The volume of a sphere is given by:
So, we need to deduct this equation. We will walk through Calculus on the concept of a solid of revolution that is a solid figure that is obtained by rotating a plane curve around some straight line (the axis of revolution<span>) that lies on the same plane. We know from calculus that:
</span>![V=\pi \int_{a}^{b}[f(x)]^{2}dx](https://tex.z-dn.net/?f=V%3D%5Cpi%20%5Cint_%7Ba%7D%5E%7Bb%7D%5Bf%28x%29%5D%5E%7B2%7Ddx)
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Then, according to the concept of solid of revolution we are going to rotate a circumference shown in the figure, then:
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Isolationg y:
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So,
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</span>![V=\pi \int_{a}^{b}[\sqrt{r^{2}-x^{2}}]^{2}dx](https://tex.z-dn.net/?f=V%3D%5Cpi%20%5Cint_%7Ba%7D%5E%7Bb%7D%5B%5Csqrt%7Br%5E%7B2%7D-x%5E%7B2%7D%7D%5D%5E%7B2%7Ddx)
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being -r and r the limits of this integral.
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Solving:
</span>![V=\pi[r^{2}x-\frac{x^{3}}{3}]\right|_{-r}^{r}](https://tex.z-dn.net/?f=V%3D%5Cpi%5Br%5E%7B2%7Dx-%5Cfrac%7Bx%5E%7B3%7D%7D%7B3%7D%5D%5Cright%7C_%7B-r%7D%5E%7Br%7D)
Finally:
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So, we need to deduct this equation. We will walk through Calculus on the concept of a solid of revolution that is a solid figure that is obtained by rotating a plane curve around some straight line (the axis of revolution<span>) that lies on the same plane. We know from calculus that:
</span>
<span>
Then, according to the concept of solid of revolution we are going to rotate a circumference shown in the figure, then:
</span>
<span>
Isolationg y:
</span>
So,
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being -r and r the limits of this integral.
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Solving:
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Finally:
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