Find the volumes of the solids whose bases are bounded by the circle with the indicated cross sections taken perpendicular to the axis

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Find the volumes of the solids whose bases are bounded by the circle with the indicated cross sections taken perpendicular to the axis

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Mộc Miên 5 years 2021-08-16T00:53:08+00:00 1 Answers 35 views 0

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  1. kimcuc
    0
    2021-08-16T00:54:14+00:00
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    Answer:

    Step-by-step explanation:

    The missing information includes finding the volume of the circle x² + y² = 4 from the x-axis of the squares and the equilateral triangles.

    x^2 + y^2 = 4

    Perpendicular to x-axis

    y = \sqrt{4 -x^2}

    Area of the square = ( \sqrt{4-x^2}+ \sqrt{4-x^2})^2

    = (2 \sqrt{4 -x^2})^2 \\ \\ = 4(4-x^2) \\ \\ = 16 - 4x^2

    Volume  V = \int ^2_{-2} (16 -4x^2) \ dx

    = \Big [ 16x - \dfrac{4x^3}{3} \Big]^2_{-2}

    = \Big [32 - \dfrac{32}{3} \Big] - \Big[-32 +\dfrac{32}{3} \Big]

    = \dfrac{128}{3}

    Area of Equilateral triangle

    = \dfrac{1}{2}\times 2 \sqrt{4-x^2}\times \sqrt{3}\sqrt{4-x^2}

    = \sqrt{3}(\sqrt{4-x^2})^2

    Volume (V )= \int ^2_{-2}\sqrt{3} (4 -x^2) \ dx

    = 4\sqrt{3} \ x - \dfrac{\sqrt{3} \ x^3 }{3} \Big|^2_{-2} \\ \\ = 16 \sqrt{3} - \dfrac{16\sqrt{3}}{3} \\ \\ = \dfrac{32\sqrt{2}}{3}

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