Question

3. ( Uern ) A figura representa um sorvete de casquinha , no qual o volume interno está preenchido por sorvete e a parte externa apresenta um volume de meia bola de sorvete:

Answer 1

Boa noite Silvio!

Solução!
$h=12cm\\\ r=6cm$

$V_{c}= \dfrac{\pi.r^{2}.h }{3} \\\\\ V_{c}= \dfrac{\pi.6^{2}.12 }{3} \\\\\ V_{c}= \dfrac{\pi.36.12 }{3} \\\\\ V_{c}= \dfrac{\pi.432 }{3} \\\\\ V_{c}= \pi.144$

Volume da esfera.

$V_{es}= \dfrac{4.\pi,r^{3} }{3}\\\\\ V_{es}= \dfrac{4.\pi,6^{3} }{3}\\\\\ V_{es}= \pi.4.72 \\\\\ V_{es}= \pi.288 \\\\\ Lembre-se ~~que~~ e~~ a ~~metade ~~da ~~esfera\\\\\\ V_{es}=\dfrac{\pi.288}{2} \\\\\ V_{es}=\pi.144cm^{3}$

$Volume~~total=V_{c}+V_{es}\\\\\\ Volume~~total=\pi.144+\pi.144\\\\\\ Volume~~total=\pi.(144+144)\\\\\\ Volume~~total=\pi.288cm^{3}$

$\boxed{ Resposta:~~Volume~~total=288\pi cm^{3}~~~~Alternativa~~C}$

Boa noite!

Bons estudos!

Answer 2

Resposta:

C

Explicação passo-a-passo:

O volume total de sorvete é formado por uma semi esfera e um cone:

$V_{T} = V_{cone} + V_{semi \: esfera} \\ V_{T} = \frac{\pi {r}^{2}h }{3} + \frac{4\pi {r}^{3} }{3} \cdot \frac{1}{2} \\ V_{T} = \frac{\pi {r}^{2}h }{3} + \frac{4\pi {r}^{3} }{6} \\ V_{T} = \frac{\pi {6}^{2} \cdot12}{3} + \frac{4\pi {6}^{3} }{6} \\ V_{T} = \frac{\pi36 \cdot12}{3} + \frac{4\pi 216 }{6} \\ V_{T} = \frac{432\pi}{3} + \frac{864\pi }{6} \\ V_{T} = 144\pi + 144\pi \\ V_{T} = 288\pi$