Class 12: Areas of Bounded Regions

$\displaystyle \text{Using integration, find the area of the following regions:}$

$\displaystyle \textbf{Question 1: }\text{Find the area of the region between the parabola }x=4y-y^{2} \\ \text{ and the line }x=2y-3.$
$\displaystyle \text{Answer:}$ $\displaystyle \text{To find the point of intersection of the parabola } x=4y-y^{2} \text{ and the line } x=2y-3$
$\displaystyle \text{Substitute } x=2y-3 \text{ in the equation of the parabola}$
$\displaystyle 2y-3=4y-y^{2}$
$\displaystyle \Rightarrow y^{2}-2y-3=0$
$\displaystyle \Rightarrow (y+1)(y-3)=0$
$\displaystyle \Rightarrow y=-1 \text{ or } y=3$
$\displaystyle \text{When } y=-1,\ x=2(-1)-3=-5$
$\displaystyle \text{When } y=3,\ x=2(3)-3=3$
$\displaystyle \text{Therefore, the points of intersection are } D(-1,-5) \text{ and } A(3,3)$
$\displaystyle \text{The area of the required region } ABCDOA$
$\displaystyle A=\int_{-1}^{3}\lvert x_{1}-x_{2}\rvert\,dy,\ \text{where } x_{1}=4y-y^{2} \text{ and } x_{2}=2y-3$
$\displaystyle =\int_{-1}^{3}(x_{1}-x_{2})\,dy\ \text{ since } x_{1}>x_{2}$
$\displaystyle =\int_{-1}^{3}\big[(4y-y^{2})-(2y-3)\big]\,dy$
$\displaystyle =\int_{-1}^{3}(-y^{2}+2y+3)\,dy$
$\displaystyle =\left[-\frac{y^{3}}{3}+y^{2}+3y\right]_{-1}^{3}$
$\displaystyle =\left(-\frac{27}{3}+9+9\right)-\left(\frac{1}{3}+1-3\right)$
$\displaystyle =9-\left(-\frac{5}{3}\right)$
$\displaystyle =\frac{32}{3}\ \text{sq. units}$

$\displaystyle \textbf{Question 2: }\text{Find the area bounded by the parabola }x=8+2y-y^{2};\ \text{the }y \\ \text{-axis and the lines } y=-1\text{ and }y=3.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{The parabola cuts the } y\text{-axis at }(0,4)\text{ and }(0,-2)$
$\displaystyle \text{Also, the points of intersection of the parabola and the lines }y=3\text{ and }y=-1\text{ are }B(5,3)\text{ and }D(5,-1)\text{ respectively}$
$\displaystyle \text{Therefore, the area of the required region }ABCDE$
$\displaystyle A=\int_{-1}^{3} x\,dy$
$\displaystyle =\int_{-1}^{3}(8+2y-y^{2})\,dy$
$\displaystyle =\left[8y+y^{2}-\frac{y^{3}}{3}\right]_{-1}^{3}$
$\displaystyle =\left\{8(3)+3^{2}-\frac{3^{3}}{3}\right\}-\left\{8(-1)+(-1)^{2}-\frac{(-1)^{3}}{3}\right\}$
$\displaystyle =\{24+9-9\}-\{-8+1+\frac{1}{3}\}$
$\displaystyle =24-\left(-7+\frac{1}{3}\right)$
$\displaystyle =24+7-\frac{1}{3}$
$\displaystyle =31-\frac{1}{3}$
$\displaystyle =\frac{92}{3}\ \text{sq. units}$

$\displaystyle \textbf{Question 3: }\text{Find the area bounded by the parabola }y^{2}=4x \text{ and the line } \\ y=2x-4.$
$\displaystyle \text{(i) By using horizontal strips}$ $\displaystyle \text{Answer:}$
$\displaystyle \text{To find the points of intersection between the parabola and the line, substitute } y=2x-4 \text{ in } y^{2}=4x.$
$\displaystyle (2x-4)^{2}=4x$
$\displaystyle \Rightarrow 4x^{2}-16x+16=4x$
$\displaystyle \Rightarrow 4x^{2}-20x+16=0$
$\displaystyle \Rightarrow x^{2}-5x+4=0$
$\displaystyle \Rightarrow (x-1)(x-4)=0$
$\displaystyle \Rightarrow x=1,4$
$\displaystyle \Rightarrow y=-2,4$
$\displaystyle \text{Therefore, the points of intersection are } C(1,-2) \text{ and } A(4,4).$
$\displaystyle \text{Using horizontal strips:}$
$\displaystyle \text{The area of the required region } ABCD$
$\displaystyle A=\int_{-2}^{4}(x_{1}-x_{2})\,dy \text{ where } x_{1}=\frac{y+4}{2} \text{ and } x_{2}=\frac{y^{2}}{4}$
$\displaystyle =\int_{-2}^{4}\left(\frac{y+4}{2}-\frac{y^{2}}{4}\right)dy$
$\displaystyle =\left[\frac{y^{2}}{4}+2y-\frac{y^{3}}{12}\right]_{-2}^{4}$
$\displaystyle =\left(\frac{4^{2}}{4}+2(4)-\frac{4^{3}}{12}\right)-\left(\frac{(-2)^{2}}{4}+2(-2)-\frac{(-2)^{3}}{12}\right)$
$\displaystyle =\left(4+8-\frac{16}{3}\right)-\left(1-4+\frac{2}{3}\right)$
$\displaystyle =12-\frac{16}{3}+3-\frac{2}{3}$
$\displaystyle =15-\frac{18}{3}$
$\displaystyle =15-6$
$\displaystyle =9 \text{ sq. units}$

$\displaystyle \text{(ii) By using vertical strips}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{To find the points of intersection between the parabola and the line let us substitute } y=2x-4 \text{ in } y^2=4x.$
$\displaystyle (2x-4)^2=4x$
$\displaystyle \Rightarrow 4x^2+16-16x=4x$
$\displaystyle \Rightarrow 4x^2-20x+16=0$
$\displaystyle \Rightarrow x^2-5x+4=0$
$\displaystyle \Rightarrow (x-1)(x-4)=0$
$\displaystyle \Rightarrow x=1,4$
$\displaystyle \Rightarrow y=-2,4$
$\displaystyle \text{Therefore, the points of intersection are } C(1,-2) \text{ and } A(4,4).$
$\displaystyle \text{Using Vertical Strips:}$
$\displaystyle \text{The area of the required region } ABCD$
$\displaystyle A=\int_{0}^{4} y_2\,dx-\int_{1}^{4} y_1\,dx \text{ where } y_2=2\sqrt{x} \text{ and } y_1=2x-4.$
$\displaystyle =\int_{0}^{4}2\sqrt{x}\,dx-\int_{1}^{4}(2x-4)\,dx$
$\displaystyle =\left[\frac{4}{3}x^{3/2}\right]_{0}^{4}-\left[x^2-4x\right]_{1}^{4}$
$\displaystyle =\left[\frac{4}{3}(4)^{3/2}-0\right]-\left[(16-16)-(1-4)\right]$
$\displaystyle =\left[\frac{32}{3}\right]-\left[0-(-3)\right]$
$\displaystyle =\frac{32}{3}-3$
$\displaystyle =\frac{23}{3}\text{ square units}$

$\displaystyle \textbf{Question 4: }\text{Find the area of the region bounded by the parabola }y^{2}=2x \\ \text{ and the straight line }x-y=4.$
$\displaystyle \text{Answer:}$ $\displaystyle y^2=2x \text{ represents a parabola opening towards the positive x-axis with focus } \left(\frac{1}{2},0\right).$
$\displaystyle x-y=4 \text{ represents a straight line passing through } (4,0) \text{ and } (0,-4).$
$\displaystyle \text{Solving } y^2=2x \text{ and } x-y=4.$
$\displaystyle y^2=2(y+4).$
$\displaystyle \Rightarrow y^2-2y-8=0.$
$\displaystyle \Rightarrow (y-4)(y+2)=0.$
$\displaystyle \Rightarrow y=4 \text{ or } y=-2.$
$\displaystyle \text{Corresponding x-values are } x=8 \text{ and } x=2.$
$\displaystyle \text{Thus, the points of intersection are } A(8,4) \text{ and } B(2,-2).$
$\displaystyle \text{Required area }=\text{Area of the shaded region } OABO.$
$\displaystyle =\int_{-2}^{4} x_{\text{line}}\,dy-\int_{-2}^{4} x_{\text{parabola}}\,dy.$
$\displaystyle =\int_{-2}^{4}(y+4)\,dy-\int_{-2}^{4}\frac{y^2}{2}\,dy.$
$\displaystyle =\left[\frac{(y+4)^2}{2}\right]_{-2}^{4}-\frac{1}{2}\left[\frac{y^3}{3}\right]_{-2}^{4}.$
$\displaystyle =\frac{1}{2}(64-4)-\frac{1}{6}(64-(-8)).$
$\displaystyle =30-12.$
$\displaystyle =18\text{ square units.}$