Respuesta :
By the divergence theorem, the flux of [tex]\vec F[/tex] across S is equal to the volume integral of [tex]\mathrm{div}(\vec F)[/tex] over the interior of S.
We have
[tex]\vec F(x,y,z) = (x^3+y^3)\,\vec\imath + (y^3+z^3)\,\vec\jmath + (z^3+x^3)\,\vec k \\\\ \implies \mathrm{div}(\vec F) = \dfrac{\partial(x^3+y^3)}{\partial x} + \dfrac{\partial(y^3+z^3)}{\partial y} + \dfrac{\partial(z^3+x^3)}{\partial z} = 3(x^2+y^2+z^2)[/tex]
so that
[tex]\displaystyle \iint_S \vec F(x,y,z)\cdot\mathrm d\vec s = \iiint_T \mathrm{div}(\vec F)\,\mathrm dV = 3 \iiint\limits_{x^2+y^2+z^2\le3} (x^2+y^2+z^2)\,\mathrm dx\,\mathrm dy\,\mathrm dz[/tex]
To compute the volume integral, convert to spherical coordinates with
x = ρ cos(θ) sin(ϕ)
y = ρ sin(θ) sin(ϕ)
z = ρ cos(ϕ)
so that
ρ ² = x ² + y ² + z ²
dx dy dz = ρ ² sin(ϕ) dρ dϕ dθ
The region T is the interior of the sphere S, given by the set
[tex]T = \left\{(\rho,\theta,\phi) \mid 0\le\rho\le3 \text{ and } 0\le\phi\le\pi \text{ and }0\le \theta\le2\pi\right\}[/tex]
So we have
[tex]\displaystyle 3 \int_0^{2\pi} \int_0^\pi \int_0^3 \rho^4 \sin(\phi) \,\mathrm d\rho \,\mathrm d\phi \,\mathrm d\theta \\\\ = 6\pi \left(\int_0^\pi \sin(\phi)\,\mathrm d\phi\right) \left(\int_0^3 \rho^4 \,\mathrm d\rho\right) = \boxed{\frac{2916\pi}5}[/tex]
The required surface integral S, and flux across F·dS; that is [tex]\dfrac{2916\pi }{5}[/tex]
Given that,
Function;[tex]F(x, y, z) = (x^3 + y^3)\vec{i} + (y^3 + z^3)\vec{j} + (z^3 + x^3)\vec{k},[/tex]
S is the sphere with center the origin and radius 3.
We have to determine,
Use the Divergence Theorem to calculate the surface integral S, F.dS that is, calculate the flux of F across S.
According to the question,
By the divergence theorem, the flux of [tex]\vec{F}[/tex]across S is equal to the volume integral of [tex]div(\vec{f})[/tex] over the interior of S.
S is the sphere with center the origin and radius 3.
Therefore,
[tex]F(x, y, z) = (x^3 + y^3)\vec{i} + (y^3 + z^3)\vec{j} + (z^3 + x^3)\vec{k},\\\\= div\vec({F}) = \dfrac{d(x^3+y^3)}{dx} + \dfrac{d(y^3+z^3)}{dx} + \dfrac{d(z^3+x^3)}{dx} = 3(x^{2} + y^{2} + z^{2} )\\\\Then,\\\\\int \int_S \vec{F}(x, y, z) . \vec{ds} = \int \int \int _T div\vec{F}dV= 3 \int \int\int (x^{2} + y^{2} + z^{2} )dx.dy.dz[/tex]
To compute the volume integral, convert to spherical co-ordinate,
[tex]x = p\ cos\theta\ sin\phi\\\\y = p \ sin\theta \ sin\phi\\\\z = p\ cos\phi\\\\[/tex]
Therefore,
[tex]p^2 = x^{2} + y^{2} +z^{2} \\\\dx.dy.dz = x^{2} \ sin\phi \ dp\ d\phi \ d\theta[/tex]
The region T is the interior of the sphere S is given by the set,
[tex]T = {[ p,\theta,\phi}] | \ (0\leq p\leq 3 \ and \ \ 0\leq \phi \leq \pi \ and \ 0\leq 0\leq 2\pi )[/tex]
Then,
[tex]= 3 \int^2_0 \int^\pi _0 \int^3_0 p^4. sin(\phi). dp.d\phi .d\theta\\\\= 6\pi ( \int^\pi _0 sin(\phi).d\phi) (\int^3_0 p^4dp\\\\= \dfrac{2916\pi }{5}[/tex]
Hence, The required surface integral S, F·dS; that is [tex]\dfrac{2916\pi }{5}[/tex]
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