Using the method of cylindrical shells, find the volume (in cubic metres) of the solid generated when the region bounded by $y = x^{2}$ , the $x$ ‑axis and the vertical lines $x = 0$ and $x = 2$ is revolved about the line $x = 3$ .

medium 3 marks Class 12 · Mathematics · Application of Integrals

Answer

8 π m^{3}

The correct answer is $8 π m^{3}$ — Therefore the volume of the solid is $8 π m^{3}$ .

Step-by-step solution

For a vertical strip at position $x$ , the radius of the shell is the horizontal distance to the axis $x = 3$ , i.e. $r = 3 - x$ .
The height of the shell is given by the function $y = x^{2}$ .
The differential volume of a shell is $d V = 2 π (r a d i u s) (h e i g h t) d x = 2 π (3 - x) x^{2} d x$ .
Integrate from $x = 0$ to $x = 2$ : $V = 2 π \int_{0}^{2} (3 x^{2} - x^{3}) d x$ .
Compute the integral: $\int_{0}^{2} (3 x^{2} - x^{3}) d x = [x^{3} - \frac{x ^{4}}{4}]_{0}^{2} = 8 - \frac{16}{4} = 8 - 4 = 4$ .
Multiply by $2 π$ : $V = 2 π \times 4 = 8 π$ .
Therefore the volume of the solid is $8 π m^{3}$ .

Common mistake: Using

r = x - 3

instead of

3 - x

, which changes the sign of the integrand and leads to a negative volume.

Concept tested: Volume by cylindrical shells