Surface Area — Question 10

Question 10 – Solution

Strategy. Apply the monotonicity of $\sqrt{1+t^2}$ to the pointwise gradient bounds, then integrate.

Step 1: Pointwise comparison From $1\le|\nabla f|\le 2$, $$\sqrt 2=\sqrt{1+1^2} \le\sqrt{1+|\nabla f|^2} \le\sqrt{1+2^2}=\sqrt 5.$$

Step 2: Integrate The graph area is $$S=\iint_D\sqrt{1+|\nabla f|^2}\,dA.$$ Integrating the pointwise inequalities and using $\operatorname{area}(D)=12$ yields $$\boxed{12\sqrt 2\le S\le 12\sqrt 5}.$$

Step 3: Sharpness The lower endpoint occurs for any plane with constant gradient magnitude $1$, for example $f(x,y)=x$. The upper endpoint occurs for a plane with constant gradient magnitude $2$, for example $f(x,y)=2x$. Indeed, $$S_{\;f=x}=12\sqrt{1+1}=12\sqrt 2,\qquad S_{\;f=2x}=12\sqrt{1+4}=12\sqrt 5.$$ Therefore no narrower interval can be guaranteed from the supplied data.