Question

Use a surface integral to find the area of the triangle t in double-struck r3 with vertices at
(1, 1, 0),

(2, 1, 2),
and
(2, 3, 3).
verify your answer by finding the lengths of the sides and using classical geometry.

Answer 1

Parameterize the triangle by

$\mathbf s(u,v)=(1-v)((1-u)\langle1,1,0\rangle+u\langle2,1,2\rangle)+v\langle2,3,3\rangle$
$\iff\mathbf s(u,v)=\langle1+u+v-uv,1+2v,2u+3v-2uv\rangle$

where $(u,v)\in[0,1]^2$. Then the surface element is

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=\sqrt{21}(1-v)\,\mathrm du\,\mathrm dv$

and the area of the triangle is

$\displaystyle\iint_{\mathcal T}\mathrm dS=\sqrt{21}\int_{v=0}^{v=1}\int_{u=0}^{u=1}(1-v)\,\mathrm du\,\mathrm dv=\frac{\sqrt{21}}2$

To confirm this result, we can determine the length of each side of the triangle - they are $\sqrt5$, $\sqrt{14}$, and $\sqrt 5$ - then apply Heron's formula. If $s$ is the semiperimeter, then $s=\dfrac{2\sqrt5+\sqrt{14}}2$, and the area is

$\sqrt{s(s-\sqrt5)(s-\sqrt5)(s-\sqrt{14})}=\dfrac{\sqrt{21}}2$

as expected.