Vector Fields

Question

Let F(x,y)=3x^2yi+(x^3+y^3)j. By integrating, find a function f so that F=del f. Don't forget your constants of integration!

Answer 1

F = ∇ƒ

= (∂ƒ/∂x)i + (∂ƒ/∂y)j

= (3x²y)i + (x³+y³)j

 

Set the corresponding unit vector components equal to each other:

• (∂ƒ/∂x)i = (3x²y)i
• (∂ƒ/∂y)j = (x³+y³)j

 

Focusing on the x-component vectors:

∂ƒ/∂x = 3x²y

∫[∂ƒ/∂x] dx

= ∫3x²y dx

= x³y + h(y) + C

 

Similarly for the y-component vectors:

∂ƒ/∂y = x³ + y³
∫[∂ƒ/∂y] dy
= ∫x³ + y³ dy
= x³y + ¼y⁴ + g(x) + C

 

Setting both results equal to each other:

x³y + h(y) + C = x³y + ¼y⁴ + g(x) + C

 

g(x) = 0

h(y) = ¼y⁴

 

The general solution:

ƒ = x³y + ¼y⁴ + C