MATH SOLVE

4 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]

[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]