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)=qx2 f(a/2)=q(a/2)2=v 6=q 22  V1=q=6/22=32=1.5=0 cm3a = 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)=v64x2 f(x)=664x2  x0=a/2=4/2=2 x1=a/2=4/2=2  V2=π {x0}{x1}f(x)dx V2=π {x0}{x1}(64x2v)dx V2=π [64 x3/3vx]{x0}{x1}  F(x)=6x64 x3/3 F1=6 x164 x13/3=6 264 23/3=8 F0=6 x064 x03/3=6 (2)64 (2)3/3=8  V2=π (F1F0)=3.1416 (8(8))50.2655=50.265 cm3f(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



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