This section discusses the translation of a graph from the \(xy\) Cartesian plane to the \(uv\) Cartesian plane and defines the Jacobian.
Theoretical discussion with Descriptive Elaboration
For any given function \(f(x,y)\), we can define x and y as a function of other variables \(g(u,v)\). To do this, we first find \(u\) and \(v\) as a function of \(x\) and \(y\) that will allow for an easier integrand. Then solve for \(x\) and \(y\) in order to translate the function so that \(x=g(u,v) \ \text{and} \ y=h(u,v)\). This translates the are region from R in the x-y plane to D in the u-v plane.
Remember:
\[I = \iint_R f(x,y) \ dA \nonumber \]
So we must find \(dA\):
\(dA\) changes from \( dxdy\) to \(|J(u,v)| \ dudv\). Each change in u (\(\Delta u\)) and change in v (\(\Delta v\)) create parallelograms that are small areas \(\Delta A\) or dA . We can find the area of each of these parallelograms (P) by taking the cross product of the two vectors that create it (\(\Delta u \ \text{and} \ \Delta v\)).
\[ \text{Area of P} = \begin{vmatrix} \vec{V_1} \times \vec{V_2} \end{vmatrix} = J(u,v) \nonumber \]
\[\begin{align*} J(u,v) &= \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix} \\ &= \dfrac{\partial x}{\partial u} \dfrac{\partial y}{\partial v} - \dfrac{\partial y}{\partial u} \dfrac{\partial x}{\partial v} \\ &= \dfrac{\partial (x,y)}{\partial (u,v)} \end{align*} \]
\(|J(u,v)|\) represents the area of the parallelogram, and it is the determinant of the Jacobian matrix, shown above. The Jacobian measures how much the transformation is changing from the region \(R\) to the region \(G\). Therefore, the translation of the integration of \(f(x,y)\) is represented by
\[ \iint_R f(x,y) \ dx \ dy = \iint_G f(g(u,v), \ h(u,v)) | J(u,v) | du \ dv . \nonumber \]
The same can be applied for triple integrals, where the translations are represented by
\[x=g(u,v,w), \; y=h(u,v,w), \; z=k(u,v,w) \nonumber \]
\[\begin{align*} J(u,v,w) &= \dfrac{\partial (x,y,z)}{\partial (u,v,w)} \\ &= \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} & \dfrac{\partial x}{\partial w} \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} & \dfrac{\partial y}{\partial w} \\ \dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} & \dfrac{\partial z}{\partial w} \end{vmatrix} \\ &= \dfrac{\partial x}{\partial u} \begin{vmatrix} \dfrac{\partial y}{\partial v} & \dfrac{\partial y}{\partial w} \\ \dfrac{\partial z}{\partial v} & \dfrac{\partial z}{\partial w} \end{vmatrix} - \dfrac{\partial x}{\partial v} \begin{vmatrix} \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial w} \\ \dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial w} \end{vmatrix} + \dfrac{\partial x}{\partial w} \begin{vmatrix} \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \\ \dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} \end{vmatrix} \\ &= \dfrac{\partial x}{\partial u} \left( \dfrac{\partial y}{\partial v} \cdot \dfrac{\partial z}{\partial w} - \dfrac{\partial z}{\partial v} \cdot \dfrac{\partial y}{\partial w} \right) - \dfrac{\partial x}{\partial v} \left( \dfrac{\partial y}{\partial u} \cdot \dfrac{\partial z}{\partial w} - \dfrac{\partial z}{\partial u} \cdot \dfrac{\partial y}{\partial w} \right) + \dfrac{\partial x}{\partial w} \left( \dfrac{\partial y}{\partial u} \cdot \dfrac{\partial z}{\partial v} - \dfrac{\partial z}{\partial u} \cdot \dfrac{\partial y}{\partial v} \right) \end{align*} \]
This method for getting the Jacobian is called cofactor expansion.
Although the introduction focused primarily on translating a Cartesian function to a different Cartesian coordinate system, the concept of the Jacobian can also be used to explain how translations into polar coordinates work as well.
For cylindrical coordinates
\[ x=r \ \text{cos} \ \theta, \ y=r \ \text{sin} \ \theta, \ z=z \nonumber \]
Therefore:
\[ J(u,v,w) = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} & \dfrac{\partial x}{\partial w} \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} & \dfrac{\partial y}{\partial w} \\ \dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} & \dfrac{\partial z}{\partial w} \end{vmatrix} = \begin{vmatrix} \text{cos} \ \theta & -r \ \text{sin} \ \theta & 0 \\ \text{sin} & r \ \text{cos} \ \theta & 0 \\ 0 & 0 & 1 \end{vmatrix} \nonumber \]
\[ \iiint_D F(x,y,z) \ dx \ dy \ dz = \iiint_G H(r, \theta , z) |r| \ dr \ d \theta \ dz \nonumber \]
For spherical coordinates
\[ J( \rho , \phi , \theta) = \begin{vmatrix} \dfrac{\partial x}{\partial \rho} & \dfrac{\partial x}{\partial \phi} & \dfrac{\partial x}{\partial \theta} \\ \dfrac{\partial y}{\partial \rho} & \dfrac{\partial y}{\partial \phi} & \dfrac{\partial y}{\partial \theta} \\ \dfrac{\partial z}{\partial \rho} & \dfrac{\partial z}{\partial \phi} & \dfrac{\partial z}{\partial \theta} \end{vmatrix} = \rho ^2 \ \text{sin} \phi \nonumber \]
\[ \iiint_D F(x,y,z) \ dx \ dy \ dz = \iiint_G H(\rho , \phi , \theta) | \rho ^2 \text{sin} \ \phi \ | \ d \rho \ d \phi \ d \theta \nonumber \]
Hence, \(dx\,dy\,dz\) becomes \(rd\,rd\,\theta\) in cylindrical coordinates and \(\rho ^2 \text{sin} \ \phi \ d \rho \ d \phi \ d \theta \) in spherical coordinates.
Example \(\PageIndex{1}\)
Use the following transformation to evaluate the integral.
\[ u = \dfrac{y}{x} \ \ \text{and} \ \ v =xy \nonumber \]
\[ \iint_R \dfrac{y}{x} dA \nonumber \]
\[ \text{Where R is bounded by:} \ \ 1 \le u \le 2 \ \ \text{and} \ \ 1 \le v \le 2 \nonumber \]
Solution
First find \(x\) and \(y\) as functions of \(u\) and \(v\):
\(u = \dfrac{y}{x} \) \(v = xy\)
\(y = xu\) \(\rightarrow\) \(x= \dfrac{v}{y} \)
\(x= \dfrac{v}{xu}\)
\(x^2 = \dfrac{v}{u}\)
\(x = \sqrt{\dfrac{v}{u}} \)
\(y = xu\)
\(y = \left( \sqrt{\dfrac{v}{u}} \right) u \)
\(y= \sqrt{vu} \)
\(x=g(u,v) = \sqrt{\dfrac{v}{u}} \ \text{and} \ y=h(u,v) = \sqrt{vu} \)
Using \(x=g(u,v) \ \text{and} \ y=h(u,v)\), find the integrand in terms of \( u \ \text{and} \ v\):
\[ \dfrac{y}{x} = \dfrac{\sqrt{vu}}{\sqrt{\dfrac{v}{u}}} = u \nonumber \]
And \( dA\) changes from \( dx dy \) to \(\left| J(u,v) \right| \ du \ dv \). The Jacobian equals:
\[ J(u,v) = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix} = \dfrac{\partial x}{\partial u} \dfrac{\partial y}{\partial v} - \dfrac{\partial y}{\partial u} \dfrac{\partial x}{\partial v} \nonumber \]
\[ \dfrac{\partial x}{\partial u} = \dfrac{1}{2} u^{- \dfrac{3}{2}} v^{\dfrac{1}{2}} \ \ \ \dfrac{\partial x}{\partial v} = \dfrac{1}{2} u^{- \dfrac{1}{2}} v^{- \dfrac{1}{2}} \ \ \ \dfrac{\partial y}{\partial u} = \dfrac{1}{2} u^{- \dfrac{1}{2}} v^{\dfrac{1}{2}} \ \ \ \dfrac{\partial y}{\partial v} = \dfrac{1}{2} u^{\dfrac{1}{2}} v^{- \dfrac{1}{2}} \nonumber \]
\[\begin{align*} J(u,v) \ &= \begin{vmatrix} \dfrac{1}{2} u^{- \dfrac{3}{2}} v^{\dfrac{1}{2}} & \dfrac{1}{2} u^{- \dfrac{1}{2}} v^{- \dfrac{1}{2}} \\ \dfrac{1}{2} u^{- \dfrac{1}{2}} v^{\dfrac{1}{2}} & \dfrac{1}{2} u^{\dfrac{1}{2}} v^{- \dfrac{1}{2}} \end{vmatrix} \\ & = \left( - \dfrac{1}{4} u^{-1} - \dfrac{1}{4} u^{-1} \right) \\ & = \dfrac{1}{2u} \end{align*} \]
Therefore, evaluate:
\[\begin{align*} &\int_1^2 \int_1^2 u \ \left( \dfrac{1}{2u} \right) \ du \ dv \\ &= \int_1^2 \int_1^2 \dfrac{1}{2} \ du \ dv \\ &= \int_1^2 \left. \dfrac{1}{2} u \right|_1^2 \ dv\\ &= \int_1^2 1 - \dfrac{1}{2} \ dv \\ & = \left. \dfrac{1}{2} v \right|_1^2 \\ & = 1 - \dfrac{1}{2} = \dfrac{1}{2} \end{align*} \]
Example \(\PageIndex{2}\)
Use the following transformation to evaluate the integral.
\[u=x- \dfrac{1}{2} y \ \text{and} \ v=y \nonumber \]
\[ \int_0^{\dfrac{1}{2}} \int_{\dfrac{y}{2}}^{\dfrac{y+4}{2}} y^3 (2x-y) e^{{2x-y}^2} \ dx \ dy \nonumber \]
Solution
First solve for \(x\) and \(y\):
\( u= x- \dfrac{1}{2} y\) \(v =y\)
\(u = x- \dfrac{1}{2} v \) \(y =v \)
\(x= u + \dfrac{1}{2} v \).
Then substitute \(x\) and \(y\) for \( g(u,v) \) and \(h(u,v)\) :
The integrand:
\[y^3 (2x-y) e^{{2x-y}^2} \ \rightarrow \ v^3 [2 (u + \dfrac{1}{2} v) - v ] e^{{[2 (u + \dfrac{1}{2} v) - v]} ^2} \nonumber \]
\[= v^3 (2u) e^{{2u}^2} \nonumber \]
\[=(2uv^3) e^{4u^2} \nonumber \]
The transformation also changes the bounds of integration
\(\begin{align*} x &= \dfrac{y+4}{2} \ \rightarrow \ u + \dfrac{1}{2} v = \dfrac{v +4}{2} \\[4pt] &= \dfrac{y}{4} \ \rightarrow \ u + \dfrac{v}{2} = \dfrac {v}{2} \end{align*} \]
\(u = \dfrac{4}{2} \)
\(u = 0 \)
\(u = 2\)
And \( dx dy \) changes to \(\left| J(u,v) \right| \ du \ dv \). The Jacobian equals:
\[ J(u,v) = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial u} \end{vmatrix} = \dfrac{\partial x}{\partial u} \dfrac{\partial y}{\partial v} - \dfrac{\partial y}{\partial u} \dfrac{\partial x}{\partial v} \nonumber \]
\[= \begin{vmatrix} 1& 5 \\ 0 & 1 \end{vmatrix} =1 \nonumber \]
Thus,
\[\begin{align*} \iint_R f(x,y) \ dx \ dy &= \iint_G f(g(u,v), \ h(u,v)) | J(u,v) | du \ dv \\ &= \int_0^{\dfrac{1}{2}} \int_0^2 2uv^3 e^{4u^2} \ (1) \ dv \ du \\ & = \int_0^{\dfrac{1}{2}} \left. \dfrac{ue^{4u^2}v^4}{2} \right|_0^2 \ du \\ & = \int_0^{\dfrac{1}{2}} 8ue^{4u^2} \ du \\ & = \left. e^{4u^2} \right|_0^{\dfrac{1}{2}} \\ & = e-1 \end{align*} \]
Example \(\PageIndex{3}\)
Find the mass of an object bounded by
\(1 \le x \le 2, \ \ 0 \le xy \le 1, \ \ 0 \le z \le 2\)
with a density that can be described by the formula \( x^2 y + 2xyz\) by using the transformation \( u = x, \ \ v=xy, \ \ w = 3z\).
Solution
Set up the integral in cartesian coordinates:
\[ \int_1^2 \int_0^{\dfrac{1}{x}} \int_0^2 x^2y + 2xyz \ dzdydx. \nonumber \]
To apply the substitution, first solve for \(x\) and \(y\) using the given transformations:
\[u=x \qquad v=xy \qquad w=3z \nonumber \]
\[ x=u \qquad y= \dfrac{v}{x} \qquad z = \dfrac{w}{3} \nonumber \]
\[ y = \dfrac{v}{u} . \nonumber \]
Then make the appropriate substitutions within the integrand:
\[x^2 y + 2xyz \rightarrow u^2 \left( \dfrac{v}{u} \right) + 2u \left( \dfrac{v}{u} \right) \left( \dfrac{w}{3} \right) \ \rightarrow \ uv + \dfrac{2vw}{3}. \nonumber \]
Next, find the new boundaries to the region we want to integrate:
\( 1 \le x \le 2 \ \rightarrow \ 1 \le u \le 2 \)
\(0 \le xy \le 1 \ \rightarrow \ 0 \le v \le 1 \)
\( 0 \le z \le 2 \ \rightarrow \ 0 \le \dfrac{w}{3} \le 2 \ \rightarrow \ 0 \le w \le 6 \).
To complete the transformation, find the Jacobian:
\[\begin{align*} J(u,v,w) &= \dfrac{\partial (x,y,z)}{\partial (u,v,w)} = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} & \dfrac{\partial x}{\partial w} \\ \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} & \dfrac{\partial y}{\partial w} \\ \dfrac{\partial z}{\partial u} & \dfrac{\partial z}{\partial v} & \dfrac{\partial z}{\partial w} \end{vmatrix} \\ &= \begin{vmatrix} 1 & 0 & 0 \\ -\dfrac{v}{u^2} & \dfrac{1}{u} & 0 \\ 0 & 0 & \dfrac{1}{3} \end{vmatrix} \\ & = \dfrac{1}{3u} . \end{align*} \]
Notice the Jacobian of a lower triangular matrix (the values above the diagonal are all zero) is the multiplication of the diagonal entries. You can confirm this with cofactor expansion.
Using all of our calculated transformations, we can compute the new integral:
\[\begin{align*} \text{Mass } &= \int_0^6 \int_0^1 \int_1^2 \left( uv + \dfrac{2vw}{3} \right) \dfrac{1}{3u} \ dudvdw \\ & = \dfrac{1}{3} \int_0^6 \int_0^1 \int_1^2 v + \dfrac{2vw}{3u} \ dudvdw \\ & = \dfrac{1}{2} \int_0^6 \int_0^1 \left. vu + \dfrac{2vw}{3}\ln|u| \right|_1^2 \ dvdw \\ & = \dfrac{1}{3} \int_0^6 \int_0^1 v + \dfrac{2vw}{3} \ln2 \ dvdw \\ & = \dfrac{1}{3} \int_0^6 \left. \dfrac{v^2}{2} + \dfrac{2w\ln2}{3} \left( \dfrac{v^2}{2}\right) \right|_0^1 \ dw \\ & = \dfrac{1}{3} \int_0^6 \dfrac{1}{2} + \dfrac{w\ln2}{3} \ dw \\ & = \dfrac{1}{3} \left[ \dfrac{1}{2} w + \dfrac{w^2 \ln2}{6} \right]_0^6 \\ & = \dfrac{1}{3} \left[ 3 + 6\ln2 \right] \\ & = 1 + 2\ln2 . \end{align*} \]