I am in a multivariable calculus class this semester and have a final soon, one of the questions is about finding the surface area of a part of a plane. I am absolutely lost on what I’m supposed to do and was wondering if anyone could help me out in teaching me how to solve the problem below
How do I find the surface area of part of a planar surface using
$$\sum(u,v) = < u,v,1-u-v >$$
where $< u,v >$ belongs to the triangle in the uv-plane with the vertices $(0,0)$,$ (1,0)$, $(0,1)$
ok,assuming $u=x$ and $v=y$,then our surface can be defined by, $$S=x\hat i+y\hat j+(1-x-y)\hat k$$ Now,$$\dfrac{\partial S}{\partial x}=\hat i-\hat k$$ $$\dfrac{\partial S}{\partial y}=\hat j-\hat k$$
so,$$\dfrac{\partial S}{\partial x} \times \dfrac{\partial S}{\partial y}= \hat i+\hat j+\hat k$$ $$\implies\biggl|\dfrac{\partial S}{\partial x} \times \dfrac{\partial S}{\partial y}\biggr|=\sqrt{3}$$ so,total surface area is, $$\int \int_{S}ds=\int_{x=0}^1\int_{y=0}^{1-x}\sqrt{3}~dy~dx $$ $$\implies \int \int_{S}ds=\boxed{\dfrac{\sqrt{3}}{2}}$$