Question

Find the area of the surface. the part of the cylinder y^2+z^2=9 that lies above the rectangle with vertices (0,0),(4,0),(0,2), and (4,2)

Answer 1

We can parameterize this part of the cylinder using

$\mathbf s(u,v)=\langleu,3\cos v,3\sin v\rangle$

with $0\le u\le4$ and $\cos^{-1}\dfrac23\le v\le\dfrac\pi2$. Then the surface element is

$\mathrm d\mathbf S=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv$
$\mathrm d\mathbf S=3\,\mathrm du\,\mathrm dv$

so the area is given by the surface integral, where $D$ denotes the part of the cylinder in question,

$\displaystyle\iint_D\mathrm d\mathbf S=3\int_{v=\cos^{-1}(2/3)}^{v=\pi/2}\int_{u=0}^{u=4}\mathrm du\,\mathrm dv=6\pi$

Probably less work would involve a simpler geometric approach, but it doesn't hurt to practice setting up a proper surface integral.