3.8.4 Integration, Long Questions (Question 7 & 8)

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


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

Question 8:
Diagram below shows the curve $y=\frac{4}{x^2}$ and the straight line y = mx + c. The straight line y = mx + c is a tangent to the curve at (2, 1).

(a) Find the value of m and of c.

(b) Calculate the area of the shaded region.

(c) It is given that the volume generated when the region bounded by the curve, the x–axis and the straight lines x = 2 and x = h is revolved through 360^o about the x-axis is $\frac{38\pi }{81}{\text{ unit}}^3.$
Find the value of h, such that h > 2.


Solution:
(a)
$\begin{array}{l}y=\frac{4}{x^2}=4x^{-2}\\ \frac{dy}{dx}=-8x^{-3}\\ =-\frac{8}{x^3}\\ \text{At }x=2,\\ \frac{dy}{dx}=-\frac{8}{2^3}\\ =-1\\ \\ \text{Equation of tangent:}\\ y-y_1=m\left(x-x_1\right)\\ y-1=-1\left(x-2\right)\\ y=-x+2+1\\ y=-x+3\\ \\ m=-1,c=3\end{array}$


(b)
$\begin{array}{l}\text{At }x\text{-axis, }y=0\\ \text{From the straight line }y=-x+3,x=3\\ \\ \text{Area of the shaded region}\\ =\text{Area under the curve}-\text{Area of triangle}\\ ={\displaystyle {\int }_2^{4}ydx-}\frac{1}{2}\times 1\times 1\\ ={\displaystyle {\int }_2^4\left(4x^{-2}\right)dx}-\frac{1}{2}\\ ={\left[\frac{4x^{-1}}{-1}\right]}_2^4-\frac{1}{2}\\ ={\left[-\frac{4}{x}\right]}_2^4-\frac{1}{2}\\ =\left[-\frac{4}{4}-\left(-\frac{4}{2}\right)\right]-\frac{1}{2}\\ =\frac{1}{2}{\text{ unit}}^2\end{array}$


(c)
$\begin{array}{l}\text{Volume generated}=\frac{38\pi }{81}\\ \pi {\displaystyle {\int }_2^h{y}^{2}dx}=\frac{38\pi }{81}\\ {\displaystyle {\int }_2^h{\left(\frac{4}{x^2}\right)}^{2}d}x=\frac{38}{81}\\ {\displaystyle {\int }_2^h\left(\frac{16}{x^4}\right)dx}=\frac{38}{81}\\ {\displaystyle {\int }_2^h\left(16x^{-4}\right)dx}=\frac{38}{81}\\ {\left[\frac{16x^{-3}}{-3}\right]}_2^h=\frac{38}{81}\\ {\left[-\frac{16}{3x^3}\right]}_2^h=\frac{38}{81}\\ -\frac{16}{3h^3}-\left(-\frac{16}{3{\left(2\right)}^3}\right)=\frac{38}{81}\\ \frac{16}{3h^3}=\frac{16}{24}-\frac{38}{81}\\ \frac{16}{3h^3}=\frac{16}{81}\\ 3h^3=81\\ h^3=27\\ h=3\end{array}$