{"id":1303,"date":"2025-10-17T13:23:49","date_gmt":"2025-10-17T13:23:49","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1303"},"modified":"2025-11-03T17:20:10","modified_gmt":"2025-11-03T17:20:10","slug":"change-of-variables-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/change-of-variables-learn-it-2\/","title":{"raw":"Change of Variables: Learn It 2","rendered":"Change of Variables: Learn It 2"},"content":{"raw":"

Jacobians<\/h2>\r\n

We mentioned earlier that transformation functions must have continuous first partial derivatives\u2014meaning [latex]g_u[\/latex], [latex]g_v[\/latex], [latex]h_u[\/latex], and [latex]h_v[\/latex] exist and are continuous. A transformation with this property is called a [latex]C^1[\/latex] transformation<\/strong> (where [latex]C[\/latex] denotes continuous).<\/p>\r\n

Let [latex]T(u,v) = (g(u,v), h(u,v))[\/latex] be a one-to-one [latex]C^1[\/latex] transformation, where [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex]. We want to understand how this transformation affects a small rectangular region [latex]S[\/latex] in the [latex]uv[\/latex]-plane, measuring [latex]\\Delta u[\/latex] units by [latex]\\Delta v[\/latex] units (see Figure 4).<\/p>\r\n\r\n[caption id=\"attachment_1421\" align=\"aligncenter\" width=\"765\"]\"On Figure 4.\u00a0A small rectangle [latex]S[\/latex] in the\u00a0[latex]uv[\/latex]-plane is transformed into a region\u00a0[latex]R[\/latex] in the\u00a0[latex]xy[\/latex]-plane.[\/caption]\r\n

<\/h3>\r\n

Since [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex], we can write the position vector of the image of point [latex](u, v)[\/latex]\u00a0as:<\/p>\r\n

[latex]\\mathbf{r}(u,v) = g(u,v)\\mathbf{i} + h(u,v)\\mathbf{j}[\/latex]<\/p>\r\n

Suppose [latex](u_0, v_0)[\/latex] is the lower left corner of rectangle [latex]S[\/latex], which maps to [latex](x_0, y_0) = T(u_0, v_0)[\/latex].<\/p>\r\n

The line [latex]v = v_0[\/latex] in the [latex]uv[\/latex]-plane maps to an image curve with vector function [latex]\\mathbf{r}(u, v_0)[\/latex]. The tangent vector at [latex](x_0, y_0)[\/latex] to this image curve is:<\/p>\r\n

[latex]\\mathbf{r}_u = g_u(u_0, v_0)\\mathbf{i} + h_u(u_0, v_0)\\mathbf{j} = \\frac{\\partial x}{\\partial u}\\mathbf{i} + \\frac{\\partial y}{\\partial u}\\mathbf{j}[\/latex]<\/p>\r\n

Similarly, the line [latex]u = u_0[\/latex] maps to an image curve with vector function [latex]\\mathbf{r}(u_0, v)[\/latex], and its tangent vector at [latex](x_0, y_0)[\/latex] is:<\/p>\r\n

[latex]\\mathbf{r}_v = g_v(u_0, v_0)\\mathbf{i} + h_v(u_0, v_0)\\mathbf{j} = \\frac{\\partial x}{\\partial v}\\mathbf{i} + \\frac{\\partial y}{\\partial v}\\mathbf{j}[\/latex]<\/p>\r\n\r\n

\r\n

Tangent Vectors and Derivatives<\/strong><\/p>\r\n

For a vector function [latex]\\mathbf{r}(u)[\/latex], the derivative [latex]\\mathbf{r}_u[\/latex] gives the tangent vector:<\/p>\r\n

[latex]{{\\bf{r}}_{u}} = {\\displaystyle\\lim_{{\\Delta}{u}{\\rightarrow}{0}}}{\\dfrac{{\\bf{r}}{({u_0} + {{\\Delta}{u}},{v_0}) - {\\bf{r}}{({u_0},{v_0})}}}{{\\Delta}{u}}} [\/latex]<\/p>\r\n

This means [latex]\\mathbf{r}(u_0 + \\Delta u) - \\mathbf{r}(u_0) \\approx \\Delta u \\mathbf{r}_u[\/latex] for small [latex]\\Delta u[\/latex].<\/p>\r\n\r\n<\/section>\r\n

From the definition of derivative, we have:<\/p>\r\n

[latex]{{\\bf{r}}_{u}} = {\\displaystyle\\lim_{{\\Delta}{u}{\\rightarrow}{0}}}{\\dfrac{{\\bf{r}}{({u_0} + {{\\Delta}{u}},{v_0}) - {\\bf{r}}{({u_0},{v_0})}}}{{\\Delta}{u}}} \\ {\\text{so}} \\ {{\\bf{r}}{({u_0} + {{\\Delta}{u}},{v_0}) - {\\bf{r}}{({u_0},{v_0})}}}{\\approx}{\\Delta}{u}{\\bf{r}}_{u}[\/latex].<\/p>\r\n

Similarly,<\/p>\r\n

[latex]{{\\bf{r}}_{v}} = {\\displaystyle\\lim_{{\\Delta}{v}{\\rightarrow}{0}}}{\\dfrac{{\\bf{r}}{({u_0},{v_0} + {{\\Delta}{v}}) - {\\bf{r}}{({u_0},{v_0})}}}{{\\Delta}{v}}} \\ {\\text{so}} \\ {\\bf{r}}{({u_0},{v_0} + {{\\Delta}{v}}) - {\\bf{r}}{({u_0},{v_0})}}{\\approx}{\\Delta}{v}{\\bf{r}}_{v}[\/latex].<\/p>\r\n\r\n

Finding the Area Element<\/h3>\r\nThe small rectangle [latex]S[\/latex] in the [latex]uv[\/latex]-plane transforms into a curved region [latex]R[\/latex] in the [latex]xy[\/latex]-plane. We can estimate the area [latex]\\Delta A[\/latex] of image [latex]R[\/latex] by finding the area of the parallelogram formed by the sides [latex]\\Delta u \\mathbf{r}_u[\/latex] and [latex]\\Delta v \\mathbf{r}_v[\/latex].\r\n\r\n
\r\n

Area of a Parallelogram Using Cross Product<\/strong><\/p>\r\n

For vectors [latex]\\mathbf{a}[\/latex] and [latex]\\mathbf{b}[\/latex] in space, the area of the parallelogram they span is [latex]|\\mathbf{a} \\times \\mathbf{b}|[\/latex].<\/p>\r\n

To compute the cross product of two-dimensional vectors, add a [latex]\\mathbf{k}[\/latex] component of zero to make them three-dimensional.<\/p>\r\n\r\n<\/section>\r\n

Using the cross product (by adding the [latex]\\mathbf{k}[\/latex] component as [latex]0[\/latex]), the area [latex]\\Delta A[\/latex] of image [latex]R[\/latex] is approximately:<\/p>\r\n

[latex]|\\Delta u \\mathbf{r}_u \\times \\Delta v \\mathbf{r}_v| = |\\mathbf{r}_u \\times \\mathbf{r}_v| \\Delta u \\Delta v[\/latex]<\/p>\r\n

In determinant form, the cross product is:<\/p>\r\n

[latex]\\mathbf{r}_u \\times \\mathbf{r}_v = \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ \\dfrac{\\partial x}{\\partial u} & \\dfrac{\\partial y}{\\partial u} & 0 \\\\ \\dfrac{\\partial x}{\\partial v} & \\dfrac{\\partial y}{\\partial v} & 0 \\end{vmatrix} = \\begin{vmatrix} \\dfrac{\\partial x}{\\partial u} & \\dfrac{\\partial y}{\\partial u} \\\\ \\dfrac{\\partial x}{\\partial v} & \\dfrac{\\partial y}{\\partial v} \\end{vmatrix} \\mathbf{k} = \\left( \\dfrac{\\partial x}{\\partial u} \\dfrac{\\partial y}{\\partial v} - \\dfrac{\\partial x}{\\partial v} \\dfrac{\\partial y}{\\partial u} \\right) \\mathbf{k}[\/latex]<\/p>\r\n

Since [latex]|\\mathbf{k}| = 1[\/latex], we have:<\/p>\r\n

[latex]\\Delta A \\approx |\\mathbf{r}_u \\times \\mathbf{r}_v| \\Delta u \\Delta v = \\left| \\dfrac{\\partial x}{\\partial u} \\dfrac{\\partial y}{\\partial v} - \\dfrac{\\partial x}{\\partial v} \\dfrac{\\partial y}{\\partial u} \\right| \\Delta u \\Delta v[\/latex]<\/p>\r\n

This determinant expression is so important that it gets its own name: the Jacobian.<\/p>\r\n\r\n

The Jacobian Determinant<\/h3>\r\n

The Jacobian<\/strong> of the [latex]C^1[\/latex] transformation [latex]T(u,v) = (g(u,v), h(u,v))[\/latex] is denoted by [latex]J(u,v)[\/latex] and is defined by the [latex]2 \\times 2[\/latex] determinant:<\/p>\r\n

[latex]J(u,v) = \\left| \\dfrac{\\partial(x,y)}{\\partial(u,v)} \\right| = \\begin{vmatrix} \\dfrac{\\partial x}{\\partial u} & \\dfrac{\\partial y}{\\partial u} \\\\ \\dfrac{\\partial x}{\\partial v} & \\dfrac{\\partial y}{\\partial v} \\end{vmatrix} = \\left( \\dfrac{\\partial x}{\\partial u} \\dfrac{\\partial y}{\\partial v} - \\dfrac{\\partial x}{\\partial v} \\dfrac{\\partial y}{\\partial u} \\right)[\/latex]<\/p>\r\n\r\n

\r\n

the Jacobian<\/h3>\r\n

For transformation [latex]T(u,v) = (x,y)[\/latex] where [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex]:<\/p>\r\n

[latex]J(u,v) = \\dfrac{\\partial(x,y)}{\\partial(u,v)} = \\begin{vmatrix} \\dfrac{\\partial x}{\\partial u} & \\dfrac{\\partial y}{\\partial u} \\\\ \\dfrac{\\partial x}{\\partial v} & \\dfrac{\\partial y}{\\partial v} \\end{vmatrix}[\/latex]<\/p>\r\n

The Jacobian tells us how the transformation scales areas: [latex]\\Delta A \\approx |J(u,v)| \\Delta u \\Delta v[\/latex]<\/p>\r\n\r\n<\/section>\r\n

Using this definition, we can write:<\/p>\r\n

[latex]\\Delta A \\approx J(u,v) \\Delta u \\Delta v = \\left| \\dfrac{\\partial(x,y)}{\\partial(u,v)} \\right| \\Delta u \\Delta v[\/latex]<\/p>\r\n

Note that the Jacobian is frequently written simply as:<\/p>\r\n

[latex]J(u,v) = \\dfrac{\\partial(x,y)}{\\partial(u,v)}[\/latex]<\/p>\r\n\r\n

\r\n

Two Ways to Arrange the Jacobian<\/strong><\/p>\r\n

The determinant can be computed in two equivalent ways:<\/p>\r\n

[latex]\\begin{vmatrix} \\dfrac{\\partial x}{\\partial u} & \\dfrac{\\partial y}{\\partial u} \\\\ \\dfrac{\\partial x}{\\partial v} & \\dfrac{\\partial y}{\\partial v} \\end{vmatrix} = \\left( \\dfrac{\\partial x}{\\partial u} \\dfrac{\\partial y}{\\partial v} - \\dfrac{\\partial x}{\\partial v} \\dfrac{\\partial y}{\\partial u} \\right) = \\begin{vmatrix} \\dfrac{\\partial x}{\\partial u} & \\dfrac{\\partial x}{\\partial v} \\\\ \\dfrac{\\partial y}{\\partial u} & \\dfrac{\\partial y}{\\partial v} \\end{vmatrix}[\/latex]<\/p>\r\n

The notation [latex]J(u,v) = \\dfrac{\\partial(x,y)}{\\partial(u,v)}[\/latex] suggests we can write the Jacobian with partials of [latex]x[\/latex] in the first row and partials of [latex]y[\/latex] in the second row.<\/p>\r\n\r\n<\/section>

Consider the transformation [latex]T(r,\\theta) = (x,y)[\/latex] where [latex]x = r\\cos\\theta[\/latex] and [latex]y = r\\sin\\theta[\/latex]. Find the Jacobian of this transformation.[reveal-answer q=\"602743587\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"602743587\"]\r\n

The transformation is [latex]{T} = {({r},{\\theta})} = {({r}{\\cos}{\\theta},{r}{\\sin}{\\theta})}[\/latex] where [latex]{x} = {r}{\\cos}{\\theta}[\/latex] and [latex]{y} = {r}{\\sin}{\\theta}[\/latex]. Thus the Jacobian is<\/p>\r\n[latex]\\hspace{4cm}\\large{\\begin{align}\r\n\r\nJ(r,\\theta)&=\\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\n\r\n&=r\\cos^2\\theta+r\\sin^2\\theta=r(\\cos^2\\theta+\\sin^2\\theta)=r \\end{align}}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

Consider the transformation [latex]T(u,v) = (x,y)[\/latex] where [latex]x = u^2 + v^2[\/latex] and [latex]y = uv[\/latex]. Find the Jacobian of this transformation.\r\n[reveal-answer q=\"113725\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"113725\"]The transformation is [latex]T(u, v)=(u^{2}-v^{2}, uv)[\/latex] where [latex]x=u^{2}-v^{2}[\/latex] and [latex]y=uv[\/latex]. Thus the Jacobian is\r\n

[latex]\\large{J(u,v)=\\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}2u&v \\\\ -2v&u\\end{vmatrix}=2u^2+2v^2}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

Jacobians<\/h2>\n

We mentioned earlier that transformation functions must have continuous first partial derivatives\u2014meaning [latex]g_u[\/latex], [latex]g_v[\/latex], [latex]h_u[\/latex], and [latex]h_v[\/latex] exist and are continuous. A transformation with this property is called a [latex]C^1[\/latex] transformation<\/strong> (where [latex]C[\/latex] denotes continuous).<\/p>\n

Let [latex]T(u,v) = (g(u,v), h(u,v))[\/latex] be a one-to-one [latex]C^1[\/latex] transformation, where [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex]. We want to understand how this transformation affects a small rectangular region [latex]S[\/latex] in the [latex]uv[\/latex]-plane, measuring [latex]\\Delta u[\/latex] units by [latex]\\Delta v[\/latex] units (see Figure 4).<\/p>\n

\"On
Figure 4.\u00a0A small rectangle [latex]S[\/latex] in the\u00a0[latex]uv[\/latex]-plane is transformed into a region\u00a0[latex]R[\/latex] in the\u00a0[latex]xy[\/latex]-plane.<\/figcaption><\/figure>\n

<\/h3>\n

Since [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex], we can write the position vector of the image of point [latex](u, v)[\/latex]\u00a0as:<\/p>\n

[latex]\\mathbf{r}(u,v) = g(u,v)\\mathbf{i} + h(u,v)\\mathbf{j}[\/latex]<\/p>\n

Suppose [latex](u_0, v_0)[\/latex] is the lower left corner of rectangle [latex]S[\/latex], which maps to [latex](x_0, y_0) = T(u_0, v_0)[\/latex].<\/p>\n

The line [latex]v = v_0[\/latex] in the [latex]uv[\/latex]-plane maps to an image curve with vector function [latex]\\mathbf{r}(u, v_0)[\/latex]. The tangent vector at [latex](x_0, y_0)[\/latex] to this image curve is:<\/p>\n

[latex]\\mathbf{r}_u = g_u(u_0, v_0)\\mathbf{i} + h_u(u_0, v_0)\\mathbf{j} = \\frac{\\partial x}{\\partial u}\\mathbf{i} + \\frac{\\partial y}{\\partial u}\\mathbf{j}[\/latex]<\/p>\n

Similarly, the line [latex]u = u_0[\/latex] maps to an image curve with vector function [latex]\\mathbf{r}(u_0, v)[\/latex], and its tangent vector at [latex](x_0, y_0)[\/latex] is:<\/p>\n

[latex]\\mathbf{r}_v = g_v(u_0, v_0)\\mathbf{i} + h_v(u_0, v_0)\\mathbf{j} = \\frac{\\partial x}{\\partial v}\\mathbf{i} + \\frac{\\partial y}{\\partial v}\\mathbf{j}[\/latex]<\/p>\n

\n

Tangent Vectors and Derivatives<\/strong><\/p>\n

For a vector function [latex]\\mathbf{r}(u)[\/latex], the derivative [latex]\\mathbf{r}_u[\/latex] gives the tangent vector:<\/p>\n

[latex]{{\\bf{r}}_{u}} = {\\displaystyle\\lim_{{\\Delta}{u}{\\rightarrow}{0}}}{\\dfrac{{\\bf{r}}{({u_0} + {{\\Delta}{u}},{v_0}) - {\\bf{r}}{({u_0},{v_0})}}}{{\\Delta}{u}}}[\/latex]<\/p>\n

This means [latex]\\mathbf{r}(u_0 + \\Delta u) - \\mathbf{r}(u_0) \\approx \\Delta u \\mathbf{r}_u[\/latex] for small [latex]\\Delta u[\/latex].<\/p>\n<\/section>\n

From the definition of derivative, we have:<\/p>\n

[latex]{{\\bf{r}}_{u}} = {\\displaystyle\\lim_{{\\Delta}{u}{\\rightarrow}{0}}}{\\dfrac{{\\bf{r}}{({u_0} + {{\\Delta}{u}},{v_0}) - {\\bf{r}}{({u_0},{v_0})}}}{{\\Delta}{u}}} \\ {\\text{so}} \\ {{\\bf{r}}{({u_0} + {{\\Delta}{u}},{v_0}) - {\\bf{r}}{({u_0},{v_0})}}}{\\approx}{\\Delta}{u}{\\bf{r}}_{u}[\/latex].<\/p>\n

Similarly,<\/p>\n

[latex]{{\\bf{r}}_{v}} = {\\displaystyle\\lim_{{\\Delta}{v}{\\rightarrow}{0}}}{\\dfrac{{\\bf{r}}{({u_0},{v_0} + {{\\Delta}{v}}) - {\\bf{r}}{({u_0},{v_0})}}}{{\\Delta}{v}}} \\ {\\text{so}} \\ {\\bf{r}}{({u_0},{v_0} + {{\\Delta}{v}}) - {\\bf{r}}{({u_0},{v_0})}}{\\approx}{\\Delta}{v}{\\bf{r}}_{v}[\/latex].<\/p>\n

Finding the Area Element<\/h3>\n

The small rectangle [latex]S[\/latex] in the [latex]uv[\/latex]-plane transforms into a curved region [latex]R[\/latex] in the [latex]xy[\/latex]-plane. We can estimate the area [latex]\\Delta A[\/latex] of image [latex]R[\/latex] by finding the area of the parallelogram formed by the sides [latex]\\Delta u \\mathbf{r}_u[\/latex] and [latex]\\Delta v \\mathbf{r}_v[\/latex].<\/p>\n

\n

Area of a Parallelogram Using Cross Product<\/strong><\/p>\n

For vectors [latex]\\mathbf{a}[\/latex] and [latex]\\mathbf{b}[\/latex] in space, the area of the parallelogram they span is [latex]|\\mathbf{a} \\times \\mathbf{b}|[\/latex].<\/p>\n

To compute the cross product of two-dimensional vectors, add a [latex]\\mathbf{k}[\/latex] component of zero to make them three-dimensional.<\/p>\n<\/section>\n

Using the cross product (by adding the [latex]\\mathbf{k}[\/latex] component as [latex]0[\/latex]), the area [latex]\\Delta A[\/latex] of image [latex]R[\/latex] is approximately:<\/p>\n

[latex]|\\Delta u \\mathbf{r}_u \\times \\Delta v \\mathbf{r}_v| = |\\mathbf{r}_u \\times \\mathbf{r}_v| \\Delta u \\Delta v[\/latex]<\/p>\n

In determinant form, the cross product is:<\/p>\n

[latex]\\mathbf{r}_u \\times \\mathbf{r}_v = \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ \\dfrac{\\partial x}{\\partial u} & \\dfrac{\\partial y}{\\partial u} & 0 \\\\ \\dfrac{\\partial x}{\\partial v} & \\dfrac{\\partial y}{\\partial v} & 0 \\end{vmatrix} = \\begin{vmatrix} \\dfrac{\\partial x}{\\partial u} & \\dfrac{\\partial y}{\\partial u} \\\\ \\dfrac{\\partial x}{\\partial v} & \\dfrac{\\partial y}{\\partial v} \\end{vmatrix} \\mathbf{k} = \\left( \\dfrac{\\partial x}{\\partial u} \\dfrac{\\partial y}{\\partial v} - \\dfrac{\\partial x}{\\partial v} \\dfrac{\\partial y}{\\partial u} \\right) \\mathbf{k}[\/latex]<\/p>\n

Since [latex]|\\mathbf{k}| = 1[\/latex], we have:<\/p>\n

[latex]\\Delta A \\approx |\\mathbf{r}_u \\times \\mathbf{r}_v| \\Delta u \\Delta v = \\left| \\dfrac{\\partial x}{\\partial u} \\dfrac{\\partial y}{\\partial v} - \\dfrac{\\partial x}{\\partial v} \\dfrac{\\partial y}{\\partial u} \\right| \\Delta u \\Delta v[\/latex]<\/p>\n

This determinant expression is so important that it gets its own name: the Jacobian.<\/p>\n

The Jacobian Determinant<\/h3>\n

The Jacobian<\/strong> of the [latex]C^1[\/latex] transformation [latex]T(u,v) = (g(u,v), h(u,v))[\/latex] is denoted by [latex]J(u,v)[\/latex] and is defined by the [latex]2 \\times 2[\/latex] determinant:<\/p>\n

[latex]J(u,v) = \\left| \\dfrac{\\partial(x,y)}{\\partial(u,v)} \\right| = \\begin{vmatrix} \\dfrac{\\partial x}{\\partial u} & \\dfrac{\\partial y}{\\partial u} \\\\ \\dfrac{\\partial x}{\\partial v} & \\dfrac{\\partial y}{\\partial v} \\end{vmatrix} = \\left( \\dfrac{\\partial x}{\\partial u} \\dfrac{\\partial y}{\\partial v} - \\dfrac{\\partial x}{\\partial v} \\dfrac{\\partial y}{\\partial u} \\right)[\/latex]<\/p>\n

\n

the Jacobian<\/h3>\n

For transformation [latex]T(u,v) = (x,y)[\/latex] where [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex]:<\/p>\n

[latex]J(u,v) = \\dfrac{\\partial(x,y)}{\\partial(u,v)} = \\begin{vmatrix} \\dfrac{\\partial x}{\\partial u} & \\dfrac{\\partial y}{\\partial u} \\\\ \\dfrac{\\partial x}{\\partial v} & \\dfrac{\\partial y}{\\partial v} \\end{vmatrix}[\/latex]<\/p>\n

The Jacobian tells us how the transformation scales areas: [latex]\\Delta A \\approx |J(u,v)| \\Delta u \\Delta v[\/latex]<\/p>\n<\/section>\n

Using this definition, we can write:<\/p>\n

[latex]\\Delta A \\approx J(u,v) \\Delta u \\Delta v = \\left| \\dfrac{\\partial(x,y)}{\\partial(u,v)} \\right| \\Delta u \\Delta v[\/latex]<\/p>\n

Note that the Jacobian is frequently written simply as:<\/p>\n

[latex]J(u,v) = \\dfrac{\\partial(x,y)}{\\partial(u,v)}[\/latex]<\/p>\n

\n

Two Ways to Arrange the Jacobian<\/strong><\/p>\n

The determinant can be computed in two equivalent ways:<\/p>\n

[latex]\\begin{vmatrix} \\dfrac{\\partial x}{\\partial u} & \\dfrac{\\partial y}{\\partial u} \\\\ \\dfrac{\\partial x}{\\partial v} & \\dfrac{\\partial y}{\\partial v} \\end{vmatrix} = \\left( \\dfrac{\\partial x}{\\partial u} \\dfrac{\\partial y}{\\partial v} - \\dfrac{\\partial x}{\\partial v} \\dfrac{\\partial y}{\\partial u} \\right) = \\begin{vmatrix} \\dfrac{\\partial x}{\\partial u} & \\dfrac{\\partial x}{\\partial v} \\\\ \\dfrac{\\partial y}{\\partial u} & \\dfrac{\\partial y}{\\partial v} \\end{vmatrix}[\/latex]<\/p>\n

The notation [latex]J(u,v) = \\dfrac{\\partial(x,y)}{\\partial(u,v)}[\/latex] suggests we can write the Jacobian with partials of [latex]x[\/latex] in the first row and partials of [latex]y[\/latex] in the second row.<\/p>\n<\/section>\n

Consider the transformation [latex]T(r,\\theta) = (x,y)[\/latex] where [latex]x = r\\cos\\theta[\/latex] and [latex]y = r\\sin\\theta[\/latex]. Find the Jacobian of this transformation.<\/p>\n