Vector Calc (1-14-2021)

Calculus Level 3

Consider a closed spherical surface S S :

x 2 + y 2 + z 2 = 1 x^2 + y^2 + z^2 = 1

There is a vector field throughout all of space:

F ⃗ = ( F x , F y , F z ) F x = y 2 z + 2 x F y = x − y + z F z = z 3 + x y \vec{F} = (F_x, F_y, F_z) \\ F_x = y^2 z + 2 x \\ F_y = x - y + z \\ F_z = z^3 + x y

Determine the absolute value of the flux of the vector field over the surface.

∣ ∬ s F ⃗ ⋅ d S ⃗ ∣ \Big|\iint_s \vec{F} \cdot \vec{dS} \Big|

Note: This one is fairly easy to do by hand


The answer is 6.702.

Steven Chase by Steven Chase , 37, USA

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1 solution

Karan Chatrath
Jan 14, 2021

Divergence theorem:

∯ S F ⃗ ⋅ d S ⃗ = ∫ ∫ ∫ V ( ∇ ⋅ F ⃗ ) d V \oiint_{S} \vec{F} \cdot \ d\vec{S}= \int \int \int_{V} (\nabla \cdot \vec{F}) \ dV

Now:

∇ ⋅ F ⃗ = 1 + 3 z 2 \nabla \cdot \vec{F} = 1 + 3z^2 In spherical coordinates:

∇ ⋅ F ⃗ = 1 + 3 r 2 cos ⁡ 2 θ \nabla \cdot \vec{F} = 1 + 3r^2\cos^2{\theta} ∴ ∫ ∫ ∫ V ( ∇ ⋅ F ⃗ ) d V = ∫ 0 2 π ∫ 0 π ∫ 0 1 ( 1 + 3 r 2 cos ⁡ 2 θ ) r 2 sin ⁡ θ d r d θ d ϕ \therefore \int \int \int_{V} (\nabla \cdot \vec{F}) \ dV = \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{1} \left( 1 + 3r^2\cos^2{\theta} \right)r^2 \ \sin{\theta} \ dr \ d\theta \ d\phi

Integrating with respect to r r first gives (simplification left out):

∫ ∫ ∫ V ( ∇ ⋅ F ⃗ ) d V = ∫ 0 2 π ∫ 0 π ( sin ⁡ θ 3 + 3 cos ⁡ 2 θ sin ⁡ θ 5 ) d θ d ϕ \int \int \int_{V} (\nabla \cdot \vec{F}) \ dV=\int_{0}^{2\pi} \int_{0}^{\pi} \left( \frac{\sin{\theta}}{3} + \frac{3\cos^2{\theta}\sin{\theta}}{5} \right) \ d\theta \ d\phi Changing the order of the integral: ∫ ∫ ∫ V ( ∇ ⋅ F ⃗ ) d V = ∫ 0 π ∫ 0 2 π ( sin ⁡ θ 3 + 3 cos ⁡ 2 θ sin ⁡ θ 5 ) d ϕ d θ \int \int \int_{V} (\nabla \cdot \vec{F}) \ dV=\int_{0}^{\pi} \int_{0}^{2\pi} \left( \frac{\sin{\theta}}{3} + \frac{3\cos^2{\theta}\sin{\theta}}{5} \right) \ d\phi \ d\theta ⟹ ∫ ∫ ∫ V ( ∇ ⋅ F ⃗ ) d V = 2 π ∫ 0 π ( sin ⁡ θ 3 + 3 cos ⁡ 2 θ sin ⁡ θ 5 ) d θ \implies \int \int \int_{V} (\nabla \cdot \vec{F}) \ dV=2\pi \int_{0}^{\pi} \left( \frac{\sin{\theta}}{3} + \frac{3\cos^2{\theta}\sin{\theta}}{5} \right) \ d\theta ⟹ ∯ S F ⃗ ⋅ d S ⃗ = ∫ ∫ ∫ V ( ∇ ⋅ F ⃗ ) d V = 32 π 15 \implies \oiint_{S} \vec{F} \cdot \ d\vec{S}=\int \int \int_{V} (\nabla \cdot \vec{F}) \ dV=\boxed{\frac{32\pi}{15}}

3 Helpful 3 Interesting 5 Brilliant 0 Confused

Nice job. As a side note, I made the divergence a pure function of z z so that the calculation could be done with a single integral.

Steven Chase - 4 months, 4 weeks ago

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