SPM 2021 Add Maths Trial Paper (Selangor) – Paper 2 (Question 6)

Question 6:
Diagram 5 shows two curves $x=\sqrt{y}\text{ and }x=\frac{y^2}{8}$ are intersecting at point A. The straight line AB is parallel to the x-axis.


Find
(a) the equation of the straight line AB, [2 marks]

(b) area, in unit², of the shaded region, [3 marks]

(c) the volume generated, in term of π, when the region P, which bounded by the curve $x=\frac{y^2}{8}$ , the straight line AB and the y-axis, is revolved 360^∘ about the y-axis. [3 marks]


Solution:
(a)

$\begin{array}{l}x=\sqrt{y}\mathrm{\dots ..}\left(1\right)\\ x=\frac{y^2}{8}\mathrm{\dots ..}\left(2\right)\\ \\ \text{Substitute (1) into (2),}\\ \sqrt{y}=\frac{y^2}{8}\\ {\left(y^{\frac{1}{2}}\right)}^2={\left(\frac{y^2}{8}\right)}^2\\ y=\frac{y^4}{64}\\ 64=\frac{y^4}{y}\\ y^3=64\\ y=4\\ \\ \text{From (1),}\\ x=\sqrt{4}=2\\ \text{Point }A=\left(2,4\right)\\ \\ \text{Equation of the straight line }AB\text{ is }y=4.\end{array}$


(b)

$\begin{array}{l}x=\sqrt{y}\Rightarrow y=x^2\\ x=\frac{y^2}{8}\Rightarrow y=\sqrt{8x}\\ \\ \text{Area of the shaded region}\\ ={\displaystyle {\int }_0^2(y_1}-y_2)dx\\ ={\displaystyle {\int }_0^2(\sqrt{8}{x}^{\frac{1}{2}}-x^2)dx}\\ ={\left[\frac{\sqrt{8}{x}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{x^3}{3}\right]}_0^2\\ ={\left[\frac{2\times 2{\left(2\right)}^{\frac{1}{2}}{x}^{\frac{3}{2}}}{3}-\frac{x^3}{3}\right]}_0^2\\ =\frac{4{\left(2\right)}^{\frac{1}{2}}{\left(2\right)}^{\frac{3}{2}}}{3}-\frac{{\left(2\right)}^3}{3}-0\\ =\frac{16}{3}-\frac{8}{3}\\ =\frac{8}{3}{\text{ unit}}^2\end{array}$


(c)

$\begin{array}{l}x=\frac{y^2}{8}\Rightarrow x^2=\frac{y^4}{64}\\ V=\pi {\displaystyle {\int }_a^b{x}^{2}dy}\\ =\frac{\pi }{64}{\displaystyle {\int }_0^4{y}^{4}dy}\\ =\frac{\pi }{64}{\left[\frac{y^5}{5}\right]}_0^4\\ =\frac{\pi }{64}\left[\frac{4^5}{5}-0\right]\\ =\frac{\pi }{64}\left[\frac{1024}{5}\right]\\ =\frac{16\pi }{5}{\text{ unit}}^3\end{array}$