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11 years ago


Dear Amit ,
Please post your query again , we are unable to understand .


11 years ago
							Hi
VOLUME OF A SPHERE( its formula can be derived using integral calculus)
At any given x, the incremental volume (δV) is given by the product of the cross-sectional area of the disk at x and its thickness (δx): $\!\delta V \approx \pi y^2 \cdot \delta x.$
The total volume is the summation of all incremental volumes: $\!V \approx \sum \pi y^2 \cdot \delta x.$
In the limit as δx approaches zero this becomes: $\!V = \int_{x=0}^{x=r} \pi y^2 dx.$
At any given x, a right-angled triangle connects x, y and r to the origin, whence it follows from Pythagorean theorem that: $\!r^2 = x^2 + y^2.$
Thus, substituting y with a function of x gives: $\!V = \int_{x=0}^{x=r} \pi (r^2 - x^2)dx.$
This can now be evaluated: $\!V = \pi \left[r^2x - \frac{x^3}{3} \right]_{x=0}^{x=r} = \pi \left(r^3 - \frac{r^3}{3} \right) = \frac{2}{3}\pi r^3.$
This volume as described is for a hemisphere. Doubling it gives the volume of a sphere as: $\!V = \frac{4}{3}\pi r^3.$

11 years ago
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