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Viefleur [7K]
3 years ago
11

Surface integrals using a parametric description. evaluate the surface integral \int \int_{s} f(x,y,z)dS using a parametric desc

ription of the surface.
f(x,y,z)=x2+y2, where S is the hemisphere x2+y2+z2=36, for z>=0
Mathematics
1 answer:
DiKsa [7]3 years ago
7 0

You can parameterize S using spherical coordinates by

\vec s(u,v)=\langle6\cos u\sin v,6\sin u\sin v,6\cos v\rangle

with 0\le u\le2\pi and 0\le v\le\frac\pi2.

Take the normal vector to S to be

\dfrac{\partial\vec s}{\partial\vec v}\times\dfrac{\partial\vec s}{\partial\vec u}=36\langle\cos u\sin^2v,\sin u\sin^2v,\cos v\sin v\rangle

(I use \vec s_v\times\vec s_u to avoid negative signs. The orientation of the normal vector doesn't matter for a scalar surface integral; you could just as easily use \vec s_u\times\vec s_v=-(\vec s_v\times\vec s_u).)

Then

f(x,y,z)=f(6\cos u\sin v,6\sin u\sin v,6\cos v)=36\sin^2v

and the integral of f over S is

\displaystyle\iint_Sf(x,y,z)\,\mathrm dS=\int_0^{\pi/2}\int_0^{2\pi}36\sin^2v\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^{\pi/2}\int_0^{2\pi}(36\sin^2v)(36\sin v)\,\mathrm du\,\mathrm dv

=\displaystyle2592\pi\int_0^{\pi/2}\sin^3v\,\mathrm dv=\boxed{1728\pi}

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