Respuesta :
Parameterize the hemisphere [tex]\mathcal S[/tex] by
[tex]\mathbf s(u,v)=\langle3\cos u\sin v,3\sin u\sin v,3\cos v\rangle[/tex]
with [tex]0\le u\le2\pi[/tex] and [tex]0\le v\le\dfrac\pi2[/tex]. Then the surface integral becomes
[tex]\displaystyle\iint_{\mathcal S}(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\cos v\sin^2v\|\mathbf s_u\times\mathbf s_v\|\,\mathrm dv\,\mathrm du[/tex]
We have
[tex]\mathbf s_u\times\mathbf s_v=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle[/tex]
[tex]\implies\|\mathbf s_u\times\mathbf s_v\|=9\sin v[/tex]
The integral reduces to
[tex]\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\cos v\sin^3v\,\mathrm dv\,\mathrm du[/tex]
Substitute [tex]t=\sin v[/tex], [tex]\mathrm dt=\cos v\,\mathrm dv[/tex]:
[tex]\displaystyle486\pi\int_{t=0}^{t=1}t^3\,\mathrm dt=\frac{243\pi}2[/tex]
- In the 3D geometric form of this hemisphere is half a sphere with such a flat surface solely on a single side and a circular bowl just on the second.
- Any Earth-drawn circle divides into two halves, known as called hemispheres.
- Four hemispheres were commonly considered north, south, east, and west.
- Parameterizing the hemisphere S:
[tex]\to \bold{s(u,v)=(3\cos u \sin v, 3\sin u \sin v, 3 \cos v)}\\\\[/tex]
with [tex]0\leq u\leq 2\pi[/tex] and [tex]0\leq v \leq \frac{\pi}{2}[/tex].
So, the surface integral:
[tex]\to \bold{\int \int_s (x^2+y^2)z dS=27\int^{u=\frac{\pi}{2}}_{u=0} \int^{v=\frac{\pi}{2}}_{v=0} \cos v \sin^2 v \ dv\ du}\\\\[/tex]
[tex]\to \bold{S_u \times S_v=(-9 \cos u \sin^2 v,-9 \sin u \sin^2 v, -9 \cos v \sin^2 v)}\\\\\to \bold{|| S_u \times S_v||= 9\sin v}\\\\[/tex]
Calculating the integral:
[tex]\to \bold{243 \int^{u=2\pi}_{u=0} \int^{v=\frac{\pi}{2}}_{v=0} \cos v \sin^3 v \ dv \ du}\\\\[/tex]
Substituting the value
[tex]\to \bold{t=\sin v}\\\\ \to \bold{dt=\cos v \ dv}:\\\\\to \bold{486 \int^{t=1}_{t=0} t^3\ dt \\ \\ \to \frac{243 \pi}{2}}= 381.70}[/tex]
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