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Artyom0805 [142]
3 years ago
14

Evaluate the surface integral. s x2z + y2z ds s is the hemisphere x2 + y2 + z2 = 9, z ≥ 0

Mathematics
1 answer:
uranmaximum [27]3 years ago
8 0

Parameterize the hemisphere \mathcal S by

\mathbf s(u,v)=\langle3\cos u\sin v,3\sin u\sin v,3\cos v\rangle

with 0\le u\le2\pi and 0\le v\le\dfrac\pi2. Then the surface integral becomes

\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

We have

\mathbf s_u\times\mathbf s_v=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle

\implies\|\mathbf s_u\times\mathbf s_v\|=9\sin v

The integral reduces to

\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\cos v\sin^3v\,\mathrm dv\,\mathrm du

Substitute t=\sin v, \mathrm dt=\cos v\,\mathrm dv:

\displaystyle486\pi\int_{t=0}^{t=1}t^3\,\mathrm dt=\frac{243\pi}2

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