Surface Area — Question 6

Question 6 – Solution

Strategy. The coordinate-plane restrictions cut a triangular patch from a plane with constant slope.

Step 1: Region

In the first octant, $x\ge 0$, $y\ge 0$, and $z=4-x-y\ge 0$. Thus the projection is $$D=\{(x,y):0\le x\le 4,\ 0\le y\le 4-x\},$$ a right triangle of area $8$.

Step 2: Surface integral Since $f_x=f_y=-1$, $$S=\iint_D\sqrt{1+(-1)^2+(-1)^2}\,dA =\sqrt 3(8)=\boxed{8\sqrt 3}.$$

Verification From $(4,0,0)$, use edge vectors $\mathbf a=\langle-4,4,0\rangle$ and $\mathbf b=\langle-4,0,4\rangle$. Then $|\mathbf a\times\mathbf b|=16\sqrt 3$, so the triangle area is $\frac 12(16\sqrt 3)=8\sqrt 3$.