Question

Help me about this integral

Answer 1

The gradient theorem applies here, because we can find a scalar function f for which ∇ f (or the gradient of f ) is equal to the underlying vector field:

`\\abla f(x,y,z)=\langle2xy,x^2-z^2,-2yz\rangle`

We have

`(\partial f)/(\partial x)=2xy\implies f(x,y,z)=x^2y+g(y,z)`

`(\partial f)/(\partial y)=x^2-z^2=x^2+(\partial g)/(\partial y)\implies(\partial g)/(\partial y)=-z^2\implies g(y,z)=-yz^2+h(z)`

`(\partial f)/(\partial z)=-2yz=-2yz+(\mathrm dh)/(\mathrm dz)\implies(\mathrm dh)/(\mathrm dz)=0\implies h(z)=C`

where C is an arbitrary constant.

So we found

`f(x,y,z)=x^2y-yz^2+C`

and by the gradient theorem,

`\displaystyle\int_((0,0,0))^((1,2,3))\\abla f\cdot\langle\mathrm dx,\mathrm dy,\mathrm dz\rangle=f(1,2,3)-f(0,0,0)=\boxed{-16}`