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Question Number 20471 by ajfour last updated on 27/Aug/17

$$Findthesurfaceareaofasolid$$$$thatiscommonpartoftwo$$$$cylinders{x}^2+y^2=a^2,y^2+z^2=a^2.$$$$\mathit{C}\mathit{o}\mathit{m}\mathit{p}\mathit{u}\mathit{t}\mathit{e}\mathit{t}\mathit{h}\mathit{e}\mathit{v}\mathit{o}\mathit{l}\mathit{u}\mathit{m}\mathit{e}\mathit{a}\mathit{l}\mathit{s}\mathit{o}.$$

Commented by ajfour last updated on 27/Aug/17

Commented by ajfour last updated on 27/Aug/17

$$S=16{\int }_0^{\pi /2}xad\theta =16a^2{\int }_0^{\pi /2}\cos \theta d\theta$$$$=16a^2.(sincex=z=a\cos \theta ).$$

Answered by ajfour last updated on 28/Aug/17

$$V=16{\int }_0^a{\int }_0^{a\cos \theta }(a\cos \theta -x)dx]dy$$$$=16{\int }_0^{\pi /2}(ax\cos \theta -\frac{x^2}{2}){\mid }_0^{a\cos \theta }\left(a\cos \theta \right)d\theta$$$$=16{\int }_0^{\pi /2}\frac{a^2\cos {}^2\theta }{2}\left(a\cos \theta \right)d\theta$$$$=8a^3{\int }_0^{\pi /2}(1-\sin {}^2\theta )d\left(\sin \theta \right)$$$$=8a^3[\sin \theta -\frac{\sin {}^3\theta }{3}]{\mid }_0^{\pi /2}$$$$\mathit{V}=\frac{16{\mathit{a}}^3}{3}.$$

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