To calculate the volume of a sphere of radius R, I considered a thin disc of radius r-R sin theta, where theta is the angle radius vector on the circumference makes with the axis of the disc.

brasocas6

brasocas6

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2022-08-16

Intuitive error in finding Volume of sphere using single integration.
To calculate the volume of a sphere of radius R, I considered a thin disc of radius r = R s i n θ, where θ is the angle radius vector on the circumference makes with the axis of the disc. However I considered the volume d V = π r 2 R d θ, R d θ being the infinitesimal thickness of the disc, which is erroneous. The volume of the sphere comes out to be 3 π 2 R 3 2 . The correct thickness is R s i n θ d θ which is found usually in other derivations gives the correct volume. However I do not understand what exactly is the issue in my approach and why the component of the curve length is considered.

Answer & Explanation

Riya Cline

Riya Cline

Beginner2022-08-17Added 17 answers

Step 1
Let the x-axis of R 3 be the axis of the disc. The disc is then bounded by the sphere of radius R and two planes x = a and x = b. We now have to express a and b in terms of θ and d θ. When b = R cos θ then a = R cos ( θ + d θ ) < b.
Step 2
The thickness of the disc is then given by b a = R ( cos θ cos ( θ + d θ ) ) R ( cos ( θ ) ) d θ = R sin θ d θ   ,
and has nothing to do with the arc length R d θ along the meridian of B. We obtain v o l ( B ) = 0 π π r 2 ( θ ) R sin θ d θ = 4 π 3 R 3   .

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