By using the substitution ${3(y - 1) = 5\sin\theta}$, prove that \(\displaystyle\displaystyle

Question

By using the substitution ${3(y - 1) = 5\sin\theta}$, prove that $\displaystyle\displaystyle \displaystyle\int_{\frac{1}{6}}^{\frac{8}{3}}{\sqrt{25-9{{\left( y-1 \right)}^{2}}}}\ \mathrm{d}y=25\left( \frac{\pi }{9}+\frac{\sqrt{3}}{24} \right)$. The curve ${C}$ has equation ${(x - 1)^2 + 9(y - 1)^2 = 25}$.[4] The region ${R}$ is bounded by the curve ${C}$ and the lines ${3y - 2x - 6 = 0}$ and ${y = \frac{1}{6}}$ as shown in the diagram below. Find the exact area of the region ${R}$.[3] The region ${S}$ is the region inside the curve ${C}$ such that ${3y - 2x - 6 \geq 0}$. Find the volume of the solid formed when the region ${S}$ is rotated about the x-axis through four right angles.[4]

Accepted Answer

(a)(i) $25\left(\dfrac\pi9+\dfrac{\sqrt3}{24}\right)$ (ii) $25\left(\dfrac\pi9+\dfrac{\sqrt3}{24}-\dfrac3{16}\right)\text{ units}^2$ (b) $55.6\text{ units}^3$

2009 AJC P1 Q14

2009 AJC P1 Q14

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