}}\\&=\int _{0}^{2\pi }\int _{0}^{a}rh\left(1-{\frac {r}{a}}\right)\,dr\,d\theta \\&=\int _{0}^{2\pi }\left[{\frac {hr^{2}}{2}}-{\frac {hr^{3}}{3a}}\right]_{r=0}^{r=a}\,d\theta \\&=h\int _{0}^{2\pi }\left(\underbrace {{\frac {a^{2}}{2}}-{\frac {a^{{\cancel {3}}2}}{3{\cancel {a}}}}} _{a^{2}/6}\right)\,d\theta \\&={\frac {(2\pi )a^{2}}{6}}\\&={\frac {1}{3}}\pi a^{2}h\end{aligned}}} ◻ {\displaystyle \Box }
證明。
首先，將球體的中心置於原點。設 S {\displaystyle S} S ′ {\displaystyle S'} 球面 ∭ S 1 d V = ∭ S ′ ρ 2 sin ϕ d ρ d ϕ d θ . {\displaystyle \iiint _{S}1\,dV=\iiint _{S'}\rho ^{2}\sin \phi \,d\rho \,d\phi \,d\theta .} ρ , ϕ , θ {\displaystyle \rho ,\phi ,\theta } 0 ≤ ρ ≤ r , 0 ≤ ϕ ≤ π {\displaystyle 0\leq \rho \leq r,\,0\leq \phi \leq \pi } 0 ≤ θ ≤ 2 π {\displaystyle 0\leq \theta \leq 2\pi } S ′ {\displaystyle S'} ∭ S ′ ρ 2 sin ϕ d ρ d ϕ d θ = ∫ 0 2 π ∫ 0 π ∫ 0 r ρ 2 sin ϕ d ρ d ϕ d θ = 1 3 ∫ 0 2 π ∫ 0 π r 3 sin ϕ d ϕ d θ = 1 3 r 3 ∫ 0 2 π − ( cos ( π ) ⏟ − 1 − cos 0 ⏟ 1 ) ⏞ 2 d θ = 2 3 r 3 ( 2 π ) = 4 3 π r 3 . {\displaystyle {\begin{aligned}\iiint _{S'}\rho ^{2}\sin \phi \,d\rho \,d\phi \,d\theta &=\int _{0}^{2\pi }\int _{0}^{\pi }\int _{0}^{r}\rho ^{2}\sin \phi \,d\rho \,d\phi \,d\theta \\&={\frac {1}{3}}\int _{0}^{2\pi }\int _{0}^{\pi }r^{3}\sin \phi \,d\phi \,d\theta \\&={\frac {1}{3}}r^{3}\int _{0}^{2\pi }\overbrace {-(\underbrace {\cos(\pi )} _{-1}-\underbrace {\cos 0} _{1})} ^{2}\,d\theta \\&={\frac {2}{3}}r^{3}(2\pi )\\&={\frac {4}{3}}\pi r^{3}.\end{aligned}}}
![{\displaystyle {\begin{aligned}\int _{e}^{e^{2}}{\frac {\sin(\ln(x))}{x}}\ dx&=\int _{e}^{e^{2}}\sin(\ln(x))\ {\Big (}{\frac {1}{x}}\ dx{\Big )}&\quad {\text{rearrangement}}\\&=\int _{1}^{2}\sin(u)\ du&\quad {\text{remember }}u=\ln(x){\text{ and }}du={\frac {1}{x}}\ dx\\&={\Big [}-\cos(u){\Big ]}_{1}^{2}&\quad {\text{integration}}\\&=\cos(1)-\cos(2)\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e1c0c4b74bd5330a19fb8c6c4dece2c202ed726d)
![{\displaystyle {\begin{aligned}\iint _{D}{\frac {y(x+y)}{x}}dx&=\int _{4}^{5}\int _{2}^{3}\left(uv\cdot \underbrace {\frac {u}{(v+1)^{2}}} _{>0{\text{ because of the bounds}}}\right)\,du\,dv\\&=\int _{4}^{5}{\frac {v}{(v+1)^{2}}}\left({\frac {3^{3}}{3}}-{\frac {2^{3}}{3}}\right)\,dv\\&={\frac {19}{3}}\int _{4+1}^{5+1}{\frac {w-1}{w^{2}}}\,dw\qquad {\text{let }}w=v+1\implies dw=dv\\&={\frac {19}{3}}\left[\ln |w|-{\frac {1}{w}}\right]_{5}^{6}\\&={\frac {19}{3}}\left(\ln 6-\ln 5+{\frac {1}{30}}\right)\\&={\frac {19}{3}}\left(\ln \left({\frac {6}{5}}\right)+{\frac {1}{30}}\right)\qquad {\text{by properties of logarithm}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c61e80c3c14e9895783b67883681771c4147ae00)
![{\displaystyle {\begin{aligned}\iiint _{C'}r\,dr\,d\theta \,dz&=\int _{0}^{2\pi }\int _{0}^{a}\int _{0}^{h\left(1-{\frac {r}{a}}\right)}r\,dz\,dr\,d\theta \qquad {\text{by generalized Fubini's theorem for triple integration}}\\&=\int _{0}^{2\pi }\int _{0}^{a}rh\left(1-{\frac {r}{a}}\right)\,dr\,d\theta \\&=\int _{0}^{2\pi }\left[{\frac {hr^{2}}{2}}-{\frac {hr^{3}}{3a}}\right]_{r=0}^{r=a}\,d\theta \\&=h\int _{0}^{2\pi }\left(\underbrace {{\frac {a^{2}}{2}}-{\frac {a^{{\cancel {3}}2}}{3{\cancel {a}}}}} _{a^{2}/6}\right)\,d\theta \\&={\frac {(2\pi )a^{2}}{6}}\\&={\frac {1}{3}}\pi a^{2}h\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/444524b85f072795032ec6b2ece4eceabc42629a)
