# 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:

$V=3432 \ \text{cm}^3 \ \\ h=3 \ \text{cm} \ \\ V=V_{1} - V_{2} \ \\ V_{1}=4/3 \pi \cdot \ (r+h)^3 \ \\ V_{2}=4/3 \pi \cdot \ r^3 \ \\ 3/4/\pi \cdot \ V=(r+h)^3 - r^3 \ \\ k=3/4/\pi \cdot \ V=3/4/3.1416 \cdot \ 3432 \doteq 819.3296 \ \\ 9r^2+27r +27=819.3296 \ \\ \ \\ 9r^2 +27r -792.33=0 \ \\ \ \\ a=9; b=27; c=-792.33 \ \\ D=b^2 - 4ac=27^2 - 4\cdot 9 \cdot (-792.33)=29252.8656 \ \\ D>0 \ \\ \ \\ r_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ -27 \pm \sqrt{ 29252.87 } }{ 18 } \ \\ r_{1,2}=-1.5 \pm 9.5019272898829 \ \\ r_{1}=8.0019272898829 \ \\ r_{2}=-11.001927289883 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 9 (r -8.0019272898829) (r +11.001927289883)=0 \ \\ r>0 \ \\ r=[ { r_{1} } ]=[ { 8.0019 } ]=8 \ \text{cm} \ \\ V_{1}=4/3 \pi \cdot \ (r+h)^3=4/3 \cdot \ 3.1416 \cdot \ (8+3)^3 \doteq 5575.2798 \ \\ V_{2}=4/3 \pi \cdot \ r^3=4/3 \cdot \ 3.1416 \cdot \ 8^3 \doteq 2144.6606 \ \\ V_{3}=V_{1}-V_{2}=5575.2798-2144.6606 \doteq 3430.6192$

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