Correct answer: \( V = \pi (\sqrt{3} + 4\ln 2 + \pi) \approx 24.0 \).

Working:

Inverse function: \( y = \arctan(x – 2) \implies x = \tan y + 2 \).

Limits: At \( x = 2 \), \( y = \arctan(0) = 0 \).

At \( x = 2 + \sqrt{3} \), \( y = \arctan(\sqrt{3}) = \frac{\pi}{3} \).

Volume by disk method: \( V = \pi \int_0^{\frac{\pi}{3}} (x)^2 \, dy \).

\( x = \tan y + 2 \), so \( x^2 = (\tan y + 2)^2 \).

Expand: \( (\tan y + 2)^2 = \tan^2 y + 4\tan y + 4 \).

Use identity: \( \tan^2 y = \sec^2 y – 1 \).

\( (\tan y + 2)^2 = \sec^2 y – 1 + 4\tan y + 4 = \sec^2 y + 4\tan y + 3 \).

Integrate: \( V = \pi \int_0^{\frac{\pi}{3}} (\sec^2 y + 4\tan y + 3) \, dy \).

Antiderivative: \( \left[ \tan y – 4\ln|\cos y| + 3y \right] \).

Evaluate at \( y = \frac{\pi}{3} \): \( \tan \frac{\pi}{3} = \sqrt{3} \), \( \cos \frac{\pi}{3} = \frac{1}{2} \), \( -\ln \frac{1}{2} = \ln 2 \), \( 3 \cdot \frac{\pi}{3} = \pi \).

\[ = \sqrt{3} + 4\ln 2 + \pi \]

Evaluate at \( y = 0 \): \( \tan 0 = 0 \), \( \ln|\cos 0| = 0 \), \( 3 \cdot 0 = 0 \).

\[ V = \pi [ (\sqrt{3} + 4\ln 2 + \pi) – 0 ] = \pi (\sqrt{3} + 4\ln 2 + \pi) \]

Numerical value: \( \approx 24.0 \) (3 significant figures).

Markscheme:

Attempts to express \( x \) in terms of \( \tan y \) (M1)

\( x = \tan y + 2 \) (A1)

Let \( V \) be the volume of the solid

Correctly uses \( V = \pi \int_{a}^{b} x^2 dy \) (M1)

Note: Award M0 for \( V = \pi \int_{a}^{b} (\arctan (x – 2))^2 dy \)

\[ V = \pi \int_{0}^{\frac{\pi}{3}} (\tan y + 2)^2 dy = \pi \int_{0}^{1.0472…} (\tan y + 2)^2 dy \] (A1)

\[ = 24.0213… \]

\[ = 24.0 \ (-\pi (4 \ln 2 + \pi + \sqrt{3})) \] (A1)

Note: GDC in degrees gives 13.3