2 Answers

Vijay Luxmi Askiitiansexpert
357 Points
13 years ago

Dear Amit ,

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

AskiitianExpert Shine
10 Points
13 years ago


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.

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