{"id":1125,"date":"2021-11-09T23:42:35","date_gmt":"2021-11-09T23:42:35","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&p=1125"},"modified":"2022-11-01T05:27:08","modified_gmt":"2022-11-01T05:27:08","slug":"using-divergence-and-curl","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/using-divergence-and-curl\/","title":{"raw":"Using Divergence and Curl","rendered":"Using Divergence and Curl"},"content":{"raw":"
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Learning Objectives<\/h3>\r\n
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  • Use the properties of curl and divergence to determine whether a vector field is conservative.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\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>\r\n\r\n<\/div>\r\n

    If\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is a vector field in [latex]\\mathbb{R}^3[\/latex], then the curl of\u00a0[latex]{\\bf{F}}[\/latex]\u00a0is 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. To give this result a physical interpretation, recall that divergence of a velocity field\u00a0[latex]{\\bf{v}}[\/latex]\u00a0at point [latex]P[\/latex] measures the tendency of the corresponding fluid to flow out of [latex]P[\/latex]. Since [latex]\\text{div curl }({\\bf{v}})=0[\/latex], the net rate of flow in vector field curl([latex]{\\bf{v}}[\/latex]) at any point is zero. Taking the curl of vector field\u00a0[latex]{\\bf{F}}[\/latex]\u00a0eliminates whatever divergence was present in\u00a0[latex]{\\bf{F}}[\/latex].<\/div>\r\n
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    Theorem: divergence of the curl<\/h3>\r\n\r\n
    \r\n\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, [latex]\\text{div curl }({\\bf{F}})=\\nabla\\cdot(\\nabla\\times{\\bf{F}})=0[\/latex].\r\n\r\n<\/div>\r\n

    Proof<\/h3>\r\n

    By the definitions of divergence and curl, and by Clairaut\u2019s theorem,<\/p>\r\n

    [latex]\\large{\\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<\/div>\r\n

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    Example: showing that a vector field is not the curl of another<\/h3>\r\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].\r\n\r\n[reveal-answer q=\"748329916\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"748329916\"]\r\n

    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<\/div>\r\n

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    try it<\/h3>\r\nIs it possible for [latex]{\\bf{G}}(x,y,z)=\\langle\\sin{x},\\cos{y},\\sin{(x,y,z)}\\rangle[\/latex] to be the curl of a vector field?\r\n\r\n[reveal-answer q=\"679340216\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"679340216\"]\r\n\r\nNo.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n