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 =  50.2655 cm³

Solution:

$$\begin{gathered} a=4\text{ cm } \\ v=6\text{ cm } \\ f(x)=q{x}^2 \\ f(a\mathrm{/}2)=q(a\mathrm{/}2{)}^2=v \\ 6=q\cdot 2^2 \\ {V}_1=q=6\mathrm{/}{2}^2={\displaystyle \frac{3}{2}}=1.5=0{\text{cm}}^3 \end{gathered}$$

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

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