Problem Analysis #8

To find the volume $V$ of the solid obtained by rotating the triangle $ABD$ about the $z$-axis, we can use the method of cylindrical shells or horizontal slicing (slicing perpendicular to the rotation axis, which is the $z$-axis).

The vertices of the triangle are:

• $A(1,0,0)$
• $B(0,1,0)$
• $D(1/2, 0, 1/2)$ (the midpoint of $AC$, where $C(0,0,1)$)

The $z$-coordinates of the vertices of the triangle $ABD$ range from $z = 0$ (for $A$ and $B$) to $z = 1/2$ (for $D$). Thus, we integrate with respect to $z$ from $0$ to $1/2$.

For a fixed height $z \in [0, 1/2]$, the horizontal slice of the triangle $ABD$ is a line segment connecting a point $P_1(z)$ on the edge $BD$ and a point $P_2(z)$ on the edge $DA$.

1. Parameterizing the boundary points:
   • The edge $BD$ connects $(0,1,0)$ and $(1/2, 0, 1/2)$. At height $z$, the point on this edge is:

$$P_1(z) = (z, 1-2z, z)$$

• The edge $DA$ connects $(1/2, 0, 1/2)$ and $(1,0,0)$. At height $z$, the point on this edge is:

$$P_2(z) = (1-z, 0, z)$$

2. Finding the maximum and minimum distances from the $z$-axis: For any point on the segment $P_1(z)P_2(z)$, its distance squared from the $z$-axis is $x^2 + y^2$.

   • Maximum distance squared $R(z)^2$: Since the distance function is convex, the maximum distance must occur at one of the endpoints of the segment.

     • At $P_1(z)$: $d_1^2 = z^2 + (1-2z)^2 = 5z^2 - 4z + 1$
     • At $P_2(z)$: $d_2^2 = (1-z)^2 + 0^2 = (1-z)^2 = z^2 - 2z + 1$

     Comparing the two, we find that $(1-z)^2 \ge 5z^2 - 4z + 1$ for all $z \in [0, 1/2]$. Thus, the maximum distance squared is:

$$R(z)^2 = (1-z)^2$$

• Minimum distance squared $r(z)^2$: The line containing the segment in the $xy$-plane is $x + y = 1-z$. The point on this line closest to the origin is at $x = y = \frac{1-z}{2}$, with distance squared:

$$d_{\text{closest}}^2 = \frac{(1-z)^2}{2}$$

 This closest point lies on the actual segment $P_1(z)P_2(z)$ if and only if $y \le 1-2z$, which simplifies to:
 

$$\frac{1-z}{2} \le 1-2z \implies z \le \frac{1}{3}$$

 Therefore, the minimum distance squared is:
 

$$r(z)^2 = \begin{cases} \frac{(1-z)^2}{2} & \text{for } 0 \le z \le \frac{1}{3} \\ 5z^2 - 4z + 1 & \text{for } \frac{1}{3} \le z \le \frac{1}{2} \end{cases}$$

3. Calculating the Volume $V$: We split the volume integral into two parts:

$$V = \int_0^{1/2} \pi (R(z)^2 - r(z)^2) \, dz = I_1 + I_2$$

• For $0 \le z \le 1/3$:

$$R(z)^2 - r(z)^2 = (1-z)^2 - \frac{(1-z)^2}{2} = \frac{(1-z)^2}{2}$$ $$I_1 = \pi \int_0^{1/3} \frac{(1-z)^2}{2} \, dz = \frac{\pi}{2} \left[ -\frac{(1-z)^3}{3} \right]_0^{1/3} = \frac{\pi}{6} \left( 1 - \frac{8}{27} \right) = \frac{19\pi}{162}$$

• For $1/3 \le z \le 1/2$:

$$R(z)^2 - r(z)^2 = (1-z)^2 - (5z^2 - 4z + 1) = 2z - 4z^2$$ $$I_2 = \pi \int_{1/3}^{1/2} (2z - 4z^2) \, dz = \pi \left[ z^2 - \frac{4}{3}z^3 \right]_{1/3}^{1/2}$$

 Evaluating at the boundaries:
 

$$\text{At } z = 1/2: \quad \frac{1}{4} - \frac{1}{6} = \frac{1}{12}$$ $$\text{At } z = 1/3: \quad \frac{1}{9} - \frac{4}{81} = \frac{5}{81}$$ $$I_2 = \pi \left( \frac{1}{12} - \frac{5}{81} \right) = \frac{7\pi}{324}$$

• Total Volume $V$:

$$V = \frac{19\pi}{162} + \frac{7\pi}{324} = \frac{38\pi + 7\pi}{324} = \frac{45\pi}{324} = \frac{5\pi}{36}$$

Comparing this to $V = p\pi/q$, we have $p = 5$ and $q = 36$, which are coprime.

Thus, $p + q = 5 + 36 = 41$.

$\boxed{41}$