**:**

**Soalan**7Rajah di bawah menunjukkan suatu lengkung $y=\frac{1}{4}{x}^{2}+3$ yang menyilang suatu garis lurus

*y*=

*x*+ 6 pada titik

*A*.

(a) Cari koordinat

*A*.

(b) ) hitung

(i) luas rantau berlorek

*M*,

(ii) isipadu kisaran, dalam sebutan π, apabila rantau berlorek

*N*diputarkan melalui 360

^{o}pada paksi-

*y*.

*Penyelesaian:***(a)**

$\begin{array}{l}y=\frac{1}{4}{x}^{2}+\mathrm{3..........}\left(1\right)\\ y=x+\mathrm{6..........}\left(2\right)\\ \text{Gantikan(2)kedalam(1),}\\ x+6=\frac{1}{4}{x}^{2}+3\\ 4x+24={x}^{2}+12\\ {x}^{2}-4x-12=0\\ \left(x+2\right)\left(x-6\right)=0\\ x=-2\text{or}x=6\text{}\left(\text{ditolak}\right)\\ \\ \text{Apabila}x=-2\\ y=-2+6=4\\ \text{Olehitu,}A=\left(-2,4\right).\end{array}$

**(b)(i)**

$\begin{array}{l}\text{Padapaksi-}x\text{,}y=0\\ \text{Dari}y=x+6,x=-6\\ \\ \text{Luaskawasanberlorek}M\\ =\text{Luassegitiga}+\text{Luasdibawahlengkung}\\ =\frac{1}{2}\times \left(6-2\right)\times 4+{\displaystyle {\int}_{-2}^{0}y\text{}dx}\\ =8+{\displaystyle {\int}_{-2}^{0}\left(\frac{1}{4}{x}^{2}+3\right)\text{}dx}\\ =8+{\left[\frac{{x}^{3}}{4\left(3\right)}+3x\right]}_{-2}^{0}\\ =8+\left[0-\left(\frac{{\left(-2\right)}^{3}}{12}+3\left(-2\right)\right)\right]\\ =8+\left[0-\left(-\frac{8}{12}-6\right)\right]\\ =8+\left[0-\left(-\frac{20}{3}\right)\right]\\ =14\frac{2}{3}{\text{unit}}^{2}\end{array}$

**(b)(ii)**

$\begin{array}{l}\text{padapaksi-}y\text{,}x=0,\text{}\\ y=\frac{1}{4}\left(0\right)+3\\ y=3\\ \\ y=\frac{1}{4}{x}^{2}+3\\ 4y={x}^{2}+12\\ {x}^{2}=4y-12\\ \\ \text{Isipadu}N\\ \text{=}\pi {\displaystyle {\int}_{3}^{4}{x}^{2}dy}\\ \text{=}\pi {\displaystyle {\int}_{3}^{4}\left(4y-12\right)dy}\\ \text{=}\pi {\displaystyle {\int}_{3}^{4}\left(2{y}^{2}-12y\right)dy}\\ =\pi {\left[\left(2{y}^{2}-12y\right)\right]}_{3}^{4}\\ =\pi \left[\left(2{\left(4\right)}^{2}-12\left(4\right)\right)-\left(2{\left(3\right)}^{2}-12\left(3\right)\right)\right]\\ =\pi \left(-16+18\right)\\ =2\pi {\text{unit}}^{3}\end{array}$