Surface Area of a Cone in the First Octant: 04-BS-1, May 2016, Q7

Question 7: Surface Area of a Cone in the First Octant 20 marks

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The cone $z=1-r$ (with $r=\sqrt{x^2+y^2}$), restricted to the first octant $x,y,z\ge0$.

Find. The surface area of that portion.

Approach. Use the surface-area formula $A=\iint_D\sqrt{1+z_x^2+z_y^2}\,dA$ over the projected region $D$, then identify $D$ from the first-octant restriction.

Determine the projected region $D$. $z\ge0\Rightarrow r\le1$. Combined with $x,y\ge0$ (first quadrant of the $xy$-plane), $D$ is the quarter-disk $0\le r\le1$, $0\le\theta\le\pi/2$.
Compute the surface-area integrand. With $z=1-\sqrt{x^2+y^2}$, $$z_x=-\frac{x}{r},\qquad z_y=-\frac{y}{r} \;\Rightarrow\; z_x^2+z_y^2=\frac{x^2+y^2}{r^2}=1.$$ So the integrand is constant: $\sqrt{1+z_x^2+z_y^2}=\sqrt2$.
Integrate over the quarter-disk. Since the integrand is constant, the surface area is simply $\sqrt2$ times the area of $D$: $$A=\sqrt2\cdot\left(\frac{\pi(1)^2}{4}\right)=\boxed{\dfrac{\pi\sqrt2}{4}}\approx1.111$$

Quantity	Result
Projected region $D$	Quarter-disk, $0\le r\le1$, $0\le\theta\le\pi/2$
Surface-area integrand	$\sqrt2$ (constant — the cone has constant slope)
Surface area $A$	$\dfrac{\pi\sqrt2}{4}\approx1.111$

Concept & Theory

Topic. Surface area of an explicit surface $z=f(x,y)$ via the formula $A=\iint_D\sqrt{1+f_x^2+f_y^2}\,dA$.

Key relations. For a right circular cone $z=h-\alpha r$, the slope magnitude $\sqrt{z_x^2+z_y^2}=\alpha$ is the same at every point (constant zenith angle), so the area element scales by the constant factor $\sqrt{1+\alpha^2}$ everywhere — the surface area is just $\sqrt{1+\alpha^2}$ times the projected area, with no integration needed beyond finding that area.

Why this works. Because $r=\sqrt{x^2+y^2}$ has partial derivatives $r_x=x/r$, $r_y=y/r$, one finds $z_x^2+z_y^2=(x^2+y^2)/r^2\equiv1$ identically — independent of position on the cone. This uniform surface-area density is a special feature of a cone with unit slope (more generally, any ruled surface of revolution with fixed slope $\alpha$ scales by the constant factor $\sqrt{1+\alpha^2}$ everywhere), and it is exactly what converts what looks like a genuine double integral into a one-step area calculation.

Common pitfall. Forgetting that "first octant" restricts the $xy$-projection to a quarter-disk rather than the full unit disk, which gives an answer twice too large if only $z\ge0$ is enforced without also requiring $x,y\ge0$, or four times too large if the radius restriction $r\le1$ is dropped entirely. Sign errors in $z_x,z_y$ individually do not affect this particular problem (since both are squared before use), but the same sign bookkeeping matters in general surface-area problems where cross terms or asymmetric integrands appear.

Source. Stewart, Calculus: Early Transcendentals, Ch. 15 (§15.6 Surface Area).

Question 7 of 8: Surface Area of a Cone in the First Octant

Question 7: Surface Area of a Cone in the First Octant 20 marks