2013 H2 Mathematics Paper 1 Question 5

Definite Integrals: Areas and Volumes

Answers

(i)

(ii)

112aπ{\frac{1}{12} a \pi}

Full solutions

(i)

(ii)

dxdθ=acos⁡θ\frac{\mathrm{d}x}{\mathrm{d}\theta} = a \cos \theta
When x=12a,{x=\frac{1}{2}a,} θ=π6{\theta=\frac{\pi}{6}}
When x=32a,{x=\frac{\sqrt{3}}{2}a,} θ=π3{\theta=\frac{\pi}{3}}
∫12a32af(x)  dx=∫12a32a1−x2a2  dx=∫π6π31−a2sin⁡2θa2acos⁡θ  dθ=a∫π6π31−sin⁡2θcos⁡θ  dθ=a∫π6π3cos⁡2θ  dθ=a∫π6π3cos⁡2θ+12  dθ=a[14sin⁡2θ+12θ]π6π3  dθ=a(14(sin⁡2π3−sin⁡π3))+12(π6−π12))=a(112π)=112aπ  ■\begin{align*} & \int_{\frac{1}{2}a}^{\frac{\sqrt{3}}{2}a} f(x) \; \mathrm{d}x \\ & = \int_{\frac{1}{2}a}^{\frac{\sqrt{3}}{2}a} \sqrt{1-\frac{x^2}{a^2}} \; \mathrm{d}x \\ & = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1-\frac{a^2 \sin^2 \theta}{a^2}} a \cos \theta \; \mathrm{d}\theta \\ & = a \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1-\sin^2 \theta} \cos \theta \; \mathrm{d}\theta \\ & = a \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \cos^2 \theta \; \mathrm{d}\theta \\ & = a \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\cos 2 \theta + 1}{2} \; \mathrm{d}\theta \\ & = a \left[ \frac{1}{4} \sin 2 \theta + \frac{1}{2} \theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{3}} \; \mathrm{d}\theta \\ & = a \Bigg( \frac{1}{4} \left( \sin \frac{2\pi}{3} - \sin \frac{\pi}{3}) \right) + \frac{1}{2} \left( \frac{\pi}{6} - \frac{\pi}{12} \right) \Bigg) \\ & = a \left( \frac{1}{12} \pi \right) \\ & = \frac{1}{12} a \pi \; \blacksquare \end{align*}