John S.

# Use the divergence theorem to evaluate I=s∫∫(4x+3y^2+Z)dS

Use the divergence theorem to evaluate I=S∫∫(4x+3y2+z)dS,
Where s is unit sphere x2+y2+z2=1.

Caleb M.

Okay Problem: "Use the divergence theorem to evaluate I=s∫∫(4x+3y2+Z)dS
Use the divergence theorem to evaluate I=S∫∫(4x+3y2+z)dS,
Where S is unit sphere x2+y2+z2=1."

Step (I.) find divergence within the integrand "∫(4x+3y2+z)"
divergence is finding the partial derivative of the vector field or with respect to it's components Div F=∂f/∂x+ ∂f/∂y +∂f/∂z
∂f/∂x(4x+3y2+z)=4x2, ∂f/∂y(4x+3y2+z)=4y2, and ∂f/∂z(4x+3y2+z)=4z2
All together "4x2 + 4y2+4z2"

Step (II.) Use the method for spherical coordinates "x2+y2+z2=1"

with your new function you need to find your limits of integration [a,b] w/ respect to the multiple differentials.
We've been given "x2+y2+z2=1" and spherical coordinates conversion goes like this "∫∫∫ρsin(φ) dρdθdφ."
luckily we can simplify it because if you look at the equation for a sphere that they gave us "x2+y2+z2=1," "x2+y2+z2" [ρ2]; so ∫∫∫[ρ2]ρsin(φ)dρdθdφ

Step (III.) Draw/map it it out on the x-y-z vector
the differentials "dρdθdφ" are describing and action. "dρ" acts as a radius which scales the sphere in size,"dθ" acts and a point where the vector is going to be drawn out, therefore; "dθ" travels around the the equator on the world, finally "dφ" act a an enlarger that scales the sphere from top to bottom; think of flux traveling from the North dipole to the south dipole of the earth. so:
dp {0≤ρ≤1
dθ {0≤θ≤2π
dφ {0≤φ≤π
Step (v.) Start the triple integration

∫∫∫5+6(sin(φ)sin(θ))ρ2(sin(φ))dρdθdφ

the result is 10π2
Report

12/31/16