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Given a sphere with radius 30cm and there is a hole of radius 3cm.

How can I complete the question in a cylindrical coordinate?

My attempt:

$Surface Area=\int{\int_D{\sqrt{f_x^2+f_y^2+1}}} dA$

I don't know how to find out the area.

YKWAN
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  • I would like to ask about the upper limit and lower limit for the integral. Thank you. – YKWAN Mar 20 '20 at 11:54
  • Assuming the cap.is missing one can consider s surface of revolution. Perhaps https://math.stackexchange.com/questions/3586722/loss-of-height-rate-in-a-half-sphere/3586755#3586755 is useful. – Peter Szilas Mar 20 '20 at 12:07
  • draw a cross section and apply Pythagora's on $R, r$ and find the angle limits – G Cab Mar 20 '20 at 12:11
  • 0<=theta<=2pi how about r? – YKWAN Mar 20 '20 at 12:16
  • Astonishingly the volume of the remaining "ring" solely depends on its height, see https://math.stackexchange.com/questions/658004/remaining-volume-of-a-sphere/658215#658215 – Michael Hoppe Mar 20 '20 at 12:30
  • I need to calculate surface area instead of volume. – YKWAN Mar 20 '20 at 12:44
  • did you draw a sketch of the cross section ? do you see the angle $\alpha = arcsin(r/R)$ ? – G Cab Mar 20 '20 at 13:48
  • sphere: x^2+y^2+z^2=30^2 cylinder: x^2+y^2=3^2 – YKWAN Mar 20 '20 at 14:20
  • https://math.stackexchange.com/questions/347180/surface-area-of-sphere-with-cylindrical-cutout

    I tried to follow his way to do so but his way is not correct

     – YKWAN Mar 20 '20 at 14:20

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