Hollow sphere

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

Correct answer:

r =  8 cm

Step-by-step explanation:

$$\begin{gathered} V=3432{\text{cm}}^3 \\ h=3\text{ cm } \\ V=V_1-V_2 \\ {V}_1=4\mathrm{/}3\pi \cdot (r+h{)}^3 \\ {V}_2=4\mathrm{/}3\pi \cdot r^3 \\ 3\mathrm{/}4\mathrm{/}\pi \cdot V=(r+h{)}^3-r^3 \\ k=3\mathrm{/}4\mathrm{/}\pi \cdot V=3\mathrm{/}4\mathrm{/}3.1416\cdot 3432\doteq 819.3296 \\ 9r^2+27r+27=819.3296 \\ 9r^2+27r-792.33=0 \\ a=9;b=27;c=-792.33 \\ D=b^2-4ac=27^2-4\cdot 9\cdot (-792.33)=29252.8656 \\ D>0 \\ {r}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{-27\pm \sqrt{29252.87}}{18}} \\ {r}_{1,2}=-1.5\pm 9.50192728988 \\ {r}_1=8.00192728988 \\ {r}_2=-11.0019272899 \\ \text{ Factored form of the equation: } \\ 9(r-8.00192728988)(r+11.0019272899)=0 \\ r>0 \\ r=[r_1]=[8.0019]=8\text{ cm } \\ {V}_1=4\mathrm{/}3\pi \cdot (r+h{)}^3=4\mathrm{/}3\cdot 3.1416\cdot (8+3{)}^3\doteq 5575.2798 \\ {V}_2=4\mathrm{/}3\pi \cdot r^3=4\mathrm{/}3\cdot 3.1416\cdot 8^3\doteq 2144.6606 \\ {V}_3=V_1-V_2=5575.2798-2144.6606\doteq 3430.6192 \end{gathered}$$

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