Respuesta :
Assuming the vector field is [tex]\mathbf f=2x^{1/3}\,\mathbf i+e^{y/7}\,\mathbf j[/tex], we want to find a scalar function [tex]f[/tex] such that [tex]\nabla f=\mathbf f[/tex]. We require
[tex]\dfrac{\partial f}{\partial x}=2x^{1/3}[/tex]
[tex]\dfrac{\partial f}{\partial y}=e^{y/7}[/tex]
Integrating both sides of the first PDE with respect to [tex]x[/tex] gives
[tex]f(x,y)=\dfrac32x^{4/3}+g(y)[/tex]
Differentiating with respect to [tex]y[/tex] gives
[tex]\dfrac{\partial f}{\partial y}=e^{y/7}=\dfrac{\mathrm dg}{\mathrm dy}[/tex]
[tex]\implies g(y)=7e^{y/7}+C[/tex]
[tex]f(x,y)=\dfrac32x^{4/3}+7e^{y/7}+C[/tex]
Such a scalar function exists, so the fundamental theorem holds and any integral along any path is determined exactly by its endpoints. In this case,
[tex]\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(0,1)-f(1,0)=7e^{1/7}-\frac{17}2[/tex]
[tex]\dfrac{\partial f}{\partial x}=2x^{1/3}[/tex]
[tex]\dfrac{\partial f}{\partial y}=e^{y/7}[/tex]
Integrating both sides of the first PDE with respect to [tex]x[/tex] gives
[tex]f(x,y)=\dfrac32x^{4/3}+g(y)[/tex]
Differentiating with respect to [tex]y[/tex] gives
[tex]\dfrac{\partial f}{\partial y}=e^{y/7}=\dfrac{\mathrm dg}{\mathrm dy}[/tex]
[tex]\implies g(y)=7e^{y/7}+C[/tex]
[tex]f(x,y)=\dfrac32x^{4/3}+7e^{y/7}+C[/tex]
Such a scalar function exists, so the fundamental theorem holds and any integral along any path is determined exactly by its endpoints. In this case,
[tex]\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(0,1)-f(1,0)=7e^{1/7}-\frac{17}2[/tex]
The value of the provided integral using the fundamental theorem of line integrals is 7e¹/⁷-17/2.
What is the fundamental theorem of line integrals?
The fundamental theorem of line integrals states that the line integral of the gradient of a function (say f) provides the overall change in the numerical quantity of the function over the curve.
The equation whose value has to be calculated is,
[tex]\int\limits_C {\vec F} \, dr[/tex]
Here, C is the quarter of the unit circle in the first quadrant, traced counterclockwise from (1,0) to (0,1).
The vector field given in the problem is,
[tex]\vec F = 2x^{ 1/3}+e^{ y/7} \vec j[/tex]
The partial derivative of the equation,
[tex]\dfrac{\partial \vec F}{\partial x} = 2x^{ 1/3}[/tex]
Integrate this partial derivative with respect to x,
[tex]F(x,y) = \dfrac{3}{2}x^{4/3}+g(y)[/tex]
The another partial derivative of the equation is,
[tex]\dfrac{\partial \vec F}{\partial y} = e^{y/7}[/tex]
Differential it with respect to y,
[tex]\dfrac{d g}{d y} = e^{y/7}\\dg = 7e^{y/7}+C[/tex]
Thus, the function,
[tex]F(x,y)=F(x,y) = \dfrac{3}{2}x^{4/3}+g(y)\\F(x,y)=F(x,y) = \dfrac{3}{2}x^{4/3}+7e^{y/7}+C[/tex]
From the fundamental theorem of line integrals, the value of the provided integral can be calculated with the following expression.
[tex]\int\limits_C {\vec F} \, dr=f(0,1)-f(1,0)\\\int\limits_C {\vec F} \, dr=7e^{1/7}-\dfrac{17}{2}[/tex]
Thus, the value of the provided integral using the fundamental theorem of line integrals is 7e¹/⁷-17/2.
Learn more about the fundamental theorem of line integrals here;
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