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Question Number 192024 by Shlock last updated on 05/May/23

Answered by a.lgnaoui last updated on 05/May/23

$$\mathbf{The}\mathbf{shaded}\mathbf{Area}\mathbf{betwen}$$$$[\mathbf{y}=0,\mathbf{y}=\sqrt{25-{\mathbf{x}}^2}-3;\mathbf{y}=\sqrt{4-{\mathbf{x}}^2}]$$$${\int }_0^4(\sqrt{25-{\mathbf{x}}^2}-\sqrt{4-{\mathbf{x}}^2}-3)\mathbf{dx}$$$$5{\int }_0^4\sqrt{1-{\left(\frac{\mathrm{x}}{5}\right)}^2}\mathrm{dx}-2\mid {\int }_0^2\sqrt{1-{\left(\frac{\mathrm{x}}{2}\right)}^2}\mid -3{\int }_0^4\mathrm{dx}$$$$=({\mathrm{I}}_1+{\mathrm{I}}_3)-{\mathrm{I}}_2$$$$\mathbf{Calcul}\mathbf{de}{\mathbf{I}}_1$$$$\mathbf{P}\mathrm{osons}\frac{\mathrm{x}}{5}=\cos \mathrm{t}\Rightarrow \sqrt{1-\cos {}^2\mathrm{t}}=\mathrm{sint}$$$$\mathrm{x}=5\mathrm{cost}[\mathrm{dx}=-5\sin \mathrm{tdt}$$$$\mathrm{x}=0\mathrm{t}=\frac{\pi }{2}\mathrm{x}=4\mathrm{t}=\arccos \left(\frac{4}{5}\right)$$$${\mathrm{I}}_1=-5\int \sin \mathrm{tcos}\mathrm{tdt}=-\frac{5}{2}{\int }_{\frac{\mathit{\pi }}{2}}^{\mathrm{s}\mathbf{rccos}\left(\frac{4}{5}\right)}\sin 2\mathrm{tdt}$$$$=\frac{-5}{4}{\left[\cos 2\mathrm{t}\right]}_{\frac{\mathit{\pi }}{2}}^{\arccos \left(\frac{4}{5}\right)}=\frac{-5}{4}[(0,4)-1]=\frac{3}{4}$$$$\mathbf{calcul}\mathbf{de}{\mathbf{I}}_2$$$${\mathbf{I}}_2=2\sqrt{1-{\left(\frac{\mathbf{x}}{2}\right)}^2}\mathbf{dx}\left(\frac{\mathbf{x}}{2}\right)=\mathbf{cos}\mathbf{t}$$$$\mathbf{x}=2\cos \mathbf{t}\mathbf{dx}=-2\mathbf{sin}\mathbf{tdt}$$$$\mathbf{x}=0\mathrm{t}=\frac{\mathit{\pi }}{2}\mathrm{x}=2\mathrm{t}=0$$$$-2{\int }_{\frac{\mathit{\pi }}{2}}^0\mathbf{sin}2\mathbf{tdt}=-{\left[\cos 2\mathrm{t}\right]}_{\frac{\mathit{\pi }}{2}}^0=-(1+1)=-2$$$$\mathbf{calcul}\mathbf{de}{\mathbf{I}}_3{\mathbf{I}}_3=3{\left[\mathrm{x}\right]}_0^4=12$$$$\mathbf{I}=\mid \frac{3}{4}-12+2\mid =\frac{3-48+8}{4}\mid =\frac{37}{4}$$$$$$

Commented by a.lgnaoui last updated on 06/May/23

$$Ithinkthereiserrorincalcul\left(integrals\right)?$$

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