By writing $x=A\left( 2-2x \right)+B$, where $A$ and $B$ are constants, find \(\displayst

Question

By writing $x=A\left( 2-2x \right)+B$, where $A$ and $B$ are constants, find $\displaystyle\displaystyle \displaystyle\int{\frac{x}{\sqrt{3+2x-{{x}^{2}}}}\mathrm{d}x}$.[3] The use of surfaces of revolution is essential in many fields of physics and engineering. In particular, the area of the surface generated when the arc of the curve with parametric equations $x=x\left( t \right)$, $y=y\left( t \right)$ from the point corresponding to the parameter $t={{t}_{1}}$ to the point corresponding to the parameter $t={{t}_{2}}$ is rotated through $2\pi $ radians about the $y$-axis is given by $A_y = 2\pi \int_{t_1}^{t_2} x(t) \sqrt{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2} \mathrm{d}t.$ The curve $C$ has parametric equations $x={{\mathrm{e}}^{t}}-t$, $y=4{{\mathrm{e}}^{\frac{1}{2}t}}$, for $0\le t\le 1$. Find the exact surface area generated when $C$ is rotated through $2\pi $ radians about the $y$-axis.[5]

Accepted Answer

(a) $-\sqrt{3+2x-x^2}+\sin^{-1}\left(\dfrac{x-1}{2}\right)+C$ (b) $2\pi\left(\dfrac{\mathrm e^2}{2}+\mathrm e-3\right)$

2023 NYJC P1 Q5

2023 NYJC P1 Q5

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