Two hemispheres

In a wooden hemisphere with a radius r = 1, a hemispherical depression with a radius r/2 was created so that the bases of both hemispheres lie in the same plane. What is the surface of the created body (including the surface of the depression)?

Correct result:

S =  10.21 cm2

Solution:

r=1 r2=r/2=1/2=12=0.5  S1=π⋅ r2=3.1416⋅ 12≐3.1416 S2=π⋅ r22=3.1416⋅ 0.52≐0.7854  S11=4π⋅ r2/2=4⋅ 3.1416⋅ 12/2≐6.2832 S22=4π⋅ r22/2=4⋅ 3.1416⋅ 0.52/2≐1.5708  S=S11+S22+S1−S2=6.2832+1.5708+3.1416−0.7854≐10.2102   Correctness test:  q=S/π/r2=10.2102/3.1416/12=134=3.25  S=134⋅ π⋅ r2=10.21 cm2r=1 \ \\ r_{2}=r/2=1/2=\dfrac{ 1 }{ 2 }=0.5 \ \\ \ \\ S_{1}=\pi \cdot \ r^2=3.1416 \cdot \ 1^2 \doteq 3.1416 \ \\ S_{2}=\pi \cdot \ r_{2}^2=3.1416 \cdot \ 0.5^2 \doteq 0.7854 \ \\ \ \\ S_{11}=4 \pi \cdot \ r^2/2=4 \cdot \ 3.1416 \cdot \ 1^2/2 \doteq 6.2832 \ \\ S_{22}=4 \pi \cdot \ r_{2}^2/2=4 \cdot \ 3.1416 \cdot \ 0.5^2/2 \doteq 1.5708 \ \\ \ \\ S=S_{11}+S_{22}+S_{1}-S_{2}=6.2832+1.5708+3.1416-0.7854 \doteq 10.2102 \ \\ \ \\ \text{ Correctness test: } \ \\ q=S/\pi/r^2=10.2102/3.1416/1^2=\dfrac{ 13 }{ 4 }=3.25 \ \\ \ \\ S=\dfrac{ 13 }{ 4 } \cdot \ \pi \cdot \ r^2=10.21 \ \text{cm}^2

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