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If C is any closed path and U is a scalar function, then the value of contour integral ∮c(∇U).dl is (Answer up to the nearest integer)
    Correct answer is '0'. Can you explain this answer?
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    If C is any closed path and U is a scalar function, then the value of...
    According to the Stoke's theorem,
    LA .dl = ∫(∇ x A).ds
    So, ∮c(∇U).dl = ∫[∇ x (∇U)].ds
    Since, ∇ x (∇U) = 0
    (curl of the gradient of a scalar field is always zero)
    So, the contour integral is zero.
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    If C is any closed path and U is a scalar function, then the value of...
    Contour Integral and Gradient

    To understand why the value of the given contour integral ∮c(∇U).dl is zero when C is any closed path and U is a scalar function, let's break down the concept of a contour integral and the role of the gradient.

    Contour Integral

    A contour integral, denoted by ∮c(), is a way to calculate the line integral of a vector field along a closed curve C. It represents the total effect of the vector field around the closed path.

    Gradient

    The gradient of a scalar function U, denoted by ∇U, is a vector that points in the direction of the steepest increase of U at each point. It represents the rate of change of U in different directions.

    Explanation of the Answer

    To prove that the value of the given contour integral is zero, we need to show that the line integral of the gradient along the closed path C is zero for any scalar function U.

    Green's Theorem

    Green's theorem relates a line integral around a closed curve C to a double integral over the region enclosed by C. It states that the line integral of a vector field F around a closed curve C is equal to the double integral of the divergence of F over the region enclosed by C.

    Applying Green's Theorem

    To apply Green's theorem, we rewrite the given contour integral as a line integral of a vector field. Let F = ∇U, where U is the scalar function. Now, the contour integral becomes: ∮c(∇U).dl = ∮cF.dl.

    Applying Green's theorem to this line integral, we get: ∮cF.dl = ∬(∇·F)dA.

    Since F = ∇U, we substitute it into the divergence term: ∮cF.dl = ∬(∇·∇U)dA.

    Simplifying the divergence term, we have: ∮cF.dl = ∬(∇²U)dA.

    Properties of ∇²U

    The Laplacian operator (∇²) applied to U gives the divergence of the gradient of U. Mathematically, ∇²U = ∇·(∇U).

    Using this property, the double integral can be further simplified: ∮cF.dl = ∬(∇·(∇U))dA.

    Divergence of a Gradient

    The divergence of a gradient (∇·(∇U)) is zero for any scalar function U. This can be proven by applying the chain rule of differentiation to the expression ∇·(∇U) = ∂²U/∂x² + ∂²U/∂y² + ∂²U/∂z².

    Since the cross-derivatives cancel out due to the symmetry of second derivatives, we are left with zero.

    Final Result

    Therefore, the double integral ∮cF.dl = ∬(∇·(∇U))dA simplifies to zero. This means that the value of the contour integral ∮c(∇U).dl is zero for any closed path C and scalar function U.

    Hence, the correct answer to
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    If C is any closed path and U is a scalar function, then the value of contour integral ∮c(∇U).dl is (Answer up to the nearest integer)Correct answer is '0'. Can you explain this answer?
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