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

Demostration of the volume of an sphere V = \frac{4 π r^{3}}{3}

x^{2} + y^{2} + z^{2} = r^{2} We divide the sphere in 8 parts . So the volume of a part is

\int_{0}^{r} \int_{0}^{\sqrt{r^{2} - x^{2}}} \sqrt{r^{2} - x^{2} - y^{2}} \partial y \partial x Lets asumme a^{2} = r^{2} - x^{2}

\int_{0}^{r} \int_{0}^{a} \sqrt{a^{2} - y^{2}} \partial y \partial x

\int_{0}^{r} a \int_{0}^{a} \sqrt{1 - {(\frac{y}{a})}^{2}} \partial y \partial x Lets assume \frac{y}{a} = \sin θ \Rightarrow \frac{\partial y}{a} = \cos θ \partial θ

\int \sqrt{1 - \sin^{2} θ} acos θ \partial θ

\int {acos}^{2} θ \partial θ

a (\frac{θ}{2} - \frac{\sin 2 θ}{4})

a (\frac{\arcsin (\frac{y}{a})}{2} - \frac{y \sqrt{a^{2} - y^{2}}}{{2 a}^{2}})

\int_{0}^{r} a^{2} (\frac{\arcsin (\frac{y}{a})}{2} - \frac{y \sqrt{a^{2} - y^{2}}}{{2 a}^{2}}) ∣_{0}^{a} \partial x

\int_{0}^{r} (\frac{a^{2} \arcsin (\frac{y}{a}) - y \sqrt{a^{2} - y^{2}}}{2}) ∣_{0}^{a} \partial x

\int_{0}^{r} \frac{π a^{2}}{4} \partial x

\int_{0}^{r} \frac{π (r^{2} - x^{2})}{4} \partial x

(\frac{6 π r^{2} x - 2 π x^{3}}{24}) ∣_{0}^{r}

\frac{π r^{3}}{6} = 1 / 8 Volume of the sphrere so \dots

V = \frac{4 π r^{3}}{3}