Respuesta :

Assuming [tex]S[/tex] does not include the plane [tex]z=0[/tex], we can parameterize the region in spherical coordinates using

[tex]\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle[/tex]

where [tex]0\le u\le2\pi[/tex] and [tex]0\le v\le\dfrac\pi/2[/tex]. We then have

[tex]x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v[/tex]
[tex](x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v[/tex]

Then the surface integral is equivalent to

[tex]\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du[/tex]

We have

[tex]\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle[/tex]
[tex]\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle[/tex]
[tex]\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle[/tex]
[tex]\implies\left\|\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}\|=9\sin v[/tex]

So the surface integral is equivalent to

[tex]\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du[/tex]
[tex]=\displaystyle486\pi\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv[/tex]
[tex]=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw[/tex]

where [tex]w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv[/tex].

[tex]=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}[/tex]
[tex]=\dfrac{243}2\pi[/tex]