divergence of the curl<\/h3>\r\nLet [latex]{\\bf{F}}=\\langle{P},Q,R\\rangle[\/latex] be a vector field in [latex]\\mathbb{R}^3[\/latex] such that the component functions all have continuous second-order partial derivatives. Then,\r\n\r\n[latex]\\text{div curl }({\\bf{F}})=\\nabla\\cdot(\\nabla\\times{\\bf{F}})=0[\/latex].<\/center><\/section>\r\nProof<\/strong><\/p>\r\n\r\n\r\n
\r\nBy the definitions of divergence and curl, and by Clairaut\u2019s theorem,<\/p>\r\n
[latex]\\begin{aligned}\r\n\\text{div curl }({\\bf{F}})&=\\text{div }[(R_y-Q_z){\\bf{i}}+(P_z-R_x){\\bf{j}}+(Q_x-P_y){\\bf{k}}] \\\\&=R_{yx}-Q_{xz}+P_{yz}-R_{yx}+Q_{zx}-P_{zy} \\\\\r\n&=0\r\n\\end{aligned}[\/latex].<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>Show that [latex]{\\bf{F}}(x,y,z)=e^x{\\bf{i}}+yz{\\bf{j}}+xz^2{\\bf{k}}[\/latex] is not the curl of another vector field. That is, show that there is no other vector\u00a0[latex]{\\bf{G}}[\/latex]\u00a0with curl [latex]{\\bf{G}}={\\bf{F}}[\/latex].[reveal-answer q=\"748329916\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"748329916\"]\r\nNotice that the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is all of [latex]\\mathbb{R}^3[\/latex] and the second-order partials of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0are all continuous. Therefore, we can apply the previous theorem to\u00a0[latex]{\\bf{F}}[\/latex].<\/p>\r\n
The divergence of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is [latex]e^x+z+2xz[\/latex]. If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0were the curl of vector field\u00a0[latex]\\bf{G}[\/latex], then [latex]\\text{div }{\\bf{F}}=\\text{div curl }{\\bf{G}}=0[\/latex]. But, the divergence of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not zero, and therefore\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not the curl of any other vector field.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n
With the next two theorems, we show that if [latex]\\mathbf{F}[\/latex] is a conservative vector field then its curl is zero, and if the domain of [latex]\\mathbf{F}[\/latex] is simply connected then the converse is also true. This gives us another way to test whether a vector field is conservative.<\/p>\r\n\r\n\r\ncurl of a conservative vector field<\/h3>\r\nIf [latex]{\\bf{F}}=\\langle{P},Q,R\\rangle[\/latex] is conservative, then [latex] \\text{curl } \\bf{F}=\\bf{0} [\/latex].\u00a0<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\r\n\r\n<\/section>\r\nProof<\/strong><\/p>\r\n\r\n\r\n
\r\nSince conservative vector fields satisfy the cross-partials property, all the cross-partials of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0are equal. Therefore,<\/p>\r\n
[latex]\\begin{aligned}\r\n\\text{curl }{\\bf{F}}&=(R_y-Q_z){\\bf{i}}+(P_z-R_x){\\bf{j}}+(Q_x-P_y){\\bf{k}} \\\\\r\n&=0\r\n\\end{aligned}[\/latex].<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>The same theorem is true for vector fields in a plane.\r\n
Since a conservative vector field is the gradient of a scalar function, the previous theorem says that [latex]\\text{curl }(\\nabla f)=0[\/latex] for any scalar function [latex]f[\/latex]. In terms of our curl notation, [latex]\\nabla\\times\\nabla(f)=0[\/latex]. This equation makes sense because the cross product of a vector with itself is always the zero vector. Sometimes equation [latex]\\nabla\\times\\nabla(f)=0[\/latex] is simplified as [latex]\\nabla\\times\\nabla=0[\/latex].<\/div>\r\n\r\ncurl test for a conservative field<\/h3>\r\nLet [latex]{\\bf{F}}=\\langle{P},Q,R\\rangle[\/latex] be a vector field in space on a simply connected domain. If [latex]\\text{curl }{\\bf{F}}=0[\/latex], then [latex]{\\bf{F}}[\/latex] is conservative.\r\n\r\n<\/section>\r\nProof<\/strong><\/p>\r\n\r\n\r\n
\r\nSince [latex]\\text{curl }{\\bf{F}}=0[\/latex], we have that [latex]R_y=Q_z[\/latex],\u00a0[latex]P_z=R_x[\/latex], and\u00a0[latex]Q_x=P_y[\/latex]. Therefore,\u00a0[latex]{\\bf{F}}[\/latex]\u00a0satisfies the cross-partials property on a simply connected domain, and\u00a0Cross-Partial Property of Conservative Fields Theorem\u00a0implies that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>\r\n
The same theorem is also true in a plane. Therefore, if [latex]\\mathbf{F}[\/latex] is a vector field in a plane or in space and the domain is simply connected, then [latex]\\mathbf{F}[\/latex] is conservative if and only if [latex]\\text{curl }\\mathbf{F}=0[\/latex].<\/p>\r\n\r\nUse the curl to determine whether [latex]{\\bf{F}}(x,y,z)=\\langle{y}z,xz,xy\\rangle[\/latex] is conservative.[reveal-answer q=\"478259405\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"478259405\"]Note that the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is all of [latex]\\mathbb{R}^3[\/latex], which is simply connected (Figure 1). Therefore, we can test whether\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative by calculating its curl.[caption id=\"attachment_1448\" align=\"aligncenter\" width=\"382\"]![\"A](\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/68\/2025\/10\/17152358\/7749bed210536f0229f8267b69c34794d0305459.webp\")
Figure 1. The curl of vector field [latex]{\\bf{F}}(x,y,z)=\\langle{y}z,xz,xy\\rangle[\/latex] is zero.[\/caption]\r\nThe curl of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is<\/p>\r\n
[latex]\\left(\\frac{\\partial}{\\partial{y}}xy-\\frac{\\partial}{\\partial{z}}xz\\right){\\bf{i}}+\\left(\\frac{\\partial}{\\partial{y}}yz-\\frac{\\partial}{\\partial{z}}xy\\right){\\bf{j}}+\\left(\\frac{\\partial}{\\partial{y}}xz-\\frac{\\partial}{\\partial{z}}yz\\right){\\bf{k}}=(x-x){\\bf{i}}+(y-y){\\bf{j}}+(z-z){\\bf{k}}=0[\/latex].<\/p>\r\n
Thus,\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n
The Laplace Operator<\/h3>\r\n
We have seen that the curl of a gradient is zero. What is the divergence of a gradient? If [latex]f[\/latex] is a function of two variables, then [latex]\\text{div }(\\nabla f)=\\nabla\\cdot(\\nabla f)=f_{xx}+f_{yy}[\/latex]. We abbreviate this \"double dot product\" as [latex]\\nabla^2[\/latex]. This operator is called the Laplace operator<\/strong>, and in this notation Laplace's equation becomes [latex]\\nabla^2f=0[\/latex]. Therefore, a harmonic function is a function that becomes zero after taking the divergence of a gradient.<\/p>\r\nSimilarly, if [latex]f[\/latex] is a function of three variables then,<\/p>\r\n
[latex]\\large{\\text{div }(\\nabla{f})=\\nabla\\cdot(\\nabla{f})=f_{xx}+f_{yy}+f_{zz}}[\/latex].<\/p>\r\n
Using this notation we get Laplace\u2019s equation for harmonic functions of three variables:<\/p>\r\n
[latex]\\large{\\nabla^2f=0}[\/latex].<\/p>\r\n
Harmonic functions arise in many applications. For example, the potential function of an electrostatic field in a region of space that has no static charge is harmonic.<\/p>\r\n\r\nIs it possible for [latex]f(x, y)=x^{2}+x-y[\/latex] to be the potential function of an electrostatic field that is located in a region of [latex]\\mathbb{R}^2[\/latex] free of static charge?[reveal-answer q=\"783691260\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"783691260\"]If [latex]f[\/latex] were such a potential function, then [latex]f[\/latex] would be harmonic. Note that [latex]f_{xx}=2[\/latex], and [latex]f_{yy}=0[\/latex], so [latex]f_{xx}+f_{yy}\\ne0[\/latex]. Therefore, [latex]f[\/latex] is not harmonic and [latex]f[\/latex] cannot represent an electrostatic potential.[\/hidden-answer]\r\n\r\n<\/section><\/div>","rendered":"Using Divergence and Curl<\/h2>\n
Now that we understand the basic concepts of divergence and curl, we can discuss their properties and establish relationships between them and conservative vector fields.<\/p>\n
If [latex]\\mathbf{F}[\/latex] is a vector field in [latex]\\mathbb{R}^3[\/latex], then the curl of [latex]\\mathbf{F}[\/latex] is also a vector field in [latex]\\mathbb{R}^3[\/latex]. Therefore, we can take the divergence of a curl. The next theorem says that the result is always zero. This result is useful because it gives us a way to show that some vector fields are not the curl of any other field.<\/p>\n
To give this result a physical interpretation, recall that divergence of a velocity field [latex]\\mathbf{v}[\/latex] at point [latex]P[\/latex] measures the tendency of the corresponding fluid to flow out of [latex]P[\/latex]. Since [latex]\\text{div curl }(\\mathbf{v})=0[\/latex], the net rate of flow in vector field curl([latex]\\mathbf{v}[\/latex]) at any point is zero. Taking the curl of vector field [latex]\\mathbf{F}[\/latex] eliminates whatever divergence was present in [latex]\\mathbf{F}[\/latex].<\/p>\n
\n\ndivergence of the curl<\/h3>\n
Let [latex]{\\bf{F}}=\\langle{P},Q,R\\rangle[\/latex] be a vector field in [latex]\\mathbb{R}^3[\/latex] such that the component functions all have continuous second-order partial derivatives. Then,<\/p>\n
[latex]\\text{div curl }({\\bf{F}})=\\nabla\\cdot(\\nabla\\times{\\bf{F}})=0[\/latex].<\/div>\n<\/section>\n\nProof<\/strong><\/p>\n
\nBy the definitions of divergence and curl, and by Clairaut\u2019s theorem,<\/p>\n
[latex]\\begin{aligned} \\text{div curl }({\\bf{F}})&=\\text{div }[(R_y-Q_z){\\bf{i}}+(P_z-R_x){\\bf{j}}+(Q_x-P_y){\\bf{k}}] \\\\&=R_{yx}-Q_{xz}+P_{yz}-R_{yx}+Q_{zx}-P_{zy} \\\\ &=0 \\end{aligned}[\/latex].<\/p>\n
[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\nShow that [latex]{\\bf{F}}(x,y,z)=e^x{\\bf{i}}+yz{\\bf{j}}+xz^2{\\bf{k}}[\/latex] is not the curl of another vector field. That is, show that there is no other vector\u00a0[latex]{\\bf{G}}[\/latex]\u00a0with curl [latex]{\\bf{G}}={\\bf{F}}[\/latex].<\/p>\n
Proof<\/strong><\/p>\r\n\r\n\r\n By the definitions of divergence and curl, and by Clairaut\u2019s theorem,<\/p>\r\n [latex]\\begin{aligned}\r\n\\text{div curl }({\\bf{F}})&=\\text{div }[(R_y-Q_z){\\bf{i}}+(P_z-R_x){\\bf{j}}+(Q_x-P_y){\\bf{k}}] \\\\&=R_{yx}-Q_{xz}+P_{yz}-R_{yx}+Q_{zx}-P_{zy} \\\\\r\n&=0\r\n\\end{aligned}[\/latex].<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section> Notice that the domain of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is all of [latex]\\mathbb{R}^3[\/latex] and the second-order partials of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0are all continuous. Therefore, we can apply the previous theorem to\u00a0[latex]{\\bf{F}}[\/latex].<\/p>\r\n The divergence of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is [latex]e^x+z+2xz[\/latex]. If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0were the curl of vector field\u00a0[latex]\\bf{G}[\/latex], then [latex]\\text{div }{\\bf{F}}=\\text{div curl }{\\bf{G}}=0[\/latex]. But, the divergence of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not zero, and therefore\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is not the curl of any other vector field.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n With the next two theorems, we show that if [latex]\\mathbf{F}[\/latex] is a conservative vector field then its curl is zero, and if the domain of [latex]\\mathbf{F}[\/latex] is simply connected then the converse is also true. This gives us another way to test whether a vector field is conservative.<\/p>\r\n\r\n Proof<\/strong><\/p>\r\n\r\n\r\n Since conservative vector fields satisfy the cross-partials property, all the cross-partials of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0are equal. Therefore,<\/p>\r\n [latex]\\begin{aligned}\r\n\\text{curl }{\\bf{F}}&=(R_y-Q_z){\\bf{i}}+(P_z-R_x){\\bf{j}}+(Q_x-P_y){\\bf{k}} \\\\\r\n&=0\r\n\\end{aligned}[\/latex].<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>The same theorem is true for vector fields in a plane.\r\n Proof<\/strong><\/p>\r\n\r\n\r\n Since [latex]\\text{curl }{\\bf{F}}=0[\/latex], we have that [latex]R_y=Q_z[\/latex],\u00a0[latex]P_z=R_x[\/latex], and\u00a0[latex]Q_x=P_y[\/latex]. Therefore,\u00a0[latex]{\\bf{F}}[\/latex]\u00a0satisfies the cross-partials property on a simply connected domain, and\u00a0Cross-Partial Property of Conservative Fields Theorem\u00a0implies that\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>\r\n The same theorem is also true in a plane. Therefore, if [latex]\\mathbf{F}[\/latex] is a vector field in a plane or in space and the domain is simply connected, then [latex]\\mathbf{F}[\/latex] is conservative if and only if [latex]\\text{curl }\\mathbf{F}=0[\/latex].<\/p>\r\n\r\n The curl of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is<\/p>\r\n [latex]\\left(\\frac{\\partial}{\\partial{y}}xy-\\frac{\\partial}{\\partial{z}}xz\\right){\\bf{i}}+\\left(\\frac{\\partial}{\\partial{y}}yz-\\frac{\\partial}{\\partial{z}}xy\\right){\\bf{j}}+\\left(\\frac{\\partial}{\\partial{y}}xz-\\frac{\\partial}{\\partial{z}}yz\\right){\\bf{k}}=(x-x){\\bf{i}}+(y-y){\\bf{j}}+(z-z){\\bf{k}}=0[\/latex].<\/p>\r\n Thus,\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is conservative.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n We have seen that the curl of a gradient is zero. What is the divergence of a gradient? If [latex]f[\/latex] is a function of two variables, then [latex]\\text{div }(\\nabla f)=\\nabla\\cdot(\\nabla f)=f_{xx}+f_{yy}[\/latex]. We abbreviate this \"double dot product\" as [latex]\\nabla^2[\/latex]. This operator is called the Laplace operator<\/strong>, and in this notation Laplace's equation becomes [latex]\\nabla^2f=0[\/latex]. Therefore, a harmonic function is a function that becomes zero after taking the divergence of a gradient.<\/p>\r\n Similarly, if [latex]f[\/latex] is a function of three variables then,<\/p>\r\n [latex]\\large{\\text{div }(\\nabla{f})=\\nabla\\cdot(\\nabla{f})=f_{xx}+f_{yy}+f_{zz}}[\/latex].<\/p>\r\n Using this notation we get Laplace\u2019s equation for harmonic functions of three variables:<\/p>\r\n [latex]\\large{\\nabla^2f=0}[\/latex].<\/p>\r\n Harmonic functions arise in many applications. For example, the potential function of an electrostatic field in a region of space that has no static charge is harmonic.<\/p>\r\n\r\n Now that we understand the basic concepts of divergence and curl, we can discuss their properties and establish relationships between them and conservative vector fields.<\/p>\n If [latex]\\mathbf{F}[\/latex] is a vector field in [latex]\\mathbb{R}^3[\/latex], then the curl of [latex]\\mathbf{F}[\/latex] is also a vector field in [latex]\\mathbb{R}^3[\/latex]. Therefore, we can take the divergence of a curl. The next theorem says that the result is always zero. This result is useful because it gives us a way to show that some vector fields are not the curl of any other field.<\/p>\n To give this result a physical interpretation, recall that divergence of a velocity field [latex]\\mathbf{v}[\/latex] at point [latex]P[\/latex] measures the tendency of the corresponding fluid to flow out of [latex]P[\/latex]. Since [latex]\\text{div curl }(\\mathbf{v})=0[\/latex], the net rate of flow in vector field curl([latex]\\mathbf{v}[\/latex]) at any point is zero. Taking the curl of vector field [latex]\\mathbf{F}[\/latex] eliminates whatever divergence was present in [latex]\\mathbf{F}[\/latex].<\/p>\n Let [latex]{\\bf{F}}=\\langle{P},Q,R\\rangle[\/latex] be a vector field in [latex]\\mathbb{R}^3[\/latex] such that the component functions all have continuous second-order partial derivatives. Then,<\/p>\n Proof<\/strong><\/p>\n By the definitions of divergence and curl, and by Clairaut\u2019s theorem,<\/p>\n [latex]\\begin{aligned} \\text{div curl }({\\bf{F}})&=\\text{div }[(R_y-Q_z){\\bf{i}}+(P_z-R_x){\\bf{j}}+(Q_x-P_y){\\bf{k}}] \\\\&=R_{yx}-Q_{xz}+P_{yz}-R_{yx}+Q_{zx}-P_{zy} \\\\ &=0 \\end{aligned}[\/latex].<\/p>\n [latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n
\r\ncurl of a conservative vector field<\/h3>\r\nIf [latex]{\\bf{F}}=\\langle{P},Q,R\\rangle[\/latex] is conservative, then [latex] \\text{curl } \\bf{F}=\\bf{0} [\/latex].\u00a0<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>\r\n\r\n<\/section>
\r\ncurl test for a conservative field<\/h3>\r\nLet [latex]{\\bf{F}}=\\langle{P},Q,R\\rangle[\/latex] be a vector field in space on a simply connected domain. If [latex]\\text{curl }{\\bf{F}}=0[\/latex], then [latex]{\\bf{F}}[\/latex] is conservative.\r\n\r\n<\/section>
\r\n Figure 1. The curl of vector field [latex]{\\bf{F}}(x,y,z)=\\langle{y}z,xz,xy\\rangle[\/latex] is zero.[\/caption]\r\nThe Laplace Operator<\/h3>\r\n
Using Divergence and Curl<\/h2>\n
divergence of the curl<\/h3>\n
\n