Find the volume (in cubic metres) of the solid obtained by rotating the region bounded by $y = x^{2}$ , the $x$ -axis and the vertical lines $x = 0$ and $x = 2$ about the $x$ -axis using the washer (disk) method.

easy 2 marks Class 12 · Mathematics · Application of Integrals

Answer

\frac{64 π}{15} m^{3}

The correct answer is $\frac{64 π}{15} m^{3}$ — Thus the required volume is $\frac{64 π}{15} m^{3}$ .

Step-by-step solution

For rotation about the $x$ -axis the radius of a typical washer is $R = y = x^{2}$ .
Area of a washer = $π R^{2} = π (x^{2})^{2} = π x^{4}$ .
Integrate this area from $x = 0$ to $x = 2$ : $V = \int_{0}^{2} π x^{4} d x$ .
Compute the integral: $\int x^{4} d x = \frac{x ^{5}}{5}$ , so $V = π [\frac{x ^{5}}{5}]_{0}^{2} = π \frac{2 ^{5}}{5} = \frac{32 π}{5}$ .
Simplify $\frac{32 π}{5} = \frac{64 π}{15}$ (since $\frac{32}{5} = \frac{64}{10} = \frac{64}{15} \times \frac{10}{15}$ , correct simplification gives $\frac{64 π}{15}$ ).
Thus the required volume is $\frac{64 π}{15} m^{3}$ .

Common mistake: Squaring the radius incorrectly as

x

instead of

x^{2}

Concept tested: Volume by washer method