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.

Correct 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 \ \text{cm} \ \\ v=6 \ \text{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 \ \text{cm}^3
f(x)=v−64x2 f(x)=6−64x2  x0=−a/2=−4/2=−2 x1=a/2=4/2=2  V2=π⋅ ∫{x0}{x1}f(x)dx V2=π⋅ ∫{x0}{x1}(64x2−v)dx V2=π⋅ [64⋅ x3/3−vx]{x0}{x1}  F(x)=6x−64⋅ x3/3 F1=6⋅ x1−64⋅ x13/3=6⋅ 2−64⋅ 23/3=8 F0=6⋅ x0−64⋅ x03/3=6⋅ (−2)−64⋅ (−2)3/3=−8  V2=π⋅ (F1−F0)=3.1416⋅ (8−(−8))=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))=50.265 \ \text{cm}^3

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