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Question Number 20939 by Hitler last updated on 08/Sep/17

$$\mathrm{Demostration}\mathrm{of}\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{an}\mathrm{sphere}\mathrm{V}=\frac{4\pi {\mathrm{r}}^3}{3}$$$${\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2={\mathrm{r}}^2\mathrm{We}\mathrm{divide}\mathrm{the}\mathrm{sphere}\mathrm{in}8\mathrm{parts}.\mathrm{So}\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{a}\mathrm{part}\mathrm{is}$$$${\int }_0^{\mathrm{r}}{\int }_0^{\sqrt{{\mathrm{r}}^2-{\mathrm{x}}^2}}\sqrt{{\mathrm{r}}^2-{\mathrm{x}}^2-{\mathrm{y}}^2}\partial \mathrm{y}\partial \mathrm{x}\mathrm{Lets}\mathrm{asumme}{\mathrm{a}}^2={\mathrm{r}}^2-{\mathrm{x}}^2$$$${\int }_0^{\mathrm{r}}{\int }_0^{\mathrm{a}}\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}\partial \mathrm{y}\partial \mathrm{x}$$$${\int }_0^{\mathrm{r}}\mathrm{a}{\int }_0^{\mathrm{a}}\sqrt{1-{\left(\frac{\mathrm{y}}{\mathrm{a}}\right)}^2}\partial \mathrm{y}\partial \mathrm{x}\mathrm{Lets}\mathrm{assume}\frac{\mathrm{y}}{\mathrm{a}}=\sin \theta \Rightarrow \frac{\partial \mathrm{y}}{\mathrm{a}}=\cos \theta \partial \theta$$$$\int \sqrt{1-{\sin }^2\theta }\mathrm{acos}\theta \partial \theta$$$$\int {\mathrm{acos}}^2\theta \partial \theta$$$$\mathrm{a}(\frac{\theta }{2}-\frac{\sin 2\theta }{4})$$$$\mathrm{a}(\frac{\arcsin \left(\frac{\mathrm{y}}{\mathrm{a}}\right)}{2}-\frac{\mathrm{y}\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}{{2\mathrm{a}}^2})$$$${\int }_0^{\mathrm{r}}{\mathrm{a}}^2(\frac{\arcsin \left(\frac{\mathrm{y}}{\mathrm{a}}\right)}{2}-\frac{\mathrm{y}\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}{{2\mathrm{a}}^2}){\mid }_0^{\mathrm{a}}\partial \mathrm{x}$$$${\int }_0^{\mathrm{r}}\left(\frac{{\mathrm{a}}^2\arcsin \left(\frac{\mathrm{y}}{\mathrm{a}}\right)-\mathrm{y}\sqrt{{\mathrm{a}}^2-{\mathrm{y}}^2}}{2}\right){\mid }_0^{\mathrm{a}}\partial \mathrm{x}$$$${\int }_0^{\mathrm{r}}\frac{\pi {\mathrm{a}}^2}{4}\partial \mathrm{x}$$$${\int }_0^{\mathrm{r}}\frac{\pi ({\mathrm{r}}^2-{\mathrm{x}}^2)}{4}\partial \mathrm{x}$$$$\left(\frac{6\pi {\mathrm{r}}^2\mathrm{x}-2\pi {\mathrm{x}}^3}{24}\right){\mid }_0^{\mathrm{r}}$$$$\frac{\pi {\mathrm{r}}^3}{6}=1/8\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{sphrere}\mathrm{so}...$$$$\mathbb{V}=\frac{4\pi {\mathrm{r}}^3}{3}$$$$$$

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