Question

solid obtained by rotation around the axis x = -1, the region delimited by x^2 - x + y = 0 and the abscissa axis

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Answer to a math question solid obtained by rotation around the axis x = -1, the region delimited by x^2 - x + y = 0 and the abscissa axis

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Eliseo
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105 Answers
Para resolver esse problema, podemos primeiro determinar os limites de integração para a rotação em torno do eixo x = -1.

A equação x^2 - x + y = 0 pode ser reescrita como y = x - x^2.

A região delimitada por essa equação e o eixo das abscissas é representada pelo gráfico abaixo:

\begin{tikzpicture}\begin{axis}[ axis lines = left, xlabel = $x$, ylabel = $y$, xmin=-1, xmax=1, ymin=0, ymax=1,]\addplot [ domain=-1:1, samples=100, color=blue,]{x-x^2};\end{axis}\end{tikzpicture}

Agora, para obter o sólido obtido pela rotação em torno do eixo x = -1, podemos usar o método do disco ou do anel.

Vamos usar o método do anel.

O raio do anel é dado pela distância entre o ponto (x, x-x^2) no gráfico e o eixo de rotação x = -1, que é r(x) = x - (-1) = x + 1.

A área do anel é dada por A(x) = π[r(x)^2 - r(x - \Delta x)^2], onde \Delta x é uma pequena variação no valor de x.

Agora, vamos calcular o volume do sólido integrando as áreas dos anéis ao longo do intervalo de x.

[passar o texto]

Portanto, o volume do sólido é dado por:

V = \int_{a}^{b} A(x) dx

V = \int_{-1}^{1} \pi[x + 1)^2 - (x + \Delta x + 1)^2] dx

V = \int_{-1}^{1} \pi[(2x + 1) \Delta x - \Delta x^2] dx

V = \pi \int_{-1}^{1} (2x + 1) \Delta x - \pi \int_{-1}^{1} \Delta x^2 dx

Agora, vamos resolver cada umas das integrais.

\pi \int_{-1}^{1} (2x + 1) \Delta x = \pi [\Delta x^2 + x^2 + x] \bigg|_{-1}^{1}

\pi [\Delta x^2 + 1 + 1] - \pi [\Delta x^2 + 1 - 1]

\pi [\Delta x^2 + 2] - \pi [\Delta x^2]

2 \pi

A segunda integral é:

\pi \int_{-1}^{1} \Delta x^2 dx = \pi \Delta x^3/3 \bigg|_{-1}^{1}

\pi [(1/3) - (-1/3)]

\pi (2/3)

Agora, vamos substituir esses valores na fórmula do volume:

V = 2 \pi - \pi (2/3)

V = 2 \pi - (2/3) \pi

V = (2 - 2/3) \pi

V = (4/3) \pi

Portanto, a resposta é:

\text{Answer: } V = (4/3) \pi

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