IIT JEE , AIEEE Forum - Mathematics , Analytical Geometry

karthik_abiram

HERE IS ANOTHER METHOD DONE BY MYSELF:---

consider the sphere be made up of several circles.the sum of the cicumference areas of the circles with varying radius from -r to +r is equal to the surface area of sphere.consider one such circle of radius r. and width dr.since it is symmetrical we can change the limits from 0 to 2R.

circumference of the circle=2

r-------------eq1---integrating to the limits ,0 to 2R we get

_{[ ]}

^{[ ]} circumference of the cicle=_[0_]

^[2R^] 2

r

=_[0_]^[2R]2

r

_2R

=2

(r²)₀ /2

=2

(4R²)/2

=4

R²

I THINK WE COULD NOT PROVE THIS BY INTEGRATION AS IT HAS NO INTEGRATOR.THE INTEGRATOR dr WHICH IS WIDTH CAN COME ONLY IN MASS AND VOLUME AND NOT IN AREA.HENCE I THINK THIS COULD NOT BE SOLVED IN THE ABOVE METHOD.I AM REALLY SORRY.PLEASE CORRECT ME IF I AM WRONG.

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