Line Integrals

[It401301]

Hello I need help understanding how to proceed with this problem:

Let F be the radial vector field F = x i + y j + z k.
Show that if x(t), a ≤ t ≤ b, is any path that lies
on the sphere x2 + y2 + z2 = c2, then

x F · ds = 0.
(Hint: If x(t) = (x(t), y(t), z(t)) lies on the sphere, then
[x(t)]2 + [y(t)]2 + [z(t)]2 = c2. Differentiate this last
equation with respect to t.)

I understand the line integral formula F(x(t)) times the magnitude of the derivative of x(t), but I don't how it applies here.

 

[HallsofIvy]

Hello I need help understanding how to proceed with this problem:

Let F be the radial vector field F = x i + y j + z k.
Show that if x(t), a ≤ t ≤ b, is any path that lies
on the sphere x2 + y2 + z2 = c2, then

x F · ds = 0.
(Hint: If x(t) = (x(t), y(t), z(t)) lies on the sphere, then
[x(t)]2 + [y(t)]2 + [z(t)]2 = c2. Differentiate this last
equation with respect to t.)

I understand the line integral formula F(x(t)) times the magnitude of the derivative of x(t), but I don't how it applies here.

The real point is that the vector F= xi+ yj+ zk is directed from (0, 0, 0) to (x, y, z) so directly out from the origin and so is perpendicular to the sphere centered at the origin and so perpendicular to any path on the sphere.

But the derivative of \(\displaystyle x^2+ y^2+ z^2= c^2\), with respect to t, is 2xx'+ 2yy'+ 2zz'= 0. That is, of course, the same as xx'+ yy'+ zz'= 0 and that can be thought of as a "dot product", \(\displaystyle <x, y, z>\cdot <x', y', z'>= 0\). From that, \(\displaystyle <x, y, z>\cdot<x', y', z'> dt= 0\) which is just \(\displaystyle F\cdot ds\). Since that is always 0, its integral is 0.

 

[It401301]

The real point is that the vector F= xi+ yj+ zk is directed from (0, 0, 0) to (x, y, z) so directly out from the origin and so is perpendicular to the sphere centered at the origin and so perpendicular to any path on the sphere.

But the derivative of \(\displaystyle x^2+ y^2+ z^2= c^2\), with respect to t, is 2xx'+ 2yy'+ 2zz'= 0. That is, of course, the same as xx'+ yy'+ zz'= 0 and that can be thought of as a "dot product", \(\displaystyle <x, y, z>\cdot <x', y', z'>= 0\). From that, \(\displaystyle <x, y, z>\cdot<x', y', z'> dt= 0\) which is just \(\displaystyle F\cdot ds\). Since that is always 0, its integral is 0.

Thanks! I thought this was the case, but the problem seemed deceptively easy.