Surface Area — Question 8

Question 8

For constants $a,b,c$, the plane $$z=ax+by+c$$ lies above the unit disk $x^2+y^2\le 1$. Its surface area there is $3\pi$.

Tasks

1. Determine all possible gradient pairs $(a,b)$.

2. Explain the role of $c$.

3. Give two distinct planes satisfying the condition and verify them.

Question 8 – Solution

Strategy. A plane has a constant area-enlargement factor determined only by the magnitude of its gradient.

Step 1: Area equation Since $f_x=a$ and $f_y=b$, $$S=\iint_D\sqrt{1+a^2+b^2}\,dA =\pi\sqrt{1+a^2+b^2}.$$ The condition $S=3\pi$ gives $$1+a^2+b^2=9,\qquad \boxed{a^2+b^2=8}.$$ Thus all possible gradients form the circle of radius $2\sqrt 2$ in the $ab$-plane.

Step 2: Vertical translation The constant $c$ does not appear in either derivative, so $$\boxed{c\text{ is arbitrary}}.$$ Changing $c$ translates the patch vertically without stretching it.

Step 3: Examples and verification Two choices are $$\boxed{z=2\sqrt 2\,x}\qquad\text{and}\qquad \boxed{z=2x+2y+5}.$$ Their gradient-square sums are respectively $8+0$ and $4+4$. Each therefore has area factor $\sqrt{1+8}=3$ and area $3\pi$ above the unit disk.