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Calculate the integral s f · ds, where s is the entire surface of the solid half ball x2 + y2 + z2 ≤ 1, z ≥ 0, and f = (x + 3y5)i + ( y + 10xz)j + (z − xy)k. (let s be oriented by the outward-pointing normal.)

Answer :

LammettHash
The divergence theorem applies since [tex]\mathcal S[/tex] is a closed surface. We have

[tex]\displaystyle\iint_{\mathcal S}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_{\mathcal V}\nabla\times\mathbf f(x,y,z)\,\mathrm dV[/tex]

where [tex]\mathcal V[/tex] is the space enclosed by the surface [tex]\mathcal S[/tex]. The divergence of the given vector field is

[tex]\nabla\times\mathbf f(x,y,z)=\dfrac{\partial(x+3y^5)}{\partial x}+\dfrac{\partial(y+10xz)}{\partial y}+\dfrac{\partial(z-xy)}{\partial z}=1+1+1=3[/tex]

So we can write the volume integral (in spherical coordinates) as

[tex]\displaystyle3\iiint_{\mathcal V}\,\mathrm dV=3\int_{\varphi=0}^{\varphi=\pi/2}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=0}^{\rho=1}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{4\pi}3[/tex]

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