Question

find the divergence A at point(2,1,-2) when A= (x^2y^2z)i + (x^3y^3z^2)j + (xyz^2)k?

Answer 1

find the divergence A at point(2,1,-2) when $A = (x^2 y^2 z)^* i + (x^3 y^3 z^2)^* j + (xyz^2)^* k$ ?

main formula is :

$$A = U ^ {*} i + V ^ {*} j + W ^ {*} k$$ $$\operatorname {d i v} (\mathrm {A}) = \mathrm {d U} / \mathrm {d x} + \mathrm {d V} / \mathrm {d y} + \mathrm {d W} / \mathrm {d z}$$

paste in our task:

$$A = \left(x ^ {2} y ^ {2} z\right) ^ {*} i + \left(x ^ {3} y ^ {3} z ^ {2}\right) ^ {*} j + \left(x y z ^ {2}\right) ^ {*} k$$ $$A = U ^ {*} i + V ^ {*} j + W ^ {*} k$$ $$U = x ^ {2} y ^ {2} z$$ $$V = x ^ {3} y ^ {3} z ^ {2}$$ $$W = x y z ^ {2}$$ $$\operatorname {d i v} (\mathrm {A}) = \frac{\mathrm{d} U}{\mathrm{d} x} + \frac{\mathrm{d} V}{\mathrm{d} y} + \frac{\mathrm{d} W}{\mathrm{d} z} = \frac{\mathrm{d} (x ^ {2} y ^ {2} z)}{\mathrm{d} x} + \frac{\mathrm{d} (x ^ {3} y ^ {3} z ^ {2})}{\mathrm{d} y} + \frac{\mathrm{d} (x y z ^ {2})}{\mathrm{d} z} = \frac{2 x y ^ {2} z}{3 x ^ {3} y ^ {2} z ^ {2}} + 2 x y z =$$

$= /$ (we must calculate divergence at the point $(2,1,-2)$ , it is mean that $x = 2$ , $y = 1$ , $z = -2$ )/=

$$= 2 ^ {*} 2 ^ {*} 1 ^ {2} (- 2) + 3 ^ {*} 2 ^ {3} 1 ^ {2} (- 2) ^ {2} + 2 ^ {*} 2 ^ {*} 1 ^ {*} (- 2) = - 8 + 9 6 - 8 = 8 0$$

Answer:

$$\operatorname {d i v} (\mathrm {a}) = 8 0.$$