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

2 thoughts on “(if F = xi + yj + zk then ss F.ds is<br />where S is the closed surface of the sphere x2 + y2 + z2 – 4​”

1. Answer:

   32π

   Step-by-step explanation:

   Surface is described by the equation

   [tex] {x}^{2} + {y}^{2} + {z}^{2} = 4 = {2}^{2} [/tex]

   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

   [tex] |f| = \sqrt{( {x}^{2} + {y}^{2} + {z}^{2} ) } = |r| [/tex]

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

   [tex]f.ds = |f| |ds| [/tex]

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

   [tex] |f| = 2[/tex]

   So finally the surface integral can be written as

   [tex]2 \times (integral \: {ds} )[/tex]

   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

   [tex]2 \times 4\pi {2}^{2} = 32\pi[/tex]

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