Problem Analysis #8

To solve the problem, we are given three fixed points in 3D space:

• $A(1,0,0)$
• $B(0,1,0)$
• $C(0,0,1)$

and $D$ is defined as the midpoint of $AC$. Thus,

$$D = \left( \frac{1+0}{2}, \frac{0+0}{2}, \frac{0+1}{2} \right) = \left( \frac{1}{2}, 0, \frac{1}{2} \right)$$

We are to compute the volume $V$ of the solid formed by rotating triangle $ABD$ about the z-axis, and express it in the form $$\frac{p\pi}{q}$$ in lowest terms. Then find $p+q$.

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Step 1: Understand the Geometry

Triangle $ABD$ lies in the plane defined by the equation:

$$x + y + z = 1$$

This can be verified by checking that all three points $A(1,0,0)$, $B(0,1,0)$, and D\left($$\frac{1}{2}$$, 0, $$\frac{1}{2}$$\right) satisfy this equation.

Now, for a fixed value of $z$, the intersection of the triangle with the horizontal plane $z = \text{const}$ is a line segment within the xy-plane (since $x + y = 1 - z$). This line segment is bounded between two points:

• One on the edge $AD$: $x = 1 - z, y = 0$
• One on the edge $BD$: $x = z, y = 1 - 2z$

Thus, for each z $$\in$$ [0, $$\frac{1}{2}$$], the line segment has endpoints:

• $P = (z, 1 - 2z)$
• $Q = (1 - z, 0)$

When this line segment is rotated about the z-axis, it sweeps out an annulus in the solid at height $z$. The outer radius is $R(z) = 1 - z$ (from point $Q$), and the inner radius depends on whether the perpendicular from the origin to the line lies within the segment or not.

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Step 2: Determine Inner Radius

The line segment in the plane $z = \text{const}$ lies along the line:

$$y = 1 - z - x$$

The minimum distance from the origin (z-axis) to this line occurs at the point where the perpendicular from the origin meets the line. That point is:

$$x = y = \frac{1 - z}{2}$$

This point lies on the segment only if x $$\geq$$ z, i.e., $$\frac{1 - z}{2}$$ $$\geq$$ z $$\Rightarrow$$ z $$\leq$$ $$\frac{1}{3}$$

So, for z $$\in$$ [0, $$\frac{1}{3}$$], the minimum radius is:

$$r_{\min}(z) = \frac{1 - z}{\sqrt{2}}$$

and for z $$\in$$ $$\frac{1}{3}, \frac{1}{2}$$, the minimum radius is:

$$r_{\min}(z) = \sqrt{z^2 + (1 - 2z)^2} = \sqrt{5z^2 - 4z + 1}$$

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Step 3: Compute the Volume

The volume of revolution is given by:

$$V = \pi \int_0^{1/2} \left[ R(z)^2 - r_{\min}(z)^2 \right] dz$$

We split the integral into two parts:

Part 1: z $$\in$$ [0, \tfrac{1}{3}]

Here, r_{$$\min$$}(z) = $$\frac{1 - z}{\sqrt{2}}$$, so:

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

Part 2: z $$\in$$ $$\tfrac{1}{3}, \tfrac{1}{2}$$

Here, r_{$$\min$$}(z)^2 = 5z^2 - 4z + 1, and:

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

Compute at bounds:

• At z = $$\frac{1}{2}$$: -$$\frac{4}{24}$$ + $$\frac{1}{4}$$ = -$$\frac{1}{6}$$ + $$\frac{1}{4}$$ = $$\frac{1}{12}$$
• At z = $$\frac{1}{3}$$: -$$\frac{4}{81}$$ + $$\frac{1}{9}$$ = $$\frac{5}{81}$$

$$I_2 = \frac{1}{12} - \frac{5}{81} = \frac{27 - 20}{324} = \frac{7}{324}$$

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Step 4: Total Volume

$$V = \pi(I_1 + I_2) = \pi \left( \frac{19}{162} + \frac{7}{324} \right) = \pi \left( \frac{38 + 7}{324} \right) = \pi \cdot \frac{45}{324} = \pi \cdot \frac{5}{36}$$

Thus, V = $$\frac{5\pi}{36}$$, so $p = 5$, $q = 36$, and:

$$p + q = 5 + 36 = \boxed{41}$$