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I tried to find the curved surface area of a hemisphere without integration. I found a glitch over there as I didn't found any mistake in my approach. Anyone who has found please do answer.

I got a handle of bucket and I formed a hemisphere by rotating it along the circumference of a imaginary semicircle of same radius 'R'(just for an idea for the figure). I found the curved surface area to be $\pi R \cdot \pi R$=$\pi ^2R^2$.

But it is actually $2\pi R^2$.Figure for help...

Prof. Meow
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1 Answers1

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It can be a mistake in calculating the circumference of individual lines on the hemisphere. Radius changes as we move outward. So, that can be avoided using Integration.

Prof. Meow
  • 101
  • In my opinion I got only this as my mistake but that is also getting avoided as we are multiplying the two circumferences and it is covering the full area of the hemisphere but not overlapping. You can also take the idea by taking a quarter circle and rotating it in along the circumference of a circle. – Prof. Meow Mar 02 '21 at 02:47
  • Using a quarter circle removes a source of confusion for some people. Now observe that $\pi^2 R^2$ is the area of the cylindrical surface you would get if you straightened the quarter-circle so that you had a segment of length $\frac12\pi R$ standing straight and perpendicular to the circle of radius $R.$ You're definitely treating all parts of the quarter-circle (except the end that's actually on the circle) as if they were farther from the rotation axis than they actually are. – David K Mar 02 '21 at 03:15
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    Yes thats what I was thinking when posting the answer of this question. – Prof. Meow Mar 02 '21 at 03:51