Long Question 7

Question 7:
Diagram below shows a curve $y=\frac{1}{4}{x}^2+3$ which intersects the straight line y = x + 6 at point A.


(a) Find the coordinates of A.
(b) Calculate
(i) the area of the shaded region M,
(ii) the volume generated, in terms of π, when the shaded region N is revolved 360^o about the y-axis.

Solution:
(a)
$\begin{array}{l}y=\frac{1}{4}{x}^2+\mathrm{3..........}\left(1\right)\\ y=x+\mathrm{6..........}\left(2\right)\\ \text{Substitute (2) into (1),}\\ x+6=\frac{1}{4}{x}^2+3\\ 4x+24=x^2+12\\ x^2-4x-12=0\\ \left(x+2\right)\left(x-6\right)=0\\ x=-2\text{ or }x=6\left(\text{rejected}\right)\\ \\ \text{When }x=-2\\ y=-2+6=4\\ \text{Therefore, }A=\left(-2,4\right).\end{array}$


(b)(i)
$\begin{array}{l}\text{At }x\text{-axis, }y=0\\ \text{From }y=x+6,x=-6\\ \\ \text{Area of region }M\\ =\text{Area of triangle}+\text{Area under the curve}\\ =\frac{1}{2}\times \left(6-2\right)\times 4+{\displaystyle {\int }_{-2}^{0}ydx}\\ =8+{\displaystyle {\int }_{-2}^0\left(\frac{1}{4}{x}^2+3\right)dx}\\ =8+{\left[\frac{x^3}{4\left(3\right)}+3x\right]}_{-2}^0\\ =8+\left[0-\left(\frac{{\left(-2\right)}^3}{12}+3\left(-2\right)\right)\right]\\ =8+\left[0-\left(-\frac{8}{12}-6\right)\right]\\ =8+\left[0-\left(-\frac{20}{3}\right)\right]\\ =14\frac{2}{3}{\text{ unit}}^2\end{array}$


(b)(ii)
$\begin{array}{l}\text{At }y\text{-axis, }x=0,\\ y=\frac{1}{4}\left(0\right)+3\\ y=3\\ \\ y=\frac{1}{4}{x}^2+3\\ 4y=x^2+12\\ x^2=4y-12\\ \\ \text{Volume of }N\\ \text{= }\pi {\displaystyle {\int }_3^4{x}^{2}dy}\\ \text{= }\pi {\displaystyle {\int }_3^4\left(4y-12\right)dy}\\ \text{= }\pi {\displaystyle {\int }_3^4\left(2y^2-12y\right)dy}\\ =\pi {\left[\left(2y^2-12y\right)\right]}_3^4\\ =\pi \left[\left(2{\left(4\right)}^2-12\left(4\right)\right)-\left(2{\left(3\right)}^2-12\left(3\right)\right)\right]\\ =\pi \left(-16+18\right)\\ =2\pi {\text{ unit}}^3\end{array}$