Question

(if F = xi + yj + zk then ss F.ds is
where S is the closed surface of the sphere x2 + y2 + z2 - 4​

Answer 1

Answer:

32π

Step-by-step explanation:

Surface is described by the equation

${x}^{2} + {y}^{2} + {z}^{2} = 4 = {2}^{2}$

Now this represents a sphere of radius 2 unit.

We have,

F=xi+yj+zk,

This field represents the direction of a radius vector with the magnitude of the radius, i.e

$|f| = \sqrt{( {x}^{2} + {y}^{2} + {z}^{2} ) } = |r|$

Now we know radius vector is always perpendicular to the surface of a sphere, so

$f.ds = |f| |ds|$

And, at the surface of the sphere value of r is 2 units, so

$|f| = 2$

So finally the surface integral can be written as

$2 \times (integral \: {ds} )$

Over the entire surface of sphere

And integral of ds over entire surface of sphere would be the surface area of sphere,

so finally the value will be

$2 \times 4\pi {2}^{2} = 32\pi$