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Evaluate the surface integral. s xyz ds, s is the cone with parametric equations x = u cos(v), y = u...
5 months ago
Q:
Evaluate the surface integral. s xyz ds, s is the cone with parametric equations x = u cos(v), y = u sin(v), z = u, 0 ≤ u ≤ 3, 0 ≤ v ≤ π 2
Accepted Solution
A:
Let's capture the parameterization of the surface [tex]\mathcal S[/tex] by the vector function
[tex]\mathbf s(u,v)=\langle u\cos v,u\sin v,u\rangle[/tex]
Then the surface element is given by
[tex]\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv[/tex]
[tex]\implies\mathrm dS=\sqrt 2u\,\mathrm du\,\mathrm dv[/tex]
So the surface integral is equivalent to
[tex]\displaystyle\iint_{\mathcal S}xyz\,\mathrm dS=\sqrt2\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=3}u^4\cos v\sin v\,\mathrm du\,\mathrm dv=\frac{243}{5\sqrt2}[/tex]