# The parabolic segment

The parabolic segment has a base a = 4 cm and a height v = 6 cm. Calculate the volume of the body that results from the rotation of this segment

a) around its base
b) around its axis.

Result

V1 =  0 cm3
V2 =  50.265 cm3

#### Solution:

$a = 4 \ cm \ \\ v = 6 \ cm \ \\ f(x) = q x^2 \ \\ f(a/2) = q (a/2)^2 = v \ \\ 6 = q \cdot \ 2^2 \ \\ \ \\ V_{1}=q = 6/2^2 = \dfrac{ 3 }{ 2 } = 1.5 = 0 \ cm^3$
$f(x) = v - \dfrac{ 6 }{ 4 } x^2 \ \\ f(x) = 6 - \dfrac{ 6 }{ 4 } x^2 \ \\ \ \\ x_{ 0 } = -a/2 = -4/2 = -2 \ \\ x_{ 1 } = a/2 = 4/2 = 2 \ \\ \ \\ V_{ 2 } = \pi \cdot \ \int_\{ x_{ 0 } \}^\{x_{ 1 }\} f(x) dx \ \\ V_{ 2 } = \pi \cdot \ \int_\{x_{ 0 }\}^\{x_{ 1 }\} (\dfrac{ 6 }{ 4 } x^2 - v) dx \ \\ V_{ 2 } = \pi \cdot \ [\dfrac{ 6 }{ 4 } \cdot \ x^3/3 - vx]_\{x_{ 0 }\}^\{x_{ 1 }\} \ \\ \ \\ F(x) = 6x - \dfrac{ 6 }{ 4 } \cdot \ x^3/3 \ \\ F_{ 1 } = 6 \cdot \ x_{ 1 } - \dfrac{ 6 }{ 4 } \cdot \ x_{ 1 }^3/3 = 6 \cdot \ 2 - \dfrac{ 6 }{ 4 } \cdot \ 2^3/3 = 8 \ \\ F_{ 0 } = 6 \cdot \ x_{ 0 } - \dfrac{ 6 }{ 4 } \cdot \ x_{ 0 }^3/3 = 6 \cdot \ (-2) - \dfrac{ 6 }{ 4 } \cdot \ (-2)^3/3 = -8 \ \\ \ \\ V_{ 2 } = \pi \cdot \ (F_{ 1 } - F_{ 0 }) = 3.1416 \cdot \ (8 - (-8)) \doteq 50.2655 = 50.265 \ cm^3$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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