Long Question 9

Question 9 (10 marks):
Diagram shows the straight line 4y = x – 2 touches the curve x = y² + 6 at point P.



Find
(a) the coordinates of P,
(b) the area of the shaded region,
(c) the volume of revolution, in terms of π, when the region bounded by the curve and the straight line x = 8 is revolved through 180^o about the x-axis.


Solution:
(a)
$\begin{array}{l}4y=x-\mathrm{2.........}(1)\\ x=y^2+\mathrm{6.........}(2)\\ \\ \text{Substitute (2) into (1):}\\ 4y=\left(y^2+6\right)-2\\ y^2-4y+4=0\\ \left(y-2\right)\left(y-2\right)=0\\ y-2=0\\ y=2\\ \\ \text{Substitute }y=2\text{ into (2):}\\ x={\left(2\right)}^2+6\\ x=10\\ \text{Thus, }P=\left(10,2\right).\end{array}$


(b)
$\begin{array}{l}\text{At x-axis, }y=0\\ \therefore 4y=x-2\\ 0=x-2\\ x=2\\ \\ \text{Area of shaded region}\\ \text{= Area of triangle}-\text{Area under the curve}\\ =\frac{1}{2}\left(10-2\right)\left(2\right)-{\displaystyle {\int }_6^{10}ydx}\\ =8-{\displaystyle {\int }_6^{10}\sqrt{x-6}dx}\leftarrow \boxed{\begin{array}{l}x=y^2+6\\ y=\sqrt{x-6}\end{array}}\\ =8-{\displaystyle {\int }_6^{10}{\left(x-6\right)}^{\frac{1}{2}}dx}\\ =8-{\left[\frac{{\left(x-6\right)}^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]}_6^{10}\\ =8-{\left[\frac{2{\left(x-6\right)}^{\frac{3}{2}}}{3}\right]}_6^{10}\\ =8-\left[\frac{2{\left(10-6\right)}^{\frac{3}{2}}}{3}-\frac{2{\left(6-6\right)}^{\frac{3}{2}}}{3}\right]\\ =8-\frac{16}{3}\\ =\frac{8}{3}{\text{ units}}^2\end{array}$


(c)
$\begin{array}{l}\text{Volume of revolution}\\ =\pi {\displaystyle {\int }_6^8{y}^{2}dx}\\ =\pi {\displaystyle {\int }_6^8\left(x-6\right)dx}\leftarrow \boxed{\begin{array}{l}x=y^2+6\\ y^2=x-6\end{array}}\\ =\pi {\left[\frac{x^2}{2}-6x\right]}_6^8\\ =\pi \left[\left(32-48\right)-\left(18-36\right)\right]\\ =2\pi {\text{ units}}^3\end{array}$