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 answer:

V₁ =  51.2 cm³
V₂ =  50.2655 cm³

Step-by-step explanation:

$$\begin{gathered} a=4\text{cm} \\ v=6\text{cm} \\ f(x)=q{x}^2 \\ f(a/2)=q(a/2{)}^2=v \\ 6=q\cdot 2^2 \\ q=6/2^2=\frac{3}{2}=1.5 \\ {V}_1=\frac{8}{15}\cdot v\cdot a^2=\frac{8}{15}\cdot 6\cdot 4^2=\frac{256}{5}{\text{cm}}^3=51.2{\text{cm}}^3 \end{gathered}$$

$$\begin{gathered} f(x)=v-\frac{6}{4}{x}^2 \\ f(x)=6-\frac{6}{4}{x}^2 \\ {x}_0=-a/2=-4/2=-2 \\ {x}_1=a/2=4/2=2 \\ {V}_2=\pi \cdot {\int }_{\{}{x}_0{\}}^{\{}{x}_1\}f(x)d_x \\ {V}_2=\pi \cdot {\int }_{\{}{x}_0{\}}^{\{}{x}_1\}(\frac{6}{4}{x}^2-v)d_x \\ {V}_2=\pi \cdot [\frac{6}{4}\cdot x^3/3-v_x{]}_{\{}{x}_0{\}}^{\{}{x}_1\} \\ F(x)=6x-\frac{6}{4}\cdot x^3/3 \\ {F}_1=6\cdot x_1-\frac{6}{4}\cdot x_1^3/3=6\cdot 2-\frac{6}{4}\cdot 2^3/3=8 \\ {F}_0=6\cdot x_0-\frac{6}{4}\cdot x_0^3/3=6\cdot (-2)-\frac{6}{4}\cdot (-2{)}^3/3=-8 \\ {V}_2=\pi \cdot (F_1-F_0)=3.1416\cdot (8-(-8))=50.2655{\text{cm}}^3 \end{gathered}$$

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