7. Using the divergence theorem compute the outward flux of the vector field3F(x, y, z) = x³i + y°j+10kover the boundary of the region between the upper hemispheres of radius 1 and 2 centeredat the origin with bases in the xy-plane.

step:-1 Divergence Theorem:- ∬F.ndS=∭divF dV=∭divF ρ2 sinϕ dρ dθ dϕ Given that F→=x3i…

Answered: 7. Using the divergence theorem compute…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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7. Using the divergence theorem compute the outward flux of the vector field
3
F(x, y, z) = x³i + y°j+10k
over the boundary of the region between the upper hemispheres of radius 1 and 2 centered
at the origin with bases in the xy-plane.

Expert Solution

Step 1

step:-1

Divergence Theorem:-

$\iint F . n d S = ∭ d i v (F) d V = ∭ d i v (F) ρ^{2} \sin ϕ d ρ d θ d ϕ$

Given that

$\vec{F} = x^{3} i + y^{3} j + 10 k d i v . (\vec{F}) = (\frac{\partial}{\partial x} i + \frac{\partial}{\partial x} j + \frac{\partial}{\partial x} k) (x^{3} i + y^{3} j + 10 k) d i v . (\vec{F}) = (3 x^{2} + 3 y^{2}) = 3 (x^{2} + y^{2}) d i v . (\vec{F}) = 3 (x^{2} + y^{2})$

Step:-2

Using spherical co-ordinates, so take

$x = ρ \sin ϕ \sin θ, y = ρ \sin ϕ \cos θ, z = ρ \cos ϕ \Rightarrow x^{2} + y^{2} = ρ^{2} \sin^{2} ϕ \sin^{2} θ + ρ^{2} \sin^{2} ϕ \cos^{2} θ = ρ^{2} \sin^{2} ϕ (\sin^{2} θ + \cos^{2} θ) \Rightarrow x^{2} + y^{2} = ρ^{2} \sin^{2} ϕ a s \sin^{2} θ + \cos^{2} θ = 1$

In this picture, we can see the upper hemisphere forms circle in x y- plane, so

$0 \leq θ \leq 2 π$

As given hemisphere of radius 1 and 2 center at origin so,

$1 \leq ρ \leq 2$

and $0 \leq ϕ \leq \frac{π}{2}$ as hemisphere,

Step:-3

Using above divergence theorem we have

$\iint F . n d S = ∭ d i v (F) ρ^{2} \sin ϕ d ρ d θ d ϕ \iint F . n d S = 3 \int_{0}^{2 π} (\int_{0}^{\frac{π}{2}} (\int_{1}^{2} ρ^{4} \sin^{3} ϕ d ρ) d ϕ) d θ u \sin g e q u a t i o n (1) \iint F . n d S = 3 \int_{0}^{2 π} (\int_{0}^{2 π} \sin^{3} ϕ [\frac{2^{5}}{5} - \frac{1^{5}}{5}] d ϕ) d θ = \frac{3 \times 31}{5 \times 4} \int_{0}^{2 π} [3 - \frac{1}{3}] d θ = \frac{3 \times 31}{5 \times 4} \times \frac{8}{3} \times 2 π \iint F . ndS = \frac{124 π}{5}$

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