Q18Mathematics I

Question

Find the minimum value of the function $x + y + z$ subject to the condition $\frac{a}{x} + \frac{b}{y} + \frac{c}{z} = 1$ .

Answer 1

In multivariable calculus, calculating the physical volume of a solid of revolution requires integrating the cross-sectional area along the axis of revolution. When dealing with complex algebraic curves like the astroid (a specific type of hypocycloid with four cusps), standard Cartesian integration becomes exceptionally tedious. The most mathematically elegant and rigorous approach relies entirely on parametric equations.

Step 1: Establishing Parametric Equations

The astroid is implicitly defined by the highly symmetrical Cartesian equation:

x^{2/3} + y^{2/3} = a^{2/3}

To parameterize this curve, we seek functions $x (θ)$ and $y (θ)$ that inherently satisfy this equation for all values of $θ$ . By utilizing the fundamental Pythagorean trigonometric identity $cos^{2} θ + sin^{2} θ = 1$ , we can precisely construct the standard parametric equations:

x = a cos^{3} θ

y = a sin^{3} θ

We can explicitly verify this parameterized mapping: $(a cos^{3} θ)^{2/3} + (a sin^{3} θ)^{2/3} = a^{2/3} cos^{2} θ + a^{2/3} sin^{2} θ = a^{2/3} (1) = a^{2/3}$ . The mapping is absolutely correct.

Step 2: Setting up the Volume Integral

The general formula for the volume $V$ of a solid generated by continuously revolving a mathematical curve $y = f (x)$ entirely around the Cartesian x-axis is rigorously defined as:

V = \int π y^{2} d x

Because the complete astroid is perfectly symmetrical across both the x-axis and the y-axis, revolving the entire curve will create the exact same total volume as revolving just the top half (which spans the first and second quadrants). However, an even simpler and less error-prone mathematical strategy is to calculate the volume generated by revolving only the portion of the curve strictly residing in the first quadrant, and then systematically multiplying that partial result by exactly 2 to account for the left half of the spindle.

In the first quadrant, the geometric curve begins on the x-axis at $x = a$ (which corresponds to the parametric angle $θ = 0$ ) and ends on the y-axis at $x = 0$ (which corresponds to $θ = π /2$ ). When integrating strictly with respect to $x$ from $x = 0$ to $x = a$ , the corresponding angular limits sweep backwards from $θ = π /2$ down to $θ = 0$ .

Before substituting into the volume integral, we must compute the exact differential element $d x$ by differentiating our parametric equation for $x$ strictly with respect to $θ$ :

x = a cos^{3} θ ⟹ d x = 3 a cos^{2} θ \cdot (- sin θ) d θ = - 3 a cos^{2} θ sin θ d θ

We now aggressively substitute all these parameterized components— $y$ , $d x$ , and the new angular limits—back into the master volume equation, remembering to apply the symmetry multiplier of 2:

V = 2 \int_{π /2}^{0} π (a sin^{3} θ)^{2} (- 3 a cos^{2} θ sin θ) d θ

Step 3: Algebraic Simplification of the Integrand

We must meticulously expand the squared term and combine all identical algebraic factors:

V = 2 \int_{π /2}^{0} π (a^{2} sin^{6} θ) (- 3 a cos^{2} θ sin θ) d θ

We pull all constant terms ( $π$ , $a^{2}$ , $- 3 a$ ) entirely out of the integration process:

V = 2 π (- 3 a^{3}) \int_{π /2}^{0} sin^{6} θ \cdot sin θ \cdot cos^{2} θ d θ

V = - 6 π a^{3} \int_{π /2}^{0} sin^{7} θ cos^{2} θ d θ

A fundamental property of definite integrals allows us to instantly reverse the boundary limits by simply multiplying the entire integral by $- 1$ . This operation neatly cancels out the existing negative sign:

V = 6 π a^{3} \int_{0}^{π /2} sin^{7} θ cos^{2} θ d θ

Step 4: Evaluating the Trigonometric Integral via Beta Functions

The remaining definite integral is a classic form that can be solved instantaneously using the absolute, generalized trigonometric formulation of the Beta function. The strict mathematical formula dictates:

\int_{0}^{π /2} sin^{p} θ cos^{q} θ d θ = \frac{Γ ( \frac{p + 1}{2} ) Γ ( \frac{q + 1}{2} )}{2Γ ( \frac{p + q + 2}{2} )}

For our specific integrand, we carefully map the exponents: $p = 7$ and $q = 2$ . Substituting these exactly into the formula yields:

\int_{0}^{π /2} sin^{7} θ cos^{2} θ d θ = \frac{Γ ( \frac{7 + 1}{2} ) Γ ( \frac{2 + 1}{2} )}{2Γ ( \frac{7 + 2 + 2}{2} )} = \frac{Γ ( 4 ) Γ ( 1.5 )}{2Γ ( 5.5 )}

We must now evaluate these specific Gamma functions analytically using the recursive property $Γ (n + 1) = n Γ (n)$ and the foundational known value $Γ (1/2) = π$ :

$Γ (4) = (4 - 1)! = 3! = 6$
$Γ (1.5) = Γ (3/2) = \frac{1}{2} Γ (1/2) = \frac{1}{2} π$
$Γ (5.5) = Γ (11/2) = \frac{9}{2} \cdot \frac{7}{2} \cdot \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot π = \frac{945}{32} π$

Substituting these rigorously computed values back into the Beta fraction:

\frac{Γ ( 4 ) Γ ( 1.5 )}{2Γ ( 5.5 )} = \frac{6 \times ( \frac{1}{2} π )}{2 \times ( \frac{945}{32} π )}

The $π$ terms perfectly cancel out mathematically. We simplify the remaining numeric fraction:

= \frac{3}{2 \times ( \frac{945}{32} )} = \frac{3}{\frac{945}{16}} = \frac{3 \times 16}{945} = \frac{48}{945}

Both the numerator and denominator are strictly divisible by exactly $3$ :

= \frac{16}{315}

Step 5: Finalizing the Volume Calculation

We substitute this exhaustively calculated fractional value back into our master volume equation from Step 3:

V = 6 π a^{3} (\frac{16}{315})

We multiply the integer constants precisely: $6 \times 16 = 96$ .

V = \frac{96 π a ^{3}}{315}

Finally, we reduce the fraction to its most elegant, simplest terms by dividing the numerator and denominator completely by their greatest common divisor, which is $3$ :

V = \frac{32 π a ^{3}}{105}

Through meticulous parametric calculus and advanced Gamma function evaluation, the total volume of the revolved astroid is rigorously proven to be exactly $\frac{32 π a ^{3}}{105}$ .