Question

Um cilindro tem altura h=4R , em que R é o raio da base do cilindro.Qual deverá ser o valor do raio de uma esfera de modo que a área total do cilindro seja igual a àrea lateral da esfera ?

Answer 1

Boa noite Monica!

Solução!

Formulas do cilindro e da esfera.


$h=4r\\\\\\ At_{cilindro}=2. \pi .r(h+r)\\\\\\\ A_{esfera}=4 \pi.r^{2}\\\\\\ Condic\~ao!\\\\\\\ At_{cilindro}=A_{esfera}\\\\\\\ 2. \pi .r(h+r) = 4 \pi.r^{2}\\\\\\\ (h+r) =\dfrac{4 \pi.r^{2}}{2. \pi .r}\\\\\\\ (h+r) =2r\\\\\\ \dfrac{(h+r)}{2}=r$


$Substituindo~~o~~valor~~de~~h!\\\\\\\ \dfrac{(4r+r)}{2}=r\\\\\\ \boxed{\dfrac{5r}{2}=r}~~ \Rightarrow~~Valor~~do~~raio!$



$Verificando~~a~~igualdade!\\\\\\ 2. \pi \frac{5r}{2} ( \frac{5r}{2} + \frac{5r}{2} ) = 4 \pi.( \frac{5r}{2}) ^{2}\\\\\\\ \pi5r( \frac{10r}{2}) = 4 \pi.( \frac{25r^{2} }{4})\\\\\\\ \pi5r( \frac{10r}{2}) = \pi.25r^{2} \\\\\\\ \pi( \frac{50r^{2} }{2}) = \pi.25r^{2} \\\\\\\ \boxed{\pi 25r^{2}= \pi.25r^{2} }~~Ok!$

Boa noite!
Bons estudos!