Long Question 5

Question 5:
In Diagram below, the straight line WY is normal to the curve   $y=\frac{1}{2}{x}^2+1$ at B (2, 4). The straight line BQ is parallel to the y–axis.


Find
(a) the value of t,
(b) the area of the shaded region,
(c) the volume generated, in terms of π, when the region bounded by the curve, the y–axis
  and the straight line y = 4 is revolved through 360° about the y-axis.
Solution:
(a)
$\begin{array}{l}y=\frac{1}{2}{x}^2+1\\ \text{Gradient of tangent, }\frac{dy}{dx}=2\left(\frac{1}{2}x\right)=x\\ \text{At point }B\text{, }\frac{dy}{dx}=2\\ \text{Gradient of normal, }{m}_2=-\frac{1}{2}\\ \frac{4-0}{2-t}=-\frac{1}{2}\\ 8=-2+t\\ t=10\end{array}$


(b)
$\begin{array}{l}\text{Area of the shaded region}\\ =\text{Area under the curve + Area of triangle }BQY\\ ={\displaystyle {\int }_0^2\left(\frac{1}{2}{x}^2+1\right)}dx+\frac{1}{2}\left(10-2\right)\left(4\right)\\ ={\left[\frac{x^3}{6}+x\right]}_0^2+16\\ =\left[\frac{8}{6}+2\right]-0+16\\ =19\frac{1}{3}{\text{ unit}}^2\end{array}$


(c)
$\begin{array}{l}\text{At }y-\text{axis, }x=0,y=\frac{1}{2}\left(0\right)+1=1\\ y=\frac{1}{2}{x}^2+1,x^2=2y-2\\ \text{Volume generated }\\ \text{= }\pi {\displaystyle \int x^{2}dy}\\ =\pi {\displaystyle {\int }_1^4\left(2y-2\right)}dy\\ =\pi {\left[y^2-2y\right]}_1^4\\ =\pi \left[\left(16-8\right)-\left(1-2\right)\right]\\ =9\pi {\text{ unit}}^3\end{array}$