Surface Area of a Cone in the First Octant: 04-BS-1, May 2013, Q5

Question 5: 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. Surface $z=f(x,y)=1-\sqrt{x^2+y^2}$ (a downward cone with apex at $(0,0,1)$, meeting the $xy$-plane at the unit circle); the region of interest is the first-octant portion ($x\ge0,\ y\ge0,\ z\ge0$).

Find. The surface area of that portion.

Figure — cone z = 1 − √(x²+y²) over the quarter-disk r≤1 in the first octant; surface area = √2 × (quarter-disk area).

Approach. The first octant restricts $x,y\ge0$; since $z\ge0$ requires $\sqrt{x^2+y^2}\le1$, the projected region $D$ is the quarter unit disk ($x,y\ge0,\ x^2+y^2\le1$). Compute the surface-area integrand $\sqrt{1+f_x^2+f_y^2}$ and integrate over $D$ in polar coordinates.

Compute the surface-area integrand. With $r=\sqrt{x^2+y^2}$, $f_x=-x/r,\ f_y=-y/r$, so $$1+f_x^2+f_y^2 = 1+\frac{x^2+y^2}{r^2}=1+1=2 \quad\Rightarrow\quad \sqrt{1+f_x^2+f_y^2}=\sqrt2\ \text{(constant)}.$$
Identify the projected region. $D=\{x\ge0,y\ge0,x^2+y^2\le1\}$, the quarter unit disk, with area $\dfrac{\pi(1)^2}{4}=\dfrac{\pi}{4}$.
Integrate. Because the integrand is constant, $$S=\iint_D \sqrt2\,dA=\sqrt2\cdot\frac{\pi}{4}=\boxed{\dfrac{\pi\sqrt2}{4}}\approx 1.111.$$

Quantity	Result
Surface area	$\dfrac{\pi\sqrt2}{4}\approx 1.111$ (square units)

Concept & Theory

Topic. Surface area of an explicit surface $z=f(x,y)$ over a plane region.

Key relations. $S=\iint_D \sqrt{1+f_x^2+f_y^2}\,dA$; polar coordinates $x=r\cos\theta,y=r\sin\theta,dA=r\,dr\,d\theta$ for circular regions.

Why this works. A cone has constant slope magnitude everywhere except at the apex, so its surface-area integrand collapses to a constant ($\sqrt2$ here) — surface area is then just the constant times the base area, exactly like the lateral-surface-area formula for a right circular cone ($\pi r \ell$ generalizes this idea). Restricting to the first octant simply restricts the base region to a quarter disk.

Common pitfall. Trying to integrate in Cartesian coordinates over the quarter disk (messy square-root limits) instead of switching to polar, where the region and the (constant) integrand are both trivial.

Source. Stewart, Calculus: Early Transcendentals, Ch. 16 (Vector Calculus, §16.6 Surface Area) and Ch. 15 (§15.4 Double Integrals in Polar Coordinates).

Question 5 of 8: Surface Area of a Cone in the First Octant

Question 5: Surface Area of a Cone in the First Octant (20 marks)