Line Integral (4-3-2020)

$\red{\vec{F} = \left(3x^2y+3z^5x^2\right)\hat{i} + \left(x^3+4y^3z^2\right)\hat{j} + \left(2y^4z+5z^4x^3\right) \hat{k}} \ ; \ \blue{d\vec{l} = dx \ \hat{i} + dy \ \hat{j} + dz \ \hat{k}}$ $\implies \vec{F} \cdot d\vec{l} = \left(3x^2y+3z^5x^2\right)dx +\left(x^3+4y^3z^2\right) dy + \left(2y^4z+5z^4x^3\right)dz = d\left(x ^3y+x^3z^5+ y^4z^2 \right)$

The following integral is to be computed between points $(0,0,0)$ and $(1,2,3)$ along the line joining them. Since the dot product is an exact differential, the answer is path independent.

$\vec{F} \cdot d\vec{l} =\int_{\mathrm{initial}}^{\mathrm{final}} d\left(x ^3y+x^3z^5+ y^4z^2 \right)$

Let: $x ^3y+x^3z^5+ y^4z^2 = t$ . Recomputing the limits and applying the substitution transforms the integral to:

$\boxed{\int_{0}^{389} dt = 389}$

First, find the curl of the vector field. It comes out to be $0$ . Hence, the field is conservative.

The potential function comes out to be $f\left(x,y,z\right)=x^{3}y+y^{4}z^{2}+z^{5}x^{3}$

Hence the line integral is simply $f\left(1,2,3\right)-f\left(0,0,0\right)\ =\ 389$