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Знайди об'єм тіла, отриманого при обертанні навколо осїабсцис фігури, обмеженої лініями: y=6x^(2),y=6x

Знайди об'єм тіла, отриманого при обертанні навколо осїабсцис фігури, обмеженої лініями:\newliney=6x2,y=6x y=6 x^{2}, y=6 x

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Q. Знайди об'єм тіла, отриманого при обертанні навколо осїабсцис фігури, обмеженої лініями:\newliney=6x2,y=6x y=6 x^{2}, y=6 x
  1. Identify bounds and region: Identify the bounds of integration and the region of interest.\newlineThe curves intersect where 6x2=6x6x^2 = 6x. Solving for xx, we get x2=xx^2 = x, so x(x−1)=0x(x - 1) = 0. This gives x=0x = 0 and x=1x = 1 as points of intersection.
  2. Set up integral for volume: Set up the integral for the volume using the disk method.\newlineThe volume VV is given by the integral from 00 to 11 of π(outer radius2−inner radius2) dx\pi(\text{outer radius}^2 - \text{inner radius}^2) \, dx. Here, the outer radius is y=6xy = 6x (from the line) and the inner radius is y=6x2y = 6x^2 (from the parabola).
  3. Express radii and plug in: Express the radii in terms of xx and plug into the volume formula.\newlineOuter radius = 6x6x, so outer radius2=36x2^2 = 36x^2.\newlineInner radius = 6x26x^2, so inner radius2=36x4^2 = 36x^4.\newlineVolume integral becomes V=π∫01(36x2−36x4)dxV = \pi\int_{0}^{1}(36x^2 - 36x^4) dx.
  4. Simplify and integrate: Simplify and integrate.\newlineV=π∫01(36x2−36x4)dx=π∫0136(x2−x4)dx.V = \pi\int_{0}^{1}(36x^2 - 36x^4) dx = \pi\int_{0}^{1}36(x^2 - x^4) dx.\newlineThis simplifies to V=36π∫01(x2−x4)dx.V = 36\pi\int_{0}^{1}(x^2 - x^4) dx.
  5. Calculate final volume: Calculate the integral. ∫01(x2−x4)dx=[x33−x55]01=(13−15)−(0−0)=215\int_{0}^{1}(x^2 - x^4) dx = \left[\frac{x^3}{3} - \frac{x^5}{5}\right]_{0}^{1} = (\frac{1}{3} - \frac{1}{5}) - (0 - 0) = \frac{2}{15}. V=36π×215=72π15=4.8πV = 36\pi \times \frac{2}{15} = \frac{72\pi}{15} = 4.8\pi.

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