NCERT Solutions Class 12 Maths (Application Of Integrals) Miscellaneous Exercise

NCERT Solutions Class 12 Maths from class 12th Students will get the answers of Chapter-8 (Application Of Integrals) Miscellaneous Exercise This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

Q1. $\begin{array}{c}\text{Find the area under the given curves and given lines:}\\ \text{(i)}y=x^2,x=1,x=2\text{and}x\text{-axis}\\ \text{(ii)}y=x^4,x=1,x=5\text{and}x\text{-axis}\end{array}$

Answer. (i) The required area is represented by the shaded area ADCBA as  $\begin{array}{cc}\text{Area}ADCBA& ={\int }^{2}ydx\\ & =\int x^{2}dx\\ & ={\left[\frac{x^3}{3}\right]}_1^2\\ & =\frac{8}{3}-\frac{1}{3}\\ & =\frac{7}{3}\text{units}\end{array}$ (ii) The required area is represented by the shaded area ADCBA as  $\begin{array}{cc}\text{Area}ADCBA& ={\int }^5{x}^{4}dx\\ & ={\left[\frac{x^5}{5}\right]}_1^5\\ & =\frac{(5{)}^5}{5}-\frac{1}{5}\\ & =(5{)}^4-\frac{1}{5}\\ & =625-\frac{1}{5}\\ & =624.8\text{units}\end{array}$

Q2. Find the area between the curve $y=x\text{and}y=x^2$

Answer. The required area is represented by the shaded area OBAO as  $\begin{array}{c}\text{The points of intersection of the curves,}y=x\text{and}y=x^2,\text{is}A(1,1)\text{.}\\ \text{We draw AC perpendicular to}x\text{-axis.}\end{array}$ $\therefore \text{Area}\left(OBAO\right)=\text{Area}\left(\Delta OCA\right)-\text{Area (OCABO)}\dots \left(i\right)$ $\begin{array}{c}=\int xdx-{\int }_0^t{x}^{2}dx\\ ={\left[\frac{x^2}{2}\right]}_0^1-{\left[\frac{x^3}{3}\right]}_0^1\\ =\frac{1}{2}-\frac{1}{3}\\ =\frac{1}{6}\text{units}\end{array}$

Q3. $\begin{array}{c}\text{Find the area of the region lying in the first quadrant and bounded by}y=4x^2,x=0,y\\ =1\text{and}y=4\end{array}$

Answer. $\begin{array}{c}\text{The area in the first quadrant bounded by}y=4x^2,x=0,y=1,\text{and}y=4\text{is}\\ \text{represented by the shaded area ABCDA as}\end{array}$  $\begin{array}{cc}\therefore \text{Area}ABCD& ={\int }_x^{1}xdx\\ & =\int \frac{\sqrt{y}}{2}dx\end{array}$ $\begin{array}{cc}& =\frac{1}{2}{\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]}^4\\ & =\frac{1}{3}\left[\right(4{)}^{\frac{3}{2}}-1]\\ & =\frac{1}{3}[8-1]\\ & =\frac{7}{3}\text{units}\end{array}$

Q4. Sketch the graph of $y=|x+3|\text{and evaluate}{\int }_{-6}^0|x+3|dx$

Answer. $\begin{array}{c}\text{The given equation is}y=|x+3|\\ \text{The corresponding values of}x\text{and}y\text{are given in the following table.}\end{array}$  On plotting these points, we obtain the graph of $y=|x+3|$ as follows.  $\begin{array}{cc}\text{It is known that,}& (x+3)\le 0\text{for}-6\le x\le -3\text{and}(x+3)\ge 0\text{for}-3\le x\le 0\\ \therefore {\int }_{-6}^0(x+3)dx& =-{\int }_{-6}^{-3}(x+3)dx+{\int }_{-3}^0(x+3)dx\\ & =-{[\frac{x^2}{2}+3x]}_{-6}^{-3}+{[\frac{x^2}{2}+3x]}_{-3}^0\end{array}$ $\begin{array}{cc}& =-[(\frac{(-3{)}^2}{2}+3(-3\left)\right)-(\frac{(-6{)}^2}{2}+3(-6\left)\right)]+[0-(\frac{(-3{)}^2}{2}+3(-3\left)\right)]\\ & =-[-\frac{9}{2}]-[-\frac{9}{2}]\\ & =9\end{array}$

Q5. Find the area bounded by curve $y=\sin x\text{between}x=0\text{and}x=2\pi$

Answer. The graph of y = sin x can be drawn as  ∴ Required area = Area OABO + Area BCDB $\begin{array}{c}={\int }_0^{\pi }\sin xdx+\left|{\int }_{\pi }^{2\pi }\sin xdx\right|\\ =[-\cos x{]}_0^x+|-\cos x{]}_{\pi }^{2\pi }|\\ =[-\cos \pi +\cos 0]+|-\cos 2\pi +\cos \pi |\\ =1+1+|(-1-1)|\\ =2+|-2|\\ =2+2=4\text{units}\end{array}$

Q6. Find the area enclosed between the parabola $y^2$ = 4ax and the line y = mx

Answer. The area enclosed between the parabola, \(y^{2} = 4ax, and the line, y = mx, is represented by the shaded area OABO as  $\begin{array}{c}\text{The points of intersection of both the curves are}(0,0)\text{and}(\frac{4d}{m^2},\frac{4a}{m})\\ \text{we draw}AC\text{perpendicular to}x\text{-axis.}\end{array}$ ∴ Area OABO = Area OCABO – Area (ΔOCA)  $\begin{array}{cc}& =\frac{4}{3}\sqrt{a}{\left(\frac{4a}{m^2}\right)}^{\frac{3}{2}}-\frac{m}{2}\left[{\left(\frac{4a}{m^2}\right)}^2\right]\\ & =\frac{32a^2}{3m^3}-\frac{m}{2}\left(\frac{16a^2}{m^4}\right)\\ & =\frac{32a^2}{3m^3}-\frac{8a^2}{m^3}\\ & =\frac{8a^2}{3m^3}\text{units}\end{array}$

Q7. Find the area enclosed by the parabola $4y=3x^2\text{and the line}2y=3x+12$

Answer. The area enclosed between the parabola $4y=3x^2\text{and the line}2y=3x+12$ is represented by the shaded area OBAO as  The points of intersection of the given curves are A (–2, 3) and (4, 12). We draw AC and BD perpendicular to x-axis. ∴ Area OBAO = Area CDBA – (Area ODBO + Area OACO) $\begin{array}{cc}& ={\int }_{-2}^4\frac{1}{2}(3x+12)dx-{\int }_{-2}^1\frac{3x^2}{4}dx\\ & =\frac{1}{2}{[\frac{3x^2}{2}+12x]}_{-2}^4-\frac{3}{4}{\left[\frac{x^3}{3}\right]}_{-2}^4\end{array}$ $\begin{array}{c}=\frac{1}{2}[24+48-6+24]-\frac{1}{4}[64+8]\\ =\frac{1}{2}\left[90\right]-\frac{1}{4}\left[72\right]\\ =45-18\\ =27\text{units}\end{array}$

Q8. $\begin{array}{c}\text{Find the area of the smaller region bounded by the ellipse}\frac{x^2}{9}+\frac{y^2}{4}=1\text{and the line}\\ \frac{x}{3}+\frac{y}{2}=1\end{array}$

Answer. The area of the smaller region bounded by the ellipse, $\frac{x^2}{9}+\frac{y^2}{4}=1\text{and the line,}\frac{x}{3}+\frac{y}{2}=1$ is represented by the shaded region BCAB as  ∴ Area BCAB = Area (OBCAO) – Area (OBAO) $\begin{array}{cc}& ={\int }_0^{3}2\sqrt{1-\frac{x^2}{9}}dx-{\int }_0^{3}2(1-\frac{x}{3})dx\\ & =\frac{2}{3}\left[{\int }_0^3\sqrt{9-x^2}dx\right]-\frac{2}{3}{\int }_0^3(3-x)dx\\ & =\frac{2}{3}{[\frac{x}{2}\sqrt{9-x^2}+\frac{9}{2}{\sin }^{-1}\frac{x}{3}]}_0^3-\frac{2}{3}{[3x-\frac{x^2}{2}]}_0^3\end{array}$ $\begin{array}{cc}& =\frac{2}{3}\left[\frac{9}{2}\left(\frac{\pi }{2}\right)\right]-\frac{2}{3}[9-\frac{9}{2}]\\ & =\frac{2}{3}[\frac{9\pi }{4}-\frac{9}{2}]\\ & =\frac{2}{3}\times \frac{9}{4}(\pi -2)\\ & =\frac{3}{2}(\pi -2)\text{units}\end{array}$

Q9. Find the area of the smaller region bounded by the Find the area of the smaller region bounded by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{and the line}\frac{x}{a}+\frac{y}{b}=1$

Answer. The area of the smaller region bounded by the ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{and the line}\frac{x}{a}+\frac{y}{b}=1$ is represented by the shaded region BCAB as  ∴ Area BCAB = Area (OBCAO) – Area (OBAO) $\begin{array}{cc}& ={\int }_0^{a}b\sqrt{1-\frac{x^2}{a^2}}dx-{\int }_b^{a}b(1-\frac{x}{a})dx\\ & =\frac{b}{a}{\int }_0^0\sqrt{a^2-x^2}dx-\frac{b}{a}{\int }_0^a(a-x)dx\end{array}$ $\begin{array}{cc}& =\frac{b}{a}[{\{\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}\}}_0^a-{\{ax-\frac{x^2}{2}\}}_0^a]\\ & =\frac{b}{a}[\left\{\frac{a^2}{2}\left(\frac{\pi }{2}\right)\right\}-\{a^2-\frac{a^2}{2}\}]\\ & =\frac{b}{a}[\frac{a^2\pi }{4}-\frac{a^2}{2}]\end{array}$

Q10. Find the area of the region enclosed by the parabola $x^2$ = y, the line y = x + 2 and x-axis

Answer. The area of the region enclosed by the parabola, $x^2$ = y, the line, y = x + 2, and x-axis is represented by the shaded region OACO as  $\begin{array}{c}\text{The point of intersection of the parabola,}{x}^2=y,\text{and the line,}y=x+2,\text{is}A(-1,1)\text{.}\\ \therefore \text{Area OABCO}=\text{Area}\left(BCA\right)+\text{Area COAC}\end{array}$ $\begin{array}{c}={\int }_2^1(x+2)dx+{\int }_{-1}^0{x}^{2}dx\\ ={[\frac{x^2}{2}+2x]}_{-2}^{-1}+{\left[\frac{x^3}{3}\right]}_{-1}^0\\ =[\frac{(-1{)}^2}{2}+2(-1)-\frac{(-2{)}^2}{2}-2(-2\left)\right]+[-\frac{(-1{)}^3}{3}]\\ =[\frac{1}{2}-2-2+4+\frac{1}{3}]\\ =\frac{5}{6}\text{units}\end{array}$

Q11. $\begin{array}{c}\text{Using the method of integration find the area bounded by the curve}|x|+|y|=1\\ \text{[Hint: the required region is bounded by lines}x+y=1,x-y=1,-x+y=1\text{and}-x\\ -y=11]\end{array}$

Answer. $\begin{array}{c}\text{The area bounded by the curve,}|x|+|y|=1\text{, is represented by the shaded region}ADCB\\ \text{as}\end{array}$  The curve intersects the axes at points A (0, 1), B (1, 0), C (0, –1), and D (–1, 0). It can be observed that the given curve is symmetrical about x-axis and y-axis. ∴ Area ADCB = 4 × Area OBAO $\begin{array}{cc}& =4{\int }_0^d(1-x)dx\\ & =4{(x-\frac{x^2}{2})}^1\\ & =4{[1-\frac{1}{2}]}_0\\ & =4\left(\frac{1}{2}\right)\\ & =2\text{units}\end{array}$

Q12. Find the area bounded by curves $\left\{\right(x,y):y\ge x^2\text{and}y=|x|\}$

Answer. The area bounded by the curves, $\left\{\right(x,y):y\ge x^2\text{and}y=|x|\}$, is represented by the shaded region as  It can be observed that the required area is symmetrical about y-axis. $\begin{array}{cc}\text{Required area}& =2\left[\text{Area}\right(OCAO)-\text{Area (OCADO )}]\\ & =2[\int xdx-\int x^{2}dx]\\ & =2[{\left[\frac{x^2}{2}\right]}_0^1-{\left[\frac{x^3}{3}\right]}_0^1]\\ & =2[\frac{1}{2}-\frac{1}{3}]\\ & =2\left[\frac{1}{6}\right]=\frac{1}{3}\text{units}\end{array}$

Q13. Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A (2, 0), B (4, 5) and C (6, 3)

Answer. The vertices of ΔABC are A (2, 0), B (4, 5), and C (6, 3).  Equation of line segment AB is $\begin{array}{c}\text{Equation of line segment}AB\text{is}\\ y-0=\frac{5-0}{4-2}(x-2)\\ 2y=5x-10\\ y=\frac{5}{2}(x-2)...\left(i\right)\\ \text{Equation of line segment BC is}\end{array}$ $\begin{array}{c}y-5=\frac{3-5}{6-4}(x-4)\\ 2y-10=-2x+8\\ 2y=-2x+18\\ y=-x+9...\left(ii\right)\\ \text{Equation of line segment ca is}\\ y-3=\frac{0-3}{2-6}(x-6)\\ -4y+12=-3x+18\\ y=\frac{3}{4}(x-2)...\left(iii\right)\end{array}$ Area (ΔABC) = Area (ABLA) + Area (BLMCB) – Area (ACMA) $\begin{array}{cc}& ={\int }_2^4\frac{5}{2}(x-2)dx+{\int }_4^6(-x+9)dx-{\int }_2^6\frac{3}{4}(x-2)dx\\ & =\frac{5}{2}{[\frac{x^2}{2}-2x]}_2^4+{[\frac{-x^2}{2}+9x]}_4^6-\frac{3}{4}{[\frac{x^2}{2}-2x]}_2^6\\ & =\frac{5}{2}[8-8-2+4]+[-18+54+8-36]-\frac{3}{4}[18-12-2+4]\\ & =5+8-\frac{3}{4}\left(8\right)\\ & =13-6\\ & =7\text{units}\end{array}$

Q14. Using the method of integration find the area of the region bounded by lines: 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0

Answer. The given equations of lines are 2x + y = 4 … (i) 3x – 2y = 6 … (ii) And, x – 3y + 5 = 0 … (iii)  The area of the region bounded by the lines is the area of ΔABC. AL and CM are the perpendiculars on x-axis. Area (ΔABC) = Area (ALMCA) – Area (ALB) – Area (CMB) $\begin{array}{cc}& ={\int }^4\left(\frac{x+5}{3}\right)dx-{\int }^2(4-2x)dx-{\int }_2^4\left(\frac{3x-6}{2}\right)dx\\ & =\frac{1}{3}{[\frac{x^2}{2}+5x]}_1^4-{[4x-x^2]}_1^2-\frac{1}{2}{[\frac{3x^2}{2}-6x]}_2^4\\ & =\frac{1}{3}[8+20-\frac{1}{2}-5]-[8-4-4+1]-\frac{1}{2}[24-24-6+12]\\ & =(\frac{1}{3}\times \frac{45}{2})-\left(1\right)-\frac{1}{2}\left(6\right)\\ & =\frac{15}{2}-1-3\\ & =\frac{15}{2}-4=\frac{15-8}{2}=\frac{7}{2}\text{units}\end{array}$

Q15. Find the area of the region $\left\{\right(x,y):y^2\le 4x,4x^2+4y^2\le 9\}$

Answer. The area bounded by the curves, $\left\{\right(x,y):y^2\le 4x,4x^2+4y^2\le 9\}$ is represented as  The points of intersection of both the curves are $(\frac{1}{2},\sqrt{2})\text{and}(\frac{1}{2},-\sqrt{2})$ The required area is given by OABCO. It can be observed that area OABCO is symmetrical about x-axis. ∴ Area OABCO = 2 × Area OBC Area OBCO = Area OMC + Area MBC $\begin{array}{cc}& ={\int }_0^{1}2\sqrt{x}dx+{\int }_{\frac{1}{2}}^{\frac{3}{2}}\frac{1}{2}\sqrt{9-4x^2}dx\\ & ={\int }_0^{1}2\sqrt{x}dx+{\int }_{\frac{1}{2}}^{\frac{3}{2}}\frac{1}{2}\sqrt{(3{)}^2-(2x{)}^2}dx\end{array}$

Q16. Area bounded by the curve y = x3, the x-axis and the ordinates x = –2 and x = 1 is $\begin{array}{c}\text{A.}-9\\ {\text{B.}}^{-\frac{15}{4}}\\ {\text{B.}}^{\frac{15}{4}}\\ \text{D.}\frac{17}{4}\end{array}$

Answer.  $\begin{array}{cc}& \text{Required area}={\int }_2^{1}ydx\\ & ={\int }_2^1{x}^{3}dx\\ & ={\left[\frac{x^4}{4}\right]}_{-2}^1\\ & =[\frac{1}{4}-\frac{(-2{)}^4}{4}]\\ & =(\frac{1}{4}-4)=-\frac{15}{4}\text{units}\\ & =(\frac{1}{4}-4)=-\frac{15}{4}\text{units}\\ & \text{Thus, the correct answer is B.}\end{array}$

Q17. $\begin{array}{c}\text{The area bounded by the curve}y=x|x|,x\text{-axis and the ordinates}x=-1\text{and}x=1\text{is}\\ \text{given by}\\ \text{[Hint:}y=x^2\text{if}x>0\text{and}y=-x^2\text{if}x<0]\end{array}$ $\begin{array}{c}\text{A.}0\\ \text{B.}\frac{1}{3}\\ \text{C.}\frac{2}{3}\\ \text{D.}\frac{4}{3}\end{array}$

Answer.  $\begin{array}{cc}& \text{Required area}={\int }_{-1}^{1}ydx\\ & ={\int }_{-1}^{\infty }x|x|dx\\ & ={\int }_{-1}^0{x}^{2}dx+{\int }_0^1{x}^{2}dx\\ & ={\left[\frac{x^3}{3}\right]}_{-1}^0+{\left[\frac{x^3}{3}\right]}_0^1\\ & =-\frac{2}{3}\text{units}\\ & =\frac{2}{3}\text{units}\end{array}$ Thus, the correct answer is C.

Q18. The area of the circle $x^2+y^2=16\text{exterior to the parabola}{y}^2=6x$ $\left(A\right)\frac{4}{3}(4\pi -\sqrt{3})\text{(B)}\frac{4}{3}(4\pi +\sqrt{3})\text{(C)}\frac{4}{3}(8\pi -\sqrt{3})\text{(D)}\frac{4}{3}(8\pi +\sqrt{3})$

Answer. The given equations are $\begin{array}{c}{x}^2+y^2=16\dots \left(i\right)\\ y^2=6x\dots \left(ii\right)\end{array}$  Area bounded by the circle and parabola $\begin{array}{cc}& =2\left[\mathrm{Arca}\right(OADO)+\mathrm{Area}(ADBA\left)\right]\\ & =2[{\int }_0^2\sqrt{16x}dx+{\int }_2^{+}\sqrt{16-x^2}dx]\\ & =2\left[\sqrt{6}{\left\{\frac{x^{\frac{3}{2}}}{2}\right\}}_0^2\right]+2{[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}{\sin }^{-1}\frac{x}{4}]}_2^4\\ & =2\sqrt{6}\times \frac{2}{3}{\left[x^{\frac{3}{2}}\right]}_0^2+2[8\cdot \frac{\pi }{2}-\sqrt{16-4}-8{\sin }^{-1}\left(\frac{1}{2}\right)]\end{array}$ $\begin{array}{c}=\frac{4\sqrt{6}}{3}\left(2\sqrt{2}\right)+2[4\pi -\sqrt{12}-8\frac{\pi }{6}]\\ =\frac{16\sqrt{3}}{3}+8\pi -4\sqrt{3}-\frac{8}{3}\pi \\ =\frac{4}{3}[4\sqrt{3}+6\pi -3\sqrt{3}-2\pi ]\\ =\frac{4}{3}[\sqrt{3}+4\pi ]\\ \text{Area of circle}=n(r{)}^2\\ =16n\text{units}\end{array}$ $\begin{array}{cc}\therefore \text{Required area}& =16\pi -\frac{4}{3}[4\pi +\sqrt{3}]\\ & =\frac{4}{3}[4\times 3\pi -4\pi -\sqrt{3}]\\ & =\frac{4}{3}(8\pi -\sqrt{3})\text{units}\\ \text{Thus, the correct answer is}C\end{array}$

Q19. The area bounded by the y-axis, y = cos x and y = sin x When $\begin{array}{c}0\le x\le \frac{\pi }{2}.\\ {\text{A.}}^{2(\sqrt{2}-1)}\\ \text{B.}\sqrt{2}-1\\ \text{C.}\sqrt{2}+1\\ \text{D.}\sqrt{2}\end{array}$

Answer. The given equations are y = cos x … (i) And, y = sin x … (ii)  Required area = Area (ABLA) + area (OBLO) $\begin{array}{cc}& ={\int }_1^{4}xdy+{\int }_0^{1}xdy\\ & ={\int }_1^d{\cos }^{-1}ydy+{\int }_0^{\frac{1}{2}}{\sin }^{-1}xdy\\ \text{Integrating by parts, we obtain}& \end{array}$ $\begin{array}{c}={[y{\cos }^{-1}y-\sqrt{1-y^2}]}_{\frac{1}{\sqrt{2}}}^1+{[x{\sin }^{-1}x+\sqrt{1-x^2}]}_0^{\frac{1}{\sqrt{2}}}\\ =\left[{\cos }^{-1}\right(1)-\frac{1}{\sqrt{2}}{\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)+\sqrt{1-\frac{1}{2}}]+[\frac{1}{\sqrt{2}}{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)+\sqrt{1-\frac{1}{2}}-1]\\ =\frac{-\pi }{4\sqrt{2}}+\frac{1}{\sqrt{2}}+\frac{\pi }{4\sqrt{2}}+\frac{1}{\sqrt{2}}-1\\ =\frac{2}{\sqrt{2}}-1\\ =\sqrt{2}-1\text{units}\\ \text{Thus, the correct answer is B.}\end{array}$ $\begin{array}{c}\text{Put}2x=t\Rightarrow dx=\frac{dt}{2}\\ \text{When}x=\frac{3}{2},t=3\text{and when}x=\frac{1}{2},t=1\\ ={\int }_0^{\frac{1}{2}}2\sqrt{x}dx+\frac{1}{4}{\int }_1^3\sqrt{(3{)}^2-(t{)}^2}dt\end{array}$ $\begin{array}{cc}& =2{\left[\frac{x^{\frac{3}{2}}}{2}\right]}^{\frac{1}{2}}+\frac{1}{4}{[\frac{t}{2}\sqrt{9-t^2}+\frac{9}{2}{\sin }^{-1}\left(\frac{t}{3}\right)]}_1^3\\ & =2\left[\frac{2}{3}{\left(\frac{1}{2}\right)}^{\frac{3}{2}}\right]+\frac{1}{4}[\{\frac{3}{2}\sqrt{9-(3{)}^2}+\frac{9}{2}{\sin }^{-1}\left(\frac{3}{3}\right)\}-\{\frac{1}{2}\sqrt{9-(1{)}^2}+\frac{9}{2}{\sin }^{-1}\left(\frac{1}{3}\right)\}]\end{array}$ $\begin{array}{cc}& =\frac{2}{3\sqrt{2}}+\frac{1}{4}[\{0+\frac{9}{2}{\sin }^{-1}(1\left)\right\}-\{\frac{1}{2}\sqrt{8}+\frac{9}{2}{\sin }^{-1}\left(\frac{1}{3}\right)\}]\\ & =\frac{\sqrt{2}}{3}+\frac{1}{4}[\frac{9\pi }{4}-\sqrt{2}-\frac{9}{2}{\sin }^{-1}\left(\frac{1}{3}\right)]\\ & =\frac{\sqrt{2}}{3}+\frac{9\pi }{16}-\frac{\sqrt{2}}{4}-\frac{9}{8}{\sin }^{-1}\left(\frac{1}{3}\right)\\ & =\frac{9\pi }{16}-\frac{9}{8}{\sin }^{-1}\left(\frac{1}{3}\right)+\frac{\sqrt{2}}{12}\end{array}$ Therefore , the required area is $[2\times (\frac{9\pi }{16}-\frac{9}{8}{\sin }^{-1}\left(\frac{1}{3}\right)+\frac{\sqrt{2}}{12})]=\frac{9\pi }{8}-\frac{9}{4}{\sin }^{-1}\left(\frac{1}{3}\right)+\frac{1}{3\sqrt{2}}$ units