Question
Consider the function:
\[ f(x) = x\sqrt{3-x^2}, \quad -\sqrt{3} \leq x \leq \sqrt{3} \]
(a) Sketch the graph of \( y = f(x) \) for \( -\sqrt{3} \leq x \leq \sqrt{3} \).
The area between the graph of \( y = f(x) \) and the \( x \)-axis is rotated through \( 360^{\circ} \) about the \( x \)-axis.
(b) (i) Write down an integral that represents this volume.
(ii) Calculate the value of this integral.
The graph of the function \( f \) is transformed to give the function \( g \), in the following way:
- It is first stretched by scale factor 2 parallel to the \( x \)-axis, keeping the \( y \)-axis invariant.
- It is then stretched by scale factor 0.5 parallel to the \( y \)-axis, keeping the \( x \)-axis invariant.
(c) Find the volume obtained when the area between the graph of \( y = g(x) \) and the \( x \)-axis is rotated through \( 360^{\circ} \) about the \( x \)-axis.
▶️ Answer/Explanation
Detailed Solution
(a) Graph of \( f(x) \)
(b) (i) Writing the integral for the volume
The volume of revolution about the \( x \)-axis is given by:
\[ V = \pi \int_{-\sqrt{3}}^{\sqrt{3}} (f(x))^2 \,dx \]
Substituting \( f(x) \):
\[ V = \pi \int_{-\sqrt{3}}^{\sqrt{3}} (x\sqrt{3-x^2})^2 \,dx \]
or alternatively, using symmetry:
\[ V = 2\pi \int_0^{\sqrt{3}} (x\sqrt{3-x^2})^2 \,dx \]
(ii) Evaluating the integral
Computing the integral:
\[ V = \frac{12\pi \sqrt{3}}{5} \]
\[ V \approx 13.1 \quad (13.0593…) \]
(c) Finding the volume for \( g(x) \)
Since \( g(x) \) is transformed as:
\[ g(x) = \frac{1}{2} \left( \frac{x}{2} \sqrt{3 – \left(\frac{x}{2}\right)^2} \right) \]
or equivalently, expressed in terms of \( f(x) \):
\[ g(x) = \frac{1}{2} f\left(\frac{x}{2}\right) \]
The limits for \( g(x) \) are stretched to \( -2\sqrt{3} \leq x \leq 2\sqrt{3} \).
The volume of revolution is given by:
\[ V_g = \pi \int_{-2\sqrt{3}}^{2\sqrt{3}} (g(x))^2 \,dx \]
Substituting \( g(x) \):
\[ V_g = \pi \int_{-2\sqrt{3}}^{2\sqrt{3}} \left(\frac{1}{2} f\left(\frac{x}{2}\right)\right)^2 dx \]
Using the transformation rule, the volume scales as:
\[ V_g = 2 \times \frac{1}{2} \times V \]
\[ V_g = \frac{1}{2} \times \frac{12\pi \sqrt{3}}{5} \]
\[ V_g = \frac{6\pi \sqrt{3}}{5} \]
\[ V_g \approx 6.53 \quad (6.52967…) \]
……………………………Markscheme……………………………….
(b) (i)
\( V = \pi \int_{-\sqrt{3}}^{\sqrt{3}} (x\sqrt{3-x^2})^2 dx \) OR \( V = 2\pi \int_{0}^{\sqrt{3}} (x\sqrt{3-x^2})^2 dx \)
(ii)
\( V = 13.1 \) (or exact value \( \frac{12\pi \sqrt{3}}{5} \))
(c)
Correct transformation to \( g(x) \), correct limits \( -2\sqrt{3} \) to \( 2\sqrt{3} \)
Final volume \( V_g = 6.53 \) (or exact value \( \frac{6\pi \sqrt{3}}{5} \))