Question

Find the area of the surface. the part of the surface y = 5x + z2 that lies between the planes x = 0, x = 1, z = 0, and z = 1.

Answer 1

Parameterize the part of the surface we care about, denoted $\mathcal S$, by

$\mathbf r(u,v)=\langle u,5u+v^2,v\rangle$

with $0\le u\le1$ and $0\le v\le1$.

Then

$\|\mathbf r_u\times\mathbf r_v\|=\|\langle5,-1,2v\rangle\|=\sqrt{4v^2+26}$

The area is given by the surface integral

$\displaystyle\iint_{\mathcal S}\mathrm dS=\int_{v=0}^{v=1}\int_{u=0}^{u=1}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_{v=0}^{v=1}\sqrt{4v^2+26}\,\mathrm dv=\sqrt{\dfrac{15}2}+\dfrac{13}2\sinh^{-1}\sqrt{\dfrac2{13}}$