[tex]\vec F[/tex] has divergence
[tex]\nabla\cdot\vec F=\dfrac{\partial(xye^z)}{\partial x}+\dfrac{\partial(xy^2z^3)}{\partial y}-\dfrac{\partial(ye^z)}{\partial z}=ye^z+2xyz^3-ye^z=2xyz^3[/tex]
By the divergence theorem, the integral of [tex]\vec F[/tex] across [tex]S[/tex] is equal to the integral of [tex]\nabla\cdot\vec F[/tex] over the interior of [tex]S[/tex]:
[tex]\displaystyle\iiint_S\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^6\int_0^32xyz^3\,\mathrm dx\,\mathrm dy\,\mathrm dz=\boxed{\frac{81}2}[/tex]
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
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