Question

The volume of the solid obtained by revolving the region bounded by the curve y = x - x² and the x-axis From x=0 to x=1 will be:
1. \(\dfrac{\pi}{24}\) cubic units
2. \(\dfrac{\pi}{15}\) cubic units
3. \(\dfrac{\pi}{30}\) cubic units
4. \(\dfrac{\pi}{25}\) cubic units

Answer 1

Correct Answer - Option 3 : \(\dfrac{\pi}{30}\) cubic units

Concept:

The volume of the solid formed by revolving the region bounded by the curve $y=f\left(x\right)$ and the $x-axis\; between$$x=a$ and $x=b$ about the $x-$axis is given by

\(\rm V =\pi\; \int_{a}^{b}y^2dx\)

Calculation:

Given:

The curve is y = (x - x2)

\(\rm V =\pi\; \int_{a}^{b}y^2dx\)

\(\rm V = \pi\;\int_{0}^{1}(x-x^2)^2dx\)

\(\rm V = \pi\;\int_{0}^{1}(x^2+x^4-2x^3)dx\)

\(\rm V = \pi[\frac{x^3}{3}+\frac{x^5}{5}-\frac{2x^4}{4}]_0^1\)

\(\rm V = \pi(\frac{(1)^3}{3}+\frac{(1)^5}{5}-\frac{2(1)^4}{4})\)

\(\rm V = \pi(\frac{1}{3}+\frac{1}{5}-\frac{1}{2})\)

\(\rm V = \pi(\frac{10+6-15}{30})\)

\(V=\frac{\pi}{30}\ unit^3\)

Hence the volume of the solid will be \(\frac{\pi}{30}\) cubic unit.