IBDP Maths AHL 5.17 Area of the region and Volumes of revolution about the x-axis or y-axis AA HL Paper 1- New Syllabus
▶️ Answer/Explanation
Volume formula: \( V = \pi \int_0^{\frac{\sqrt{\pi}}{2}} [f(x)]^2 \, dx \). M1
\( [f(x)]^2 = \left( \sqrt{x \sin(x^2)} \right)^2 = x \sin(x^2) \). A1
So: \( V = \pi \int_0^{\frac{\sqrt{\pi}}{2}} x \sin(x^2) \, dx \).
Substitute: \( t = x^2 \), \( dt = 2x \, dx \), \( x \, dx = \frac{dt}{2} \). Limits: \( x = 0 \to t = 0 \), \( x = \frac{\sqrt{\pi}}{2} \to t = \frac{\pi}{4} \). M1
Integral: \( \pi \int_0^{\frac{\pi}{4}} \sin(t) \cdot \frac{dt}{2} = \frac{\pi}{2} \int_0^{\frac{\pi}{4}} \sin(t) \, dt \). A1
Integrate: \( \frac{\pi}{2} \left[ -\cos(t) \right]_0^{\frac{\pi}{4}} = \frac{\pi}{2} \left( -\cos \frac{\pi}{4} + \cos 0 \right) = \frac{\pi}{2} \left( -\frac{1}{\sqrt{2}} + 1 \right) = \frac{\pi}{2} \cdot \frac{\sqrt{2} – 1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\pi (2 – \sqrt{2})}{4} \). A1
[5 marks]
▶️ Answer/Explanation
Set \( \cos x = \sin 2x \), use \( \sin 2x = 2 \sin x \cos x \): M1
\( \cos x = 2 \sin x \cos x \), so \( \cos x (1 – 2 \sin x) = 0 \).
Solve: \( \cos x = 0 \), \( x = \frac{\pi}{2} \) (point B); or \( \sin x = \frac{1}{2} \), so \( x = \frac{\pi}{6} \), \( \frac{5\pi}{6} \) in \( [0, \pi] \). A1
Thus, point A: \( x = \frac{\pi}{6} \), point C: \( x = \frac{5\pi}{6} \). A1
[3 marks]
Area of R: \( \int_{\frac{\pi}{2}}^{\frac{5\pi}{6}} (\cos x – \sin 2x) \, dx \), since \( \cos x \geq \sin 2x \) in \( \left[ \frac{\pi}{2}, \frac{5\pi}{6} \right] \). M1
Use \( \sin 2x = 2 \sin x \cos x \): \( \int (\cos x – 2 \sin x \cos x) \, dx \). A1
Antiderivative: \( \sin x + \frac{1}{2} \cos 2x \). A1
Evaluate: \( \left[ \sin x + \frac{1}{2} \cos 2x \right]_{\frac{\pi}{2}}^{\frac{5\pi}{6}} = \left( \frac{1}{2} + \frac{1}{4} \right) – \left( 1 – \frac{1}{2} \right) = \frac{3}{4} – \frac{1}{2} = \frac{1}{4} \). A1
[4 marks]