75 points. Will give certified if work is shown and the answer is correct.

Determine the difference between the volumes of two dwarf planets. Where one planet has a radius of 832 mi; while the other has a radius of 829 mi (express your volumes and final answers in terms of [tex] \pi [/tex] )

Respuesta :

Volume of the first dwarf planet (r₁ = 832 mi):

[tex]V_1=\dfrac{4}{3}\cdot\pi\cdot r_1^3=\dfrac{4}{3}\cdot\pi\cdot 832^3=\dfrac{2303721472}{3}\pi\approx7.679\cdot10^8\pi\,\text{mi}^3[/tex]

Volume of the second dwarf planet (r₂ = 829 mi):

[tex]V_2=\dfrac{4}{3}\cdot\pi\cdot r_2^3=\dfrac{4}{3}\cdot\pi\cdot 829^3=\dfrac{2278891156}{3}\pi\approx7.5963\cdot10^8\pi\,\text{mi}^3[/tex]

So difference between the volumes is:

[tex]V_1-V_2\approx7.679\cdot10^8\pi-7.5963\cdot10^8\pi=0.0827\cdot10^8\pi=\boxed{8270000\pi\,\text{mi}^3}[/tex]

or if we want exact value (we use (a³-b³) = (a-b)(a²+ab+b²) ):

[tex]V_1-V_2=\dfrac{4}{3}\cdot\pi\cdot r_1^3-\dfrac{4}{3}\cdot\pi\cdot r_2^3=\dfrac{4}{3}\pi(r_1^3-r_2^3)=\dfrac{4}{3}\pi(832^3-829^3)=\\\\\\=\dfrac{4}{3}\pi(832-829)(832^2+832\cdot829+829^2)=\\\\\\=\dfrac{4}{3}\pi\cdot3(692224+689728+687241)=4\pi\cdot2069193=\boxed{8276772\pi\,\text{mi}^3}[/tex]
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