Axial section

The diagonal of the axial section of the rotating cylinder is 6 cm, and its surface is 30 cm². Calculate the radius of the base.

Final Answer:

r =  0.7291 cm

Step-by-step explanation:

u=6 cm S=30 cm2  S = 2π r2 + 2π⋅ r⋅ h  u2 = (2r)2+h2 u2 = 4r2+h2 r2 = (u2−h2)/4  h = u2 −4r2​   S = 2π r2 + 2π⋅ r⋅ u2 −4r2​  2πS​ = r2 + r⋅ u2 −4r2​ = r(r+(u−2r)(u+2r)​)  S = 2π ⋅ (u2−h2)/4 + π⋅ u2−h2​⋅ h 30 = 2π ⋅ (62−h2)/4 + π⋅ 62−h2​⋅ h  h1​=−1.2987 h2​=5.8201 h>0  h=h2​=5.8201 cm  r=u2−h2​/2=62−5.82012​/2≐0.7291 cm   Verifying Solution:  S2​=2π⋅ r2+2π⋅ r⋅ h=2⋅ 3.1416⋅ 0.72912+2⋅ 3.1416⋅ 0.7291⋅ 5.8201≐30.003 cm2 u2​=4⋅ r2+h2​=4⋅ 0.72912+5.82012​=6 cm

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