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

Question 8: 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. Cone $z=1-\sqrt{x^2+y^2}$ (apex at $(0,0,1)$, base circle $x^2+y^2=1$ at $z=0$); first-octant restriction $x,y,z\ge0$.

Find. The surface area of the first-octant portion.

Approach. Compute $\sqrt{1+z_x^2+z_y^2}$, which is constant on a $45^\circ$ cone, then multiply by the area of the projected region: $z\ge0$ restricts to $r\le1$, and $x,y\ge0$ restricts to the first quadrant, together giving a quarter-disk of radius $1$.

1. Constant slope factor. With $r=\sqrt{x^2+y^2}$, $z_x=-x/r$, $z_y=-y/r$, so $$z_x^2+z_y^2=\frac{x^2+y^2}{r^2}=1\ \Longrightarrow\ \sqrt{1+z_x^2+z_y^2}=\sqrt2\ \text{(constant, independent of position on the cone).}$$
2. Projected region. The first octant requires $x\ge0,\,y\ge0$ (first quadrant of the $xy$-plane) and $z\ge0\Rightarrow r\le1$ (inside the unit circle). The projection is therefore the quarter-disk $0\le r\le1,\ 0\le\theta\le\tfrac\pi2$, with area $\tfrac14\pi(1)^2=\tfrac\pi4$.
3. Surface area. Since the slope factor is constant, the surface-area integral collapses to a product: $$\text{Area}=\sqrt2\times\frac\pi4=\frac{\pi\sqrt2}4.$$

$$\text{Area}=\boxed{\dfrac{\pi\sqrt2}4}\approx1.111$$