If f = 2x³ + 3y² + 4z, the value of line integral $\underset{C}{\int }\mathrm{gradf}.\mathrm{dr}$ evaluated over contour C formed by the segments (-3, -3, 2) → (2, -3, 2) → (2, 6, 2) → (2, 6, -1) is ________.

Concept:

The gradient at a point in a curve is defined as the Tangent to that point.

$\Delta f(x,y,z)=\frac{\delta f(x,y,z)}{\delta x}+\frac{\delta f(x,y,z)}{\delta y}+\frac{\delta f(x,y,z)}{\delta z}$

The gradient in all coordinate system is defined below:Calculation:

f = 2x³ + 3y² + 4z

grad $\mathrm{f}=i\frac{\partial f}{\partial x}+j\frac{\partial f}{\partial y}+k\frac{\partial f}{\partial z}$

$=6x^{2}i+6yj+4k$

$\underset{c}{\int }gradf.dr=\underset{c}{\int }\left(6x^{2}i+6yj+4k\right)\left(dxi+dyj+dzk\right)$

$=\underset{c}{\int }\left(6x^{2}dx+6ydy+4dz\right)$

C: (-3, -3, 2) → (2, -3, 2) → (2, 6, 2) → (2, 6, -1)

⇒ (-3, -3, 2) → (2, 6, -1)

$=\underset{\left(-3,-3,2\right)}{\overset{\left(2,6,-1\right)}{\int }}\left(6x^{2}dx+6ydy+4dz\right)$

$={\left[\frac{6x^3}{3}\right]}_{-3}^2+{\left[\frac{6y^2}{2}\right]}_{-3}^6+{\left[4z\right]}_2^{-1}$

= 2 [8 + 27] + 3[36 - 9] + 4 [-1 -2]

= 70 + 81 – 12 = 139