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Question Number 153918 by joki last updated on 12/Sep/21

$$\mathrm{the}\mathrm{base}\mathrm{of}\mathrm{an}\mathrm{object}\mathrm{is}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{with}$$$$\mathrm{radius}1.\mathrm{suppose}\mathrm{that}\mathrm{all}\mathrm{section}\mathrm{of}\mathrm{the}\mathrm{object}\mathrm{are}$$$$\mathrm{perpendicular}\mathrm{to}\mathrm{a}\mathrm{diameter}\mathrm{of}\mathrm{a}\mathrm{square}.\mathrm{determine}$$$$\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{object}?$$

Answered by talminator2856791 last updated on 12/Sep/21

$$$$$$\equiv 2{\int }_0^{\frac{\pi }{2}}{(2\cdot \sin \left(x\right))}^{2}dx$$$$=2\pi$$$$$$

Commented by alisiao last updated on 12/Sep/21

$$=\left(2{\int }_0^{\frac{\mathit{\pi }}{2}}{\left(\frac{{\mathit{e}}^{\mathit{i}\mathit{x}}-{\mathit{e}}^{-\mathit{i}\mathit{x}}}{\mathit{i}}\right)}^2\mathit{d}\mathit{x}\right)$$$$$$$$=(-2{\int }_0^{\frac{\mathit{\pi }}{2}}({\mathit{e}}^{2\mathit{i}\mathit{x}}-2+{\mathit{e}}^{-2\mathit{i}\mathit{x}})\mathit{d}\mathit{x})$$$$$$$$\text{Missing left or extra right}$$$$$$$$\text{Missing left or extra right}$$$$$$$$=[-2(0-\mathit{\pi })]=2\mathit{\pi }$$$$$$$$⟨\mathit{M}.\mathit{T}⟩$$

Commented by alisiao last updated on 12/Sep/21

$$=8{\int }_0^{\frac{\mathit{\pi }}{2}}{\mathit{s}\mathit{i}\mathit{n}}^2\left(\mathit{x}\right)\mathit{d}\mathit{x}=8{\int }_0^{\frac{\mathit{\pi }}{2}}\frac{1}{2}(1-\mathit{c}\mathit{o}\mathit{s}\left(2\mathit{x}\right))\mathit{d}\mathit{x}$$$$$$$$=4{(\mathit{x}-\frac{1}{2}\mathit{s}\mathit{i}\mathit{n}\left(2\mathit{x}\right))}_0^{\frac{\mathit{\pi }}{2}}=4\left(\frac{\mathit{\pi }}{2}\right)=2\mathit{\pi }$$$$$$$$⟨\mathit{M}.\mathit{T}⟩$$

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