## Calculus: Early Transcendentals 8th Edition

The area is $4.28$ to 2 decimal places
$9x^2-4y^2=36\\ 4y^2=9x^2-36\\ y^2=\frac{9}{4}x^2-9\\ y=\sqrt{\frac{9}{4}x^2-9}\\$ When $y=0$ $9x^2=36\\ x=2\\$ Therefore the limits are $x=2$ and $x=3$ $\displaystyle{I=\int_2^3 \sqrt{\frac{9}{4}x^2-9}\ dx}\\ \displaystyle{I=\int_2^3 \sqrt{\frac{9}{4}x^2-9\times\frac{4}{4}}\ dx}\\ \displaystyle{I=\frac{3}{2}\int_2^3 \sqrt{x^2-4}\ dx}\\$ $\displaystyle \left[\begin{array}{ll} x=2\sec\theta & x^2=4\sec^2\theta \\ & \\ \frac{dx}{d\theta}=2\sec\theta\tan\theta & dx=2\sec\theta\tan\theta d\theta \end{array}\right]$ Integration by substitution $\displaystyle{I=\frac{3}{2}\int_0^{0.84} \sqrt{4\sec^2\theta-4}\times2\sec\theta\tan\theta\ d\theta}\\ \displaystyle{I=\frac{3}{2}\int_0^{0.84} 2\tan\theta\times2\sec\theta\tan\theta\ d\theta}\\ \displaystyle{I=6\int_0^{0.84} \sec\theta\tan^2\theta\ d\theta}\\ \displaystyle{I=6\int_0^{0.84} \sec^3\theta-\sec\theta\ d\theta}\\$ Integrating $\sec^3\theta$ $\displaystyle \left[\begin{array}{ll} u=\sec\theta & dv=\sec^2\theta \\ & \\ du=\sec\theta\tan\theta & v=\tan\theta \end{array}\right]$ Integration by parts $\displaystyle{\int\sec^3\theta\ d\theta=\sec\theta\tan\theta-\int\sec\theta\tan^2\theta\ d\theta}\\ \displaystyle{\int\sec^3\theta\ d\theta=\sec\theta\tan\theta-\int\sec\theta\left(\sec^2\theta-1\right)\ d\theta}\\ \displaystyle{\int\sec^3\theta\ d\theta=\sec\theta\tan\theta-\int\sec^3\theta-\sec\theta\ d\theta}\\ \displaystyle{\int\sec^3\theta\ d\theta=\sec\theta\tan\theta-\int\sec^3\theta\ d\theta+\int \sec\theta\ d\theta}\\ \displaystyle{2\int\sec^3\theta\ d\theta=\sec\theta\tan\theta+\int \sec\theta\ d\theta}\\ \displaystyle{2\int\sec^3\theta\ d\theta=\sec\theta\tan\theta+\ln\left|\sec \theta+\tan \theta\right|+C}\\ \displaystyle{\int\sec^3\theta\ d\theta=\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln\left|\sec \theta+\tan \theta\right|+C}\\$ $\displaystyle{I=6\left[\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln\left|\sec \theta+\tan \theta\right|\right]_0^{0.84}-6\int_0^{0.84} \sec\theta\ d\theta}\\ \displaystyle{I=6\left[\frac{1}{2}\sec\theta\tan\theta-\frac{1}{2}\ln\left|\sec \theta+\tan \theta\right|\right]_0^{0.84}}\\ \displaystyle{I=3\left[\sec\theta\tan\theta-\ln\left|\sec \theta+\tan \theta\right|\right]_0^{0.84}}\\ \displaystyle{I=2.14}\ \ \left(2 d.p.\right)$ The final area should be twice this (positive and negative region): $A\approx2.14*2\approx4.28$