Hollow sphere

The volume of the hollow ball is 3432 cm3. What is its internal radius when the wall thickness is 3 cm?

Correct result:

r =  8 cm

Solution:

V=3432 cm3 h=3 cm V=V1−V2 V1=4/3π⋅ (r+h)3 V2=4/3π⋅ r3 3/4/π⋅ V=(r+h)3−r3 k=3/4/π⋅ V=3/4/3.1416⋅ 3432≐819.3296 9r2+27r+27=819.3296  9r2+27r−792.33=0  a=9;b=27;c=−792.33 D=b2−4ac=272−4⋅9⋅(−792.33)=29252.8656 D>0  r1,2=−b±D2a=−27±29252.8718 r1,2=−1.5±9.50192728988 r1=8.00192728988 r2=−11.0019272899   Factored form of the equation:  9(r−8.00192728988)(r+11.0019272899)=0 r>0 r=[r1]=[8.0019]=8=8 cm V1=4/3π⋅ (r+h)3=4/3⋅ 3.1416⋅ (8+3)3≐5575.2798 V2=4/3π⋅ r3=4/3⋅ 3.1416⋅ 83≐2144.6606 V3=V1−V2=5575.2798−2144.6606≐3430.6192

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