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.

Parameterize the triangle by

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

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

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

and the area of the triangle is

[tex]\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[/tex]

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

[tex]\sqrt{s(s-\sqrt5)(s-\sqrt5)(s-\sqrt{14})}=\dfrac{\sqrt{21}}2[/tex]

as expected.