Answer to Question #149124 in Math for Usman

Question

The velocity components in a two dimensional flow are :
U = 8x^2 y - 8/3 y^3 And
V = -8xy^3 + 8/3 x^3.
Show that these velocity components represents a possible case of an irrotational flow.

Answer 1

$$u=8x^2y-\dfrac{8}{3}y^3, v=-8xy^3+\dfrac{8}{3}x^3$$

The fluid is irrotational, if $\dfrac{\partial u}{\partial y}=\dfrac{\partial v}{\partial x}.$

$$\dfrac{\partial u}{\partial y}=8x^2-8y^2, \dfrac{\partial v}{\partial x}=-8y^3+8x^2$$

$$\dfrac{\partial u}{\partial y}\not=\dfrac{\partial v}{\partial x}$$

Therefore, it is not a possible irrotational flow.

If

$$u=8x^2y-\dfrac{8}{3}y^3, v=-8xy^2+\dfrac{8}{3}x^3$$

The fluid is irrotational, if $\dfrac{\partial u}{\partial y}=\dfrac{\partial v}{\partial x}.$

$$\dfrac{\partial u}{\partial y}=8x^2-8y^2, \dfrac{\partial v}{\partial x}=-8y^2+8x^2$$

$$\dfrac{\partial u}{\partial y}=\dfrac{\partial v}{\partial x}$$

Therefore, these velocity components represents a possible case of an irrotational flow.