Question

$\rm Deduza~a~f\acute{o}rmula~do~volume~do~cone$

Answer 1

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$\sf a=tg(m)=\dfrac{R}{h}\\\sf y=ax\implies \boxed{\boxed{\sf y=\dfrac{R}{h}\cdot x}}$

$\sf observe~o~anexo.\\\sf considere~a~func_{\!\!,}\tilde{a}o~y=ax~e~o~intervalo~de~0~a~h.\\\sf Ao~ girarmos~o~tri\hat{a}ngulo~ret\hat{a}ngulo~em~torno~do~eixo~x~teremos~o~referido~cone.\\\sf usando~o~m\acute{e}todo~dos~an\acute{e}is~circulares~temos:$

$\displaystyle\sf V=\pi\int_{a}^{b}f(x)^2~dx\\\displaystyle\sf V=\pi\cdot\int_{0}^{h}\left(\dfrac{R}{h}\cdot x\right)^2~dx\\\displaystyle\sf V=\pi\cdot\int_{0}^{h}\dfrac{R^2}{h^2}\cdot x^2~dx\\\displaystyle\sf V=\pi\cdot\dfrac{R^2}{h^2}\int_{0}^{h} x^2~dx\\\displaystyle\sf V=\pi\cdot\dfrac{R^2}{h^2}\cdot\dfrac{1}{3}x^3\Bigg|_{0}^{h}$

$\sf V=\pi\cdot\dfrac{R^2}{\diagup\!\!\!\!h^2}\cdot\dfrac{1}{3}\cdot \diagup\!\!\!\!h^3\\\huge\boxed{\boxed{\boxed{\boxed{\boxed{\sf V=\dfrac{1}{3}\cdot\pi\cdot R^2\cdot h}}}}}\\\boxed{\begin{array}{c}\sf{V\longrightarrow volume~do~cone}\\\sf R\longrightarrow raio~do~cone\\\sf h\longrightarrow altura~do~cone\end{array}}$