What is the equation for line integrals involving the gradient of a function?

In summary, the conversation discusses the use of the formula ##\nabla fg = f\nabla g + g\nabla f## and how it can be applied to calculate the value of ##\displaystyle\oint_C \vec \nabla (fg) \cdot d \vec r## for any closed curve. The speaker also suggests that this formula holds true for any closed curve in R and that ##\displaystyle\oint_C f\nabla g \cdot dr = -\displaystyle\oint_C g\nabla f \cdot dr##. The logic presented is deemed to be correct.
  • #1
WMDhamnekar
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TL;DR Summary
Let f (x, y) and g(x, y) be continuously differentiable real-valued functions in a region R.
Show that ##\displaystyle\oint_C f\nabla g \cdot dr = -\displaystyle\oint_C g\nabla f \cdot dr## for any closed curve in R
I don't have any idea to answer this question. So, any math help will be accepted.
I know ##\nabla fg = f\nabla g + g\nabla f \rightarrow (1) ## But I don't understand to how to use (1) here?
 
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  • #2
What will ## \displaystyle\oint_C \vec \nabla (fg) \cdot d \vec r ## become?
 
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  • #3
drmalawi said:
What will ## \displaystyle\oint_C \vec \nabla (fg) \cdot d \vec r ## become?
I think ## \displaystyle\oint_C \nabla (fg) \cdot dr = \displaystyle\oint_C f\nabla g \cdot dr + \displaystyle\oint_C g\nabla f \cdot dr ## for any closed curve.

But, for any closed curve in R , ##\displaystyle\oint_C \nabla (fg) \cdot dr = 0,## Hence, we can deduce that ##\displaystyle\oint_C f\nabla g \cdot dr = -\displaystyle\oint_C g\nabla f \cdot dr ##
Is that logic correct?
 
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  • #4
WMDhamnekar said:
I think ∮C∇(fg)⋅dr=∮Cf∇g⋅dr+∮Cg∇f⋅dr for any closed curve.
This is valid for all lines.
 
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1. What is a line integral?

A line integral is a mathematical concept used in vector calculus to calculate the total value of a function along a given curve or path. It is represented by the symbol ∫ and is also known as a path integral.

2. How is a line integral different from a regular integral?

A line integral is different from a regular integral in that it calculates the value of a function along a specific path, rather than over a specific interval. This means that the limits of integration are not fixed and can vary depending on the path chosen.

3. What are the applications of line integrals?

Line integrals have many applications in physics and engineering, particularly in the fields of electromagnetism and fluid mechanics. They are used to calculate work, flux, and circulation, among other physical quantities.

4. How is a line integral evaluated?

A line integral is evaluated by breaking down the curve or path into small segments and approximating the value of the function at each segment. These values are then summed up to get an approximate value for the line integral. As the number of segments increases, the accuracy of the evaluation also increases.

5. Can line integrals be calculated in multiple dimensions?

Yes, line integrals can be calculated in multiple dimensions. In two dimensions, they are known as double integrals and in three dimensions, they are known as triple integrals. These integrals are used to calculate the volume under a surface in higher dimensions.

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