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dabicvietboi
12.04.2021 •
Mathematics
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Ответ:
In attached diagram, we can see 2 circumferences with radius r, and separated centers by r/2.
The solution is:
V = (11/12)×π×r³
Let´s call circumferences 1 and 2, by symmetry, rotating area A will produce a volume V₁ identical a V₂, Obtained by rotating area B ( both around x-axis), then the whole volume V will be:
V = 2× V₁
V₁ = ∫π×y²×dx (1)
Now
( x - r/2)² + y² = r² the equation of circumference 1
y² = r² - ( x - r/2)²
Plugging this value in equation (1)
V₁ = ∫π×[ r² - ( x - r/2)²]×dx with integrations limits 0 ≤ x ≤ r/2
V₁ = π×∫ ( r² - x² + (r/2)² - r×x )×dx
V₁ = π× [ r²×x - x³/3 + (r/2)²×x - (1/2) ×r×x²] evaluate between 0 and r/2
V₁ = π× [(5/4)×r²×x - x³/3 - (1/2) ×r×x²]
V₁ = π× [(5/4)×r² × ( r/2 - 0 ) - (1/3)×(r/2)³ - (1/2) ×r×(r/2)²]
V₁ = π× [ (5/8)×r³ - r³/24 - r³/8]
V₁ = π× (11/24)×r³
Then
V = 2× V₁
V = 2×π×11/24)×r³
V = (11/12)×π×r³
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