Use the divergence theorem to calculate the surface integral s f · ds; that is, calculate the flux of f across s. f(x, y, z) = xyezi + xy2z3j − yezk, s is the surface of the box bounded by the coordinate plane and the planes x = 3, y = 6, and z = 1.

The Divergence Theorem, establishes that

∫∫S) F× dS = ∫∫∫v) (Div. F ) × dv

Solution is:

∫∫∫V) 2xyz³dx dy dz = 36

∫∫S) F× dS  = 36

We can either calculate   ∫∫S) F× dS or ∫∫∫v) (Div. F ) × dv wichever is easier

In order to answer that question we first find Div F.

Div F = ( δF/δx i + δF/δy j + δF/δz k ) × F

Div F = ( δF/δx i + δF/δy j + δF/δz k ) × ( xye∧z i + xy²z³ j - ye∧z k )

Div F = ye∧z + 2xyz³ - ye∧z

Div F = 2xyz³

In rectangular coordinates:   dv = dx×dy×dz Then

∫∫∫V) 2xyz³dx dy dz          Integration limits are :

x    from 0 to 3

y    from 0 to 6

z    from 0  to 1

∫∫∫V) 2xyz³dx dy dz = 2× ∫ ₀³x dx ∫₀⁶y dy ∫₀¹z³ dz

∫∫∫V) 2xyz³dx dy dz = 2 × x²/2 |₀³ × y²/2 |₀⁶ × (1/4) z⁴|₀¹

∫∫∫V) 2xyz³dx dy dz = 8 × 18 × (1/4)

∫∫∫V) 2xyz³dx dy dz = 36

Related  Link : link

has divergence

By the divergence theorem, the integral of across is equal to the integral of over the interior of :

the answer is d on edge

step-by-step explanation:

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v