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Question Number 120964 by bramlexs22 last updated on 04/Nov/20

Commented by bramlexs22 last updated on 04/Nov/20

$$\mathrm{the}\mathrm{line}\mathrm{x}=3$$

Commented by liberty last updated on 04/Nov/20

Commented by liberty last updated on 04/Nov/20

$$\mathrm{Area}=2\underset{2}{\stackrel{3}{\int }}\mathrm{y}\mathrm{dx}$$$$\mathrm{with}\mathrm{y}=\sqrt{\frac{{9\mathrm{x}}^2-36}{4}}=\frac{3}{2}\sqrt{{\mathrm{x}}^2-4}$$$$\mathrm{Area}=3\underset{2}{\stackrel{3}{\int }}\sqrt{{\mathrm{x}}^2-4}\mathrm{dx}$$$$\mathrm{let}\mathrm{x}=2\sec z\to \mathrm{dx}=2\sec \mathrm{z}\tan \mathrm{z}\mathrm{dz}$$$$\mathrm{I}=3\int \sqrt{4(\sec {}^2\mathrm{z}-1)}2\sec \mathrm{z}\tan \mathrm{z}\mathrm{dz}$$$$\mathrm{I}=12\int \tan {}^2\mathrm{z}\sec \mathrm{z}\mathrm{dz}=12\int (\sec {}^3\mathrm{z}-\sec \mathrm{z})\mathrm{dz}$$$$\mathrm{it}\mathrm{easy}\mathrm{to}\mathrm{solve}$$

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