{"id":722,"date":"2025-10-09T15:34:33","date_gmt":"2025-10-09T15:34:33","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=722"},"modified":"2025-11-03T17:02:34","modified_gmt":"2025-11-03T17:02:34","slug":"change-of-variables-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/change-of-variables-learn-it-1\/","title":{"raw":"Change of Variables: Learn It 1","rendered":"Change of Variables: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Find how a region changes when you use a different set of variables<\/li>\r\n \t<li>Calculate the Jacobian for a transformation of variables<\/li>\r\n \t<li>Apply change of variables in double integrals<\/li>\r\n \t<li>Apply change of variables in triple integrals<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Change of Variables in Multiple Integrals<\/h2>\r\n<p class=\"whitespace-normal break-words\">When evaluating integrals, making an appropriate substitution can transform a difficult integral into a simpler one. You've already seen this technique in single-variable calculus with [latex]u[\/latex]-substitution.<\/p>\r\n\r\n<section class=\"textbox recall\" aria-label=\"Recall\">\r\n<p class=\"whitespace-normal break-words\"><strong>Substitution Rule for Single Integrals<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For [latex]\\int_a^b f(x) dx[\/latex], we substitute [latex]u = g(x)[\/latex] to get:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\int_a^b f(x) dx = \\int_c^d f(g(u)) g'(u) du[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">where [latex]x = g(u)[\/latex], [latex]dx = g'(u) du[\/latex], and the new limits satisfy [latex]c = g(a)[\/latex] and [latex]d = g(b)[\/latex].<\/p>\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">Consider the integral [latex]\\int_2^3 x(x^2 - 4)^5 dx[\/latex]. We can substitute [latex]u = g(x) = x^2 - 4[\/latex]. Then [latex]du = 2x , dx[\/latex], so [latex]x , dx = \\frac{1}{2} du[\/latex]. The limits change to [latex]u = g(2) = 2^2 - 4 = 0[\/latex] and [latex]u = g(3) = 9 - 4 = 5[\/latex]. The integral becomes [latex]\\int_0^5 \\frac{1}{2} u^5 du[\/latex], which is much simpler to evaluate.<\/p>\r\n<p class=\"whitespace-normal break-words\">You've also used this idea when transforming double integrals from rectangular to polar coordinates, or triple integrals from rectangular to cylindrical or spherical coordinates. In each case, the goal is to make computations simpler through an appropriate change of variables.<\/p>\r\n<p class=\"whitespace-normal break-words\">For double integrals, when we substitute [latex]x = h(r, \\theta) = r\\cos\\theta[\/latex], [latex]y = g(r, \\theta) = r\\sin\\theta[\/latex], and [latex]dA = dx , dy = r , dr , d\\theta[\/latex], we get:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\iint_R f(x,y) dA = \\iint_S f(r\\cos\\theta, r\\sin\\theta) r , dr , d\\theta[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">where region [latex]R[\/latex] in the [latex]xy[\/latex]-plane is replaced by region [latex]S[\/latex] in polar coordinates.<\/p>\r\n<p class=\"whitespace-normal break-words\">The function we use to change variables is called a <strong>transformation<\/strong> or <strong>mapping<\/strong>. This section develops the general theory for changing variables in multiple integrals.<\/p>\r\n\r\n<h3 data-type=\"title\">Planar Transformations<\/h3>\r\n<p class=\"whitespace-normal break-words\">A <strong>planar transformation<\/strong> [latex]T[\/latex] transforms a region [latex]G[\/latex] in one plane into a region [latex]R[\/latex] in another plane through a change of variables. Both [latex]G[\/latex] and [latex]R[\/latex] are subsets of [latex]\\mathbb{R}^2[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Figure 1 shows region [latex]G[\/latex] in the [latex]uv[\/latex]-plane transformed into region [latex]R[\/latex] in the [latex]xy[\/latex]-plane by the change of variables [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex]. We can also write this as [latex]x = x(u,v)[\/latex] and [latex]y = y(u,v)[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_1418\" align=\"aligncenter\" width=\"715\"]<img class=\"size-full wp-image-1418\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28042718\/5-7-1.jpeg\" alt=\"On the left-hand side of this figure, there is a region G with point (u, v) given in the Cartesian u v-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with x = g(u, v) and y = h(u, v). On the right-hand side of this figure there is a region R with point (x, y) given in the Cartesian xy- plane.\" width=\"715\" height=\"273\" \/> Figure 1.\u00a0The transformation of a region\u00a0[latex]G[\/latex] in the\u00a0[latex]uv[\/latex]-plane into a region\u00a0[latex]R[\/latex] in the\u00a0[latex]xy[\/latex]-plane.[\/caption]\r\n<p class=\"whitespace-normal break-words\">We typically assume that each transformation function has continuous first partial derivatives. This means [latex]g_u[\/latex], [latex]g_v[\/latex], [latex]h_u[\/latex], and [latex]h_v[\/latex] all exist and are continuous. The reason for this requirement will become clear as we develop the theory.<\/p>\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\r\n<p class=\"whitespace-normal break-words\"><strong>Notation for Transformations<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">We write [latex]T: G \\to R[\/latex] to indicate that transformation [latex]T[\/latex] maps region [latex]G[\/latex] to region [latex]R[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">The notation [latex]T(u,v) = (x,y)[\/latex] means that point [latex](u,v)[\/latex] in [latex]G[\/latex] maps to point [latex](x,y)[\/latex] in [latex]R[\/latex].<\/p>\r\n\r\n<\/section>\r\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">One-to-One Transformations<\/h3>\r\n<p class=\"whitespace-normal break-words\">A transformation [latex]T: G \\to R[\/latex] defined as [latex]T(u,v) = (x,y)[\/latex] is <strong>one-to-one<\/strong> if no two different points in [latex]G[\/latex] map to the same point in [latex]R[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">To prove that [latex]T[\/latex] is one-to-one, we assume [latex]T(u_1, v_1) = T(u_2, v_2)[\/latex] and show that this implies [latex](u_1, v_1) = (u_2, v_2)[\/latex].<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>one-to-one transformation<\/h3>\r\n<p class=\"whitespace-normal break-words\">A transformation [latex]T: G \\to R[\/latex] is <strong>one-to-one<\/strong> if:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]T(u_1, v_1) = T(u_2, v_2) \\implies (u_1, v_1) = (u_2, v_2)[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">In other words, different input points always produce different output points.<\/p>\r\n\r\n<\/section>When transformation [latex]T[\/latex] is one-to-one on domain [latex]G[\/latex], the inverse transformation [latex]T^{-1}[\/latex] exists with domain [latex]R[\/latex] such that [latex]T^{-1} \\circ T[\/latex] and [latex]T \\circ T^{-1}[\/latex] are identity functions.\r\n<h3 id=\"fs-id1167794227719\">Domain, Range, and Image<\/h3>\r\n<p class=\"whitespace-normal break-words\">In Figure 1, the mapping [latex]T(u,v) = (x,y)[\/latex] relates [latex]x[\/latex] and [latex]y[\/latex] to [latex]u[\/latex] and [latex]v[\/latex] through the equations [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Region [latex]G[\/latex] is the <strong>domain<\/strong> of [latex]T[\/latex], while region [latex]R[\/latex] is the <strong>range<\/strong> of [latex]T[\/latex], also called the <strong>image<\/strong> of [latex]G[\/latex] under transformation [latex]T[\/latex].<\/p>\r\n\r\n<section class=\"textbox connectIt\" aria-label=\"Connect It\">\r\n<p class=\"whitespace-normal break-words\"><strong>Forward vs. Inverse Problems<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">In practice, we encounter two types of problems:<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Forward problem:<\/strong> Given region [latex]G[\/latex] and transformation [latex]T[\/latex], find the image region [latex]R[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Inverse problem:<\/strong> Given region [latex]R[\/latex] and transformation [latex]T[\/latex], find the pre-image region [latex]G[\/latex].<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">Both perspectives are useful when setting up integrals with change of variables.<\/p>\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Suppose a transformation [latex]T[\/latex] is defined as [latex]{T}{({r},{\\theta})} = {({x},{y})}[\/latex] where [latex]{x} = {r}{\\cos}{\\theta}, {y} = {r}{\\sin}{\\theta}[\/latex]. Find the image of the polar rectangle [latex]{G} = {\\left \\{{({r},{\\theta})}{\\mid}{0} \\leq {r} \\leq {{1},{0}} \\leq {\\theta} \\leq {{\\pi}\/{2}} \\right \\}}[\/latex] in the [latex]{r}{\\theta}[\/latex]-plane to a region [latex]R[\/latex] in the [latex]xy[\/latex]-plane. Show that [latex]T[\/latex] is a one-to-one transformation in [latex]G[\/latex] and find [latex]{{T}^{-1}}{({x},{y})}[\/latex].[reveal-answer q=\"375348660\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"375348660\"]Since [latex]r[\/latex] varies from 0 to 1 in the [latex]{r}{\\theta}[\/latex]-plane, we have a circular disc of radius 0 to 1 in the [latex]xy[\/latex]-plane. Because [latex]{\\theta}[\/latex] varies from 0 to [latex]{\\pi}\/{2}[\/latex] in the [latex]{r}{\\theta}[\/latex]-plane, we end up getting a quarter circle of radius 1 in the first quadrant of the [latex]xy[\/latex]-plane (Figure 2). Hence [latex]R[\/latex] is a quarter circle bounded by [latex]x^{2}+y^{2}=1[\/latex] in the first quadrant.\r\n\r\n[caption id=\"attachment_1419\" align=\"aligncenter\" width=\"715\"]<img class=\"size-full wp-image-1419\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28042750\/5-7-2.jpeg\" alt=\"On the left-hand side of this figure, there is a rectangle G with a marked subrectangle given in the first quadrant of the Cartesian r theta-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with x = r cos theta and y = r sin theta. On the right-hand side of this figure there is a quarter circle R with a marked subannulus (analogous to the rectangle in the other graph) given in the Cartesian x y-plane.\" width=\"715\" height=\"276\" \/> Figure 2.\u00a0A rectangle in the\u00a0[latex]r\\theta[\/latex]-plane is mapped into a quarter circle in the\u00a0[latex]xy[\/latex]-plane.[\/caption]\r\n<p id=\"fs-id1167793240536\">In order to show that [latex]T[\/latex] is a one-to-one transformation, assume [latex]{T}{({r_1},{{\\theta}_1})} = {T}{({r_2},{{\\theta}_2})}[\/latex] and show as a consequence that [latex]{({r_1},{{\\theta}_1})} = {({r_2},{{\\theta}_2})}[\/latex]. In this case, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{T}{({r_1},{{\\theta}_1})} &amp; = {T}{({r_2},{{\\theta}_2})} \\\\\r\n{({x_1},{y_1})} &amp; = {({x_2},{y_2})} \\\\\r\n{({r_1}{\\cos}{{\\theta}_{1}},{r_1}{\\sin}{{\\theta}_{1}})} &amp; = {({r_2}{\\cos}{{\\theta}_{2}},{r_2}{\\sin}{{\\theta}_{2}})} \\\\\r\n{r_1}{\\cos}{{\\theta}_{1}} &amp; = {r_2}{\\cos}{{\\theta}_{2}} \\\\\r\n{r_1}{\\sin}{{\\theta}_{1}} &amp; = {r_2}{\\sin}{{\\theta}_{2}}\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167793456224\">Dividing, we obtain<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\frac{{r_1}{\\cos}{{\\theta}_{1}}}{{r_1}{\\sin}{{\\theta}_{1}}}} &amp; = {{\\frac{{r_2}{\\cos}{{\\theta}_{2}}}{{r_2}{\\sin}{{\\theta}_{2}}}}} \\\\\r\n{\\frac{{\\cos}{{\\theta}_{1}}}{{\\sin}{{\\theta}_{1}}}} &amp; = {\\frac{{\\cos}{{\\theta}_{2}}}{{\\sin}{{\\theta}_{2}}}} \\\\\r\n{\\tan}{{\\theta}_{1}} &amp; = {\\tan}{{\\theta}_{2}} \\\\\r\n{{\\theta}_{1}} &amp; = {{\\theta}_{2}}\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167793953463\">since the tangent function is one-one function in the interval [latex]{0} \\leq {\\theta} \\leq {{\\pi}\/{2}}[\/latex]. Also, since [latex]{0} &lt; {r} \\leq {1}[\/latex], we have [latex]{{r}_{1}} = {{r}_{2}}, {{\\theta}_{1}} = {{\\theta}_{2}}[\/latex]. Therefore, [latex]{({{r}_{1}}, {{\\theta}_{1}})} = {({{r}_{2}}, {{\\theta}_{2}})}[\/latex] and [latex]T[\/latex] is a one-to-one transformation from [latex]G[\/latex] into\u00a0[latex]R[\/latex].<\/p>\r\n<p id=\"fs-id1167793936005\">To find [latex]{{T}^{-1}}{({x},{y})}[\/latex] solve for [latex]{r},{\\theta}[\/latex] in terms of [latex]x, y[\/latex]. We already know that [latex]{{r}^{2}} = {{x}^{2}} + {{y}^{2}}[\/latex] and [latex]{\\tan}{\\theta} = {\\frac{y}{x}}[\/latex]. Thus [latex]{{T}^{-1}}{({x},{y})} = {({r},{\\theta})}[\/latex] is defined as [latex]{r} = {\\sqrt{{x^{2}}+{y^{2}}}}[\/latex] and\u00a0[latex]{\\theta} = {{\\tan}^{-1}}{({\\frac{y}{x}})}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Let the transformation [latex]T[\/latex] be defined by [latex]T(u, v)=(x, y)[\/latex] where [latex]x=u^{2}+v^{2}[\/latex] and [latex]y=uv[\/latex]. Find the image of the triangle in the [latex]uv[\/latex]-plane with vertices [latex](0, 0)[\/latex],\u00a0[latex](0, 1)[\/latex], and\u00a0[latex](1, 1)[\/latex].[reveal-answer q=\"274529843\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"274529843\"]The triangle and its image are shown in\u00a0Figure 3. To understand how the sides of the triangle transform, call the side that joins [latex](0, 0)[\/latex] and [latex](0, 1)[\/latex] side [latex]A[\/latex], the side that joins [latex](0, 0)[\/latex] and [latex](1, 1)[\/latex] side [latex]B[\/latex], and the side that joins [latex](1, 1)[\/latex] and\u00a0[latex](0, 1)[\/latex] side\u00a0[latex]C[\/latex].\r\n\r\n[caption id=\"attachment_1420\" align=\"aligncenter\" width=\"899\"]<img class=\"size-full wp-image-1420\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28042830\/5-7-3.jpeg\" alt=\"On the left-hand side of this figure, there is a triangular region given in the Cartesian uv-plane with boundaries A, B, and C represented by the v axis, the line u = v, and the line v = 1, respectively. Then there is an arrow from this graph to the right-hand side of the figure marked with x = u squared minus v squared and y = u v. On the right-hand side of this figure there is a complex region given in the Cartesian x y-plane with boundaries A\u2019, B\u2019, and C\u2019 given by the x axis, y axis, and a line curving from (negative 1, 0) through (0, 1), namely x = y squared minus 1, respectively.\" width=\"899\" height=\"348\" \/> Figure 3.\u00a0A triangular region in the\u00a0[latex]uv[\/latex]-plane is transformed into an image in the\u00a0[latex]xy[\/latex]-plane.[\/caption]\r\n<p id=\"fs-id1167793822404\">For the side [latex]A[\/latex]: [latex]u=0[\/latex],\u00a0[latex]{0} \\leq {v} \\leq {1}[\/latex] transforms to [latex]x=-v^{2}[\/latex], [latex]y=0[\/latex] so this is the side [latex]{A}^{\\prime}[\/latex] that joins [latex](-1, 0)[\/latex] and\u00a0[latex](0, 0)[\/latex].<\/p>\r\n<p id=\"fs-id1167793951154\">For the side [latex]B[\/latex]: [latex]u=v[\/latex], [latex]{0} \\leq {u} \\leq {1}[\/latex] transforms to [latex]x=0[\/latex], [latex]y=u^{2}[\/latex] so this is the side [latex]{B}^{\\prime}[\/latex] that joins [latex](0, 0)[\/latex] and\u00a0[latex](0, 1)[\/latex].<\/p>\r\n<p id=\"fs-id1167793929050\">For the side [latex]C[\/latex]: [latex]{0} \\leq {u} \\leq {1}[\/latex], [latex]v=1[\/latex] transforms to [latex]x=u^{2}-1[\/latex], [latex]y=u[\/latex] (hence [latex]x=y^{2}-1[\/latex]) so this is the side [latex]{C}{\\prime}[\/latex] that makes the upper half of the parabolic arc joining [latex](-1, 0)[\/latex] and\u00a0[latex](0, 1)[\/latex].<\/p>\r\n<p id=\"fs-id1167793844608\">All the points in the entire region of the triangle in the [latex]uv[\/latex]-plane are mapped inside the parabolic region in the [latex]xy[\/latex]-plane.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Find how a region changes when you use a different set of variables<\/li>\n<li>Calculate the Jacobian for a transformation of variables<\/li>\n<li>Apply change of variables in double integrals<\/li>\n<li>Apply change of variables in triple integrals<\/li>\n<\/ul>\n<\/section>\n<h2 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Change of Variables in Multiple Integrals<\/h2>\n<p class=\"whitespace-normal break-words\">When evaluating integrals, making an appropriate substitution can transform a difficult integral into a simpler one. You&#8217;ve already seen this technique in single-variable calculus with [latex]u[\/latex]-substitution.<\/p>\n<section class=\"textbox recall\" aria-label=\"Recall\">\n<p class=\"whitespace-normal break-words\"><strong>Substitution Rule for Single Integrals<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For [latex]\\int_a^b f(x) dx[\/latex], we substitute [latex]u = g(x)[\/latex] to get:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\int_a^b f(x) dx = \\int_c^d f(g(u)) g'(u) du[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">where [latex]x = g(u)[\/latex], [latex]dx = g'(u) du[\/latex], and the new limits satisfy [latex]c = g(a)[\/latex] and [latex]d = g(b)[\/latex].<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">Consider the integral [latex]\\int_2^3 x(x^2 - 4)^5 dx[\/latex]. We can substitute [latex]u = g(x) = x^2 - 4[\/latex]. Then [latex]du = 2x , dx[\/latex], so [latex]x , dx = \\frac{1}{2} du[\/latex]. The limits change to [latex]u = g(2) = 2^2 - 4 = 0[\/latex] and [latex]u = g(3) = 9 - 4 = 5[\/latex]. The integral becomes [latex]\\int_0^5 \\frac{1}{2} u^5 du[\/latex], which is much simpler to evaluate.<\/p>\n<p class=\"whitespace-normal break-words\">You&#8217;ve also used this idea when transforming double integrals from rectangular to polar coordinates, or triple integrals from rectangular to cylindrical or spherical coordinates. In each case, the goal is to make computations simpler through an appropriate change of variables.<\/p>\n<p class=\"whitespace-normal break-words\">For double integrals, when we substitute [latex]x = h(r, \\theta) = r\\cos\\theta[\/latex], [latex]y = g(r, \\theta) = r\\sin\\theta[\/latex], and [latex]dA = dx , dy = r , dr , d\\theta[\/latex], we get:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\iint_R f(x,y) dA = \\iint_S f(r\\cos\\theta, r\\sin\\theta) r , dr , d\\theta[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">where region [latex]R[\/latex] in the [latex]xy[\/latex]-plane is replaced by region [latex]S[\/latex] in polar coordinates.<\/p>\n<p class=\"whitespace-normal break-words\">The function we use to change variables is called a <strong>transformation<\/strong> or <strong>mapping<\/strong>. This section develops the general theory for changing variables in multiple integrals.<\/p>\n<h3 data-type=\"title\">Planar Transformations<\/h3>\n<p class=\"whitespace-normal break-words\">A <strong>planar transformation<\/strong> [latex]T[\/latex] transforms a region [latex]G[\/latex] in one plane into a region [latex]R[\/latex] in another plane through a change of variables. Both [latex]G[\/latex] and [latex]R[\/latex] are subsets of [latex]\\mathbb{R}^2[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Figure 1 shows region [latex]G[\/latex] in the [latex]uv[\/latex]-plane transformed into region [latex]R[\/latex] in the [latex]xy[\/latex]-plane by the change of variables [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex]. We can also write this as [latex]x = x(u,v)[\/latex] and [latex]y = y(u,v)[\/latex].<\/p>\n<figure id=\"attachment_1418\" aria-describedby=\"caption-attachment-1418\" style=\"width: 715px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1418\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28042718\/5-7-1.jpeg\" alt=\"On the left-hand side of this figure, there is a region G with point (u, v) given in the Cartesian u v-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with x = g(u, v) and y = h(u, v). On the right-hand side of this figure there is a region R with point (x, y) given in the Cartesian xy- plane.\" width=\"715\" height=\"273\" \/><figcaption id=\"caption-attachment-1418\" class=\"wp-caption-text\">Figure 1.\u00a0The transformation of a region\u00a0[latex]G[\/latex] in the\u00a0[latex]uv[\/latex]-plane into a region\u00a0[latex]R[\/latex] in the\u00a0[latex]xy[\/latex]-plane.<\/figcaption><\/figure>\n<p class=\"whitespace-normal break-words\">We typically assume that each transformation function has continuous first partial derivatives. This means [latex]g_u[\/latex], [latex]g_v[\/latex], [latex]h_u[\/latex], and [latex]h_v[\/latex] all exist and are continuous. The reason for this requirement will become clear as we develop the theory.<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\n<p class=\"whitespace-normal break-words\"><strong>Notation for Transformations<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">We write [latex]T: G \\to R[\/latex] to indicate that transformation [latex]T[\/latex] maps region [latex]G[\/latex] to region [latex]R[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">The notation [latex]T(u,v) = (x,y)[\/latex] means that point [latex](u,v)[\/latex] in [latex]G[\/latex] maps to point [latex](x,y)[\/latex] in [latex]R[\/latex].<\/p>\n<\/section>\n<h3 class=\"text-lg font-bold text-text-100 mt-1 -mb-1.5\">One-to-One Transformations<\/h3>\n<p class=\"whitespace-normal break-words\">A transformation [latex]T: G \\to R[\/latex] defined as [latex]T(u,v) = (x,y)[\/latex] is <strong>one-to-one<\/strong> if no two different points in [latex]G[\/latex] map to the same point in [latex]R[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">To prove that [latex]T[\/latex] is one-to-one, we assume [latex]T(u_1, v_1) = T(u_2, v_2)[\/latex] and show that this implies [latex](u_1, v_1) = (u_2, v_2)[\/latex].<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>one-to-one transformation<\/h3>\n<p class=\"whitespace-normal break-words\">A transformation [latex]T: G \\to R[\/latex] is <strong>one-to-one<\/strong> if:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]T(u_1, v_1) = T(u_2, v_2) \\implies (u_1, v_1) = (u_2, v_2)[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">In other words, different input points always produce different output points.<\/p>\n<\/section>\n<p>When transformation [latex]T[\/latex] is one-to-one on domain [latex]G[\/latex], the inverse transformation [latex]T^{-1}[\/latex] exists with domain [latex]R[\/latex] such that [latex]T^{-1} \\circ T[\/latex] and [latex]T \\circ T^{-1}[\/latex] are identity functions.<\/p>\n<h3 id=\"fs-id1167794227719\">Domain, Range, and Image<\/h3>\n<p class=\"whitespace-normal break-words\">In Figure 1, the mapping [latex]T(u,v) = (x,y)[\/latex] relates [latex]x[\/latex] and [latex]y[\/latex] to [latex]u[\/latex] and [latex]v[\/latex] through the equations [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Region [latex]G[\/latex] is the <strong>domain<\/strong> of [latex]T[\/latex], while region [latex]R[\/latex] is the <strong>range<\/strong> of [latex]T[\/latex], also called the <strong>image<\/strong> of [latex]G[\/latex] under transformation [latex]T[\/latex].<\/p>\n<section class=\"textbox connectIt\" aria-label=\"Connect It\">\n<p class=\"whitespace-normal break-words\"><strong>Forward vs. Inverse Problems<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">In practice, we encounter two types of problems:<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Forward problem:<\/strong> Given region [latex]G[\/latex] and transformation [latex]T[\/latex], find the image region [latex]R[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Inverse problem:<\/strong> Given region [latex]R[\/latex] and transformation [latex]T[\/latex], find the pre-image region [latex]G[\/latex].<\/p>\n<p class=\"whitespace-pre-wrap break-words\">Both perspectives are useful when setting up integrals with change of variables.<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Suppose a transformation [latex]T[\/latex] is defined as [latex]{T}{({r},{\\theta})} = {({x},{y})}[\/latex] where [latex]{x} = {r}{\\cos}{\\theta}, {y} = {r}{\\sin}{\\theta}[\/latex]. Find the image of the polar rectangle [latex]{G} = {\\left \\{{({r},{\\theta})}{\\mid}{0} \\leq {r} \\leq {{1},{0}} \\leq {\\theta} \\leq {{\\pi}\/{2}} \\right \\}}[\/latex] in the [latex]{r}{\\theta}[\/latex]-plane to a region [latex]R[\/latex] in the [latex]xy[\/latex]-plane. Show that [latex]T[\/latex] is a one-to-one transformation in [latex]G[\/latex] and find [latex]{{T}^{-1}}{({x},{y})}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q375348660\">Show Solution<\/button><\/p>\n<div id=\"q375348660\" class=\"hidden-answer\" style=\"display: none\">Since [latex]r[\/latex] varies from 0 to 1 in the [latex]{r}{\\theta}[\/latex]-plane, we have a circular disc of radius 0 to 1 in the [latex]xy[\/latex]-plane. Because [latex]{\\theta}[\/latex] varies from 0 to [latex]{\\pi}\/{2}[\/latex] in the [latex]{r}{\\theta}[\/latex]-plane, we end up getting a quarter circle of radius 1 in the first quadrant of the [latex]xy[\/latex]-plane (Figure 2). Hence [latex]R[\/latex] is a quarter circle bounded by [latex]x^{2}+y^{2}=1[\/latex] in the first quadrant.<\/p>\n<figure id=\"attachment_1419\" aria-describedby=\"caption-attachment-1419\" style=\"width: 715px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1419\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28042750\/5-7-2.jpeg\" alt=\"On the left-hand side of this figure, there is a rectangle G with a marked subrectangle given in the first quadrant of the Cartesian r theta-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with x = r cos theta and y = r sin theta. On the right-hand side of this figure there is a quarter circle R with a marked subannulus (analogous to the rectangle in the other graph) given in the Cartesian x y-plane.\" width=\"715\" height=\"276\" \/><figcaption id=\"caption-attachment-1419\" class=\"wp-caption-text\">Figure 2.\u00a0A rectangle in the\u00a0[latex]r\\theta[\/latex]-plane is mapped into a quarter circle in the\u00a0[latex]xy[\/latex]-plane.<\/figcaption><\/figure>\n<p id=\"fs-id1167793240536\">In order to show that [latex]T[\/latex] is a one-to-one transformation, assume [latex]{T}{({r_1},{{\\theta}_1})} = {T}{({r_2},{{\\theta}_2})}[\/latex] and show as a consequence that [latex]{({r_1},{{\\theta}_1})} = {({r_2},{{\\theta}_2})}[\/latex]. In this case, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {T}{({r_1},{{\\theta}_1})} & = {T}{({r_2},{{\\theta}_2})} \\\\  {({x_1},{y_1})} & = {({x_2},{y_2})} \\\\  {({r_1}{\\cos}{{\\theta}_{1}},{r_1}{\\sin}{{\\theta}_{1}})} & = {({r_2}{\\cos}{{\\theta}_{2}},{r_2}{\\sin}{{\\theta}_{2}})} \\\\  {r_1}{\\cos}{{\\theta}_{1}} & = {r_2}{\\cos}{{\\theta}_{2}} \\\\  {r_1}{\\sin}{{\\theta}_{1}} & = {r_2}{\\sin}{{\\theta}_{2}}  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167793456224\">Dividing, we obtain<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\frac{{r_1}{\\cos}{{\\theta}_{1}}}{{r_1}{\\sin}{{\\theta}_{1}}}} & = {{\\frac{{r_2}{\\cos}{{\\theta}_{2}}}{{r_2}{\\sin}{{\\theta}_{2}}}}} \\\\  {\\frac{{\\cos}{{\\theta}_{1}}}{{\\sin}{{\\theta}_{1}}}} & = {\\frac{{\\cos}{{\\theta}_{2}}}{{\\sin}{{\\theta}_{2}}}} \\\\  {\\tan}{{\\theta}_{1}} & = {\\tan}{{\\theta}_{2}} \\\\  {{\\theta}_{1}} & = {{\\theta}_{2}}  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167793953463\">since the tangent function is one-one function in the interval [latex]{0} \\leq {\\theta} \\leq {{\\pi}\/{2}}[\/latex]. Also, since [latex]{0} < {r} \\leq {1}[\/latex], we have [latex]{{r}_{1}} = {{r}_{2}}, {{\\theta}_{1}} = {{\\theta}_{2}}[\/latex]. Therefore, [latex]{({{r}_{1}}, {{\\theta}_{1}})} = {({{r}_{2}}, {{\\theta}_{2}})}[\/latex] and [latex]T[\/latex] is a one-to-one transformation from [latex]G[\/latex] into\u00a0[latex]R[\/latex].<\/p>\n<p id=\"fs-id1167793936005\">To find [latex]{{T}^{-1}}{({x},{y})}[\/latex] solve for [latex]{r},{\\theta}[\/latex] in terms of [latex]x, y[\/latex]. We already know that [latex]{{r}^{2}} = {{x}^{2}} + {{y}^{2}}[\/latex] and [latex]{\\tan}{\\theta} = {\\frac{y}{x}}[\/latex]. Thus [latex]{{T}^{-1}}{({x},{y})} = {({r},{\\theta})}[\/latex] is defined as [latex]{r} = {\\sqrt{{x^{2}}+{y^{2}}}}[\/latex] and\u00a0[latex]{\\theta} = {{\\tan}^{-1}}{({\\frac{y}{x}})}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Let the transformation [latex]T[\/latex] be defined by [latex]T(u, v)=(x, y)[\/latex] where [latex]x=u^{2}+v^{2}[\/latex] and [latex]y=uv[\/latex]. Find the image of the triangle in the [latex]uv[\/latex]-plane with vertices [latex](0, 0)[\/latex],\u00a0[latex](0, 1)[\/latex], and\u00a0[latex](1, 1)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q274529843\">Show Solution<\/button><\/p>\n<div id=\"q274529843\" class=\"hidden-answer\" style=\"display: none\">The triangle and its image are shown in\u00a0Figure 3. To understand how the sides of the triangle transform, call the side that joins [latex](0, 0)[\/latex] and [latex](0, 1)[\/latex] side [latex]A[\/latex], the side that joins [latex](0, 0)[\/latex] and [latex](1, 1)[\/latex] side [latex]B[\/latex], and the side that joins [latex](1, 1)[\/latex] and\u00a0[latex](0, 1)[\/latex] side\u00a0[latex]C[\/latex].<\/p>\n<figure id=\"attachment_1420\" aria-describedby=\"caption-attachment-1420\" style=\"width: 899px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1420\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28042830\/5-7-3.jpeg\" alt=\"On the left-hand side of this figure, there is a triangular region given in the Cartesian uv-plane with boundaries A, B, and C represented by the v axis, the line u = v, and the line v = 1, respectively. Then there is an arrow from this graph to the right-hand side of the figure marked with x = u squared minus v squared and y = u v. On the right-hand side of this figure there is a complex region given in the Cartesian x y-plane with boundaries A\u2019, B\u2019, and C\u2019 given by the x axis, y axis, and a line curving from (negative 1, 0) through (0, 1), namely x = y squared minus 1, respectively.\" width=\"899\" height=\"348\" \/><figcaption id=\"caption-attachment-1420\" class=\"wp-caption-text\">Figure 3.\u00a0A triangular region in the\u00a0[latex]uv[\/latex]-plane is transformed into an image in the\u00a0[latex]xy[\/latex]-plane.<\/figcaption><\/figure>\n<p id=\"fs-id1167793822404\">For the side [latex]A[\/latex]: [latex]u=0[\/latex],\u00a0[latex]{0} \\leq {v} \\leq {1}[\/latex] transforms to [latex]x=-v^{2}[\/latex], [latex]y=0[\/latex] so this is the side [latex]{A}^{\\prime}[\/latex] that joins [latex](-1, 0)[\/latex] and\u00a0[latex](0, 0)[\/latex].<\/p>\n<p id=\"fs-id1167793951154\">For the side [latex]B[\/latex]: [latex]u=v[\/latex], [latex]{0} \\leq {u} \\leq {1}[\/latex] transforms to [latex]x=0[\/latex], [latex]y=u^{2}[\/latex] so this is the side [latex]{B}^{\\prime}[\/latex] that joins [latex](0, 0)[\/latex] and\u00a0[latex](0, 1)[\/latex].<\/p>\n<p id=\"fs-id1167793929050\">For the side [latex]C[\/latex]: [latex]{0} \\leq {u} \\leq {1}[\/latex], [latex]v=1[\/latex] transforms to [latex]x=u^{2}-1[\/latex], [latex]y=u[\/latex] (hence [latex]x=y^{2}-1[\/latex]) so this is the side [latex]{C}{\\prime}[\/latex] that makes the upper half of the parabolic arc joining [latex](-1, 0)[\/latex] and\u00a0[latex](0, 1)[\/latex].<\/p>\n<p id=\"fs-id1167793844608\">All the points in the entire region of the triangle in the [latex]uv[\/latex]-plane are mapped inside the parabolic region in the [latex]xy[\/latex]-plane.<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":22,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-722","chapter","type-chapter","status-publish","hentry"],"part":434,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/722","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/722\/revisions"}],"predecessor-version":[{"id":1958,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/722\/revisions\/1958"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/434"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/722\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=722"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=722"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=722"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=722"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}