Question

Determine o volume do sólido de revolução gerado pela função f(x)=cos(x), no intervalo de integração π≤x≤2π. represente graficamente o sólido.

Answer 1

O volume do sólido de revolução gerado pela rotação do gráfico de y = f(x) em torno do eixo x, no intervalo [a,b], é definido pela integral

$\boxed{\boxed{V=\displaystyle\pi\int_{a}^{b}[f(x)]^{2}dx}}$
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Temos a = π, b = 2π e f(x) = cos(x). Portanto:

$V=\displaystyle\pi\int_{\pi}^{2\pi}[cos(x)]^{2}dx\\\\\\V=\pi\int_{\pi}^{2\pi}cos^{2}(x)dx$
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Sabemos que

$cos(2x)=cos^{2}(x)-sen^{2}(x)\\\\cos(2x)=cos^{2}(x)-(1-cos^{2}(x))\\\\cos(2x)=cos^{2}(x)-1+cos^{2}(x)\\\\cos(2x)=2cos^{2}(x)-1\\\\\boxed{\boxed{cos^{2}(x)=\dfrac{1}{2}(cos(2x)+1)}}$

Então:

$V=\pi\displaystyle\int_{\pi}^{2\pi}cos^{2}(x)dx\\\\\\V=\pi\int_{\pi}^{2\pi}\dfrac{1}{2}\cdot(cos(2x)+1)dx\\\\\\V=\dfrac{\pi}{2}\int_{\pi}^{2\pi}cos(2x)+1dx\\\\\\V=\dfrac{\pi}{2}\int_{\pi}^{2\pi}cos(2x)dx+\dfrac{\pi}{2}\int_{\pi}^{2\pi}1dx\\\\\\\dfrac{\pi}{2}\int_{\pi}^{2\pi}cos(2x)dx+\dfrac{\pi}{2}\int_{\pi}^{2\pi}dx\\\\\\V=\dfrac{\pi}{2}\int_{\pi}^{2\pi}cos(2x)dx+\dfrac{\pi}{2}[x]_{\pi}^{2\pi}$

Se trocarmos 2x por u:

$du=2dx~~~\therefore~~~\boxed{\boxed{dx=\dfrac{1}{2}du}}$

Se x = π -----> u = 2π
Se x = 2π ---> u = 4π

Então:

$V=\displaystyle\dfrac{\pi}{2}\int_{2\pi}^{4\pi}\dfrac{1}{2}cos(u)du+\dfrac{\pi}{2}(2\pi-\pi)\\\\\\V=\dfrac{\pi}{4}\int_{2\pi}^{4\pi}cos(u)du+\dfrac{\pi}{2}\cdot\pi\\\\\\V=\dfrac{\pi}{4}[sen(u)]_{2\pi}^{4\pi}+\dfrac{\pi^{2}}{2}\\\\\\V=\dfrac{\pi}{4}(sen( \pi)-sen(2\pi))+\dfrac{\pi^{2}}{2}\\\\\\V=\dfrac{\pi}{4}(0-0)+\dfrac{\pi^{2}}{2}\\\\\\\boxed{\boxed{V=\dfrac{\pi^{2}}{2}~~uv}}$