Question

O volume gerado pela rotação em torno do eixo dos X do grafico de uma função y= f(x) num intervalo [ A, B ] e dado por
$V = \pi . \int\limits^b_a {y} ^{2} \, dx$
Sendo assim calcule o volume do solido de revolução gerado pela função y=2x e torno dos X, no intervalo X ∈ [ 1, 2 ].

Alternativas:

a) $y= \frac{32}{3} \pi$
b) $y= \frac{2}{3} \pi$
c) $y= \frac{14}{3} \pi$
d) $y= \frac{28}{3} \pi$

Answer 1

Boa tarde Pirata!

Solução!

Formula para calcular o volume!

$\boxed{V= \pi \displaystyle \int_{a}^{b}[f(x)]^{2}dx}$


$a=1\\\\ b=2\\\\ f(x)=2x$

$V= \pi \displaystyle \int_{1}^{2}[2x^{1} ]^{2}dx\\\\\\\\\\\ V= \pi \displaystyle \int_{1}^{2}(4x^{2} )dx\\\\\\\\\\ V= \pi \displaystyle \int_{1}^{2}\frac{(4x)}{2+1} ^{2+1} dx= \frac{4x^{3} }{3} \\\\\\\\\\ V= \pi \Bigg|_{1}^{2} \left ( \frac{ 4x^{3} }{3} \right )-\left ( \frac{4 x^{3} }{3} \right )\\\\\\\\\\ V= \pi \Bigg|_{1}^{2} \left ( \frac{ 4(2)^{3} }{3} \right )-\left ( \frac{4 (1)^{3} }{3} \right )$

$V= \pi \Bigg|_{1}^{2} \left ( \dfrac{ 4(8) }{3} \right )-\left ( \dfrac{4 (1) }{3} \right )\\\\\\\ V= \pi \Bigg|_{1}^{2} \left ( \dfrac{ 32 }{3} \right )-\left ( \dfrac{4 }{3} \right )\\\\\\\ V= \pi \Bigg|_{1}^{2} \left ( \dfrac{ 32 }{3} - \dfrac{4}{3} \right )\\\\\\ V= \pi \Bigg|_{1}^{2} \left ( \dfrac{ 28 }{3} \right )\\\\\\ V= \dfrac{ 28 \pi }{3}$



$\boxed{Resposta:V= \pi \displaystyle \int_{1}^{2}[2x]^{2}dx= \frac{28 \pi }{3}~~\boxed{Alternativa~~D}}$


Boa tarde!
Bons estudos!

Answer 2

$\boxed{\begin{array}{c}V=\pi\cdot \displaystyle\int_a^b y^2\,dx \end{array}}$

$y=2x\,,$ e $x\in[1,\,2].$


Logo, o volume do sólido de revolução é dado por

$V=\pi\cdot \displaystyle\int_1^2 y^2\,dx\\\\\\ V=\pi\cdot \int_1^2 (2x)^2\,dx\\\\\\ =\pi\cdot \int_1^2 4x^2\,dx\\\\\\ =\pi\cdot \left.\left(\dfrac{4x^3}{3} \right )\right|_1^2\\\\\\ =\pi\cdot \left(\dfrac{4\cdot 2^3}{3}-\dfrac{4\cdot 1^3}{3} \right )\\\\\\ =\pi\cdot \left(\dfrac{4\cdot 8}{3}-\dfrac{4\cdot 1}{3} \right )$

$=\pi\cdot \left(\dfrac{32}{3}-\dfrac{4}{3} \right )\\\\\\ =\pi\cdot \dfrac{32-4}{3}\\\\\\ \therefore~~\boxed{\begin{array}{c}V=\dfrac{28}{3}\,\pi\mathrm{~u.v.} \end{array}}$


Resposta: alternativa $\text{d) }V=\dfrac{28}{3}\,\pi.$