Surface Area — Question 1

Question 1 – Solution

Strategy. Parametrize the graph by its projection and use the tangent-parallelogram area scale.

Step 1: Derive the element For a graph $z=f(x,y)$, $$\mathbf r(x,y)=\langle x,y,f(x,y)\rangle,\quad \mathbf r_x=\langle 1,0,f_x\rangle,\quad \mathbf r_y=\langle 0,1,f_y\rangle.$$

Thus $$|\mathbf r_x\times\mathbf r_y| =|\,\langle-f_x,-f_y,1\rangle\,| =\sqrt{1+f_x^2+f_y^2}.$$

Step 2: Apply to the plane Here $f_x=2$ and $f_y=-1$, so $$S=\int_0^3\int_0^2\sqrt{1+2^2+(-1)^2}\,dy\,dx =\sqrt 6(3)(2)=\boxed{6\sqrt 6}.$$

Verification Every small projected area is enlarged by the same factor $\sqrt 6$. Since the rectangle has area $6$, the surface area must be $6\sqrt 6$.