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Question Number 215995 by universe last updated on 25/Jan/25

$$\int \int \underset{D}{\int }\sqrt{x^2+y^2+z^2}\mathrm{dv}=?$$$$\mathrm{D}=x^2+y^2+z^2<z$$

Answered by mr W last updated on 25/Jan/25

$$D=x^2+y^2+{(z-a)}^2⩽a^{2}witha=\frac{1}{2}$$$$x=\rho \cos \theta \cos \varphi$$$$y=\rho \cos \theta \sin \varphi$$$$z=a+\rho \sin \theta$$$$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\int }_0^a\sqrt{{\rho }^2{\cos }^2\theta +{(a+\rho \sin \theta )}^2}2\pi \rho \cos \theta \rho d\rho d\theta$$$$=2\pi {\int }_0^a{\rho }^2{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos \theta \sqrt{{\rho }^2+2a\rho \sin \theta +a^2}d\theta d\rho$$$$=\frac{2\pi }{3a}{\int }_0^a\rho {\left[{({\rho }^2+2a\rho \sin \theta +a^2)}^{\frac{3}{2}}\right]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}d\rho$$$$=\frac{2\pi }{3a}{\int }_0^a\rho [{({\rho }^2+2a\rho +a^2)}^{\frac{3}{2}}-{({\rho }^2-2a\rho +a^2)}^{\frac{3}{2}}]d\rho$$$$=\frac{2\pi }{3a}{\int }_0^a\rho [{(a+\rho )}^3-{(a-\rho )}^3]d\rho$$$$=\frac{2\pi }{3a}{\int }_0^a\rho (6a^2\rho +2{\rho }^3)d\rho$$$$=\frac{2\pi }{3a}{[2a^2{\rho }^3+\frac{2{\rho }^5}{5}]}_0^a$$$$=\frac{2\pi }{3}(2+\frac{2}{5})a^4$$$$=\frac{8\pi a^4}{5}$$$$=\frac{\pi }{10}$$

Commented by universe last updated on 25/Jan/25

$$thankyousir$$

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