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anonymous

  • 5 years ago

A solid generated when region in the first quadrant is enclosed by curves y=2x^2 and y^2=4x is revolved around the x-axis. What is the volume? A. Pi/5 B. 2Pi/5 C. 8Pi/5 D. 6Pi/5

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  1. anonymous
    • 5 years ago
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    solve for intersections sub the first eqn into the second equation 4x^4= 4x x^4 - x =0 x(x^3-1) =0 x=0, x=1 for real solutions

  2. anonymous
    • 5 years ago
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  3. anonymous
    • 5 years ago
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    See the picture this is equal to the volumeof rotation below the curve y^2 = 4x minus the volume of rotation found by rotating the lower curve ( ie y=2x^2) Remember Volume of curve about x axis form x=a to x=b is \[V = \int\limits_{a}^{b} y^2 dx\]

  4. anonymous
    • 5 years ago
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    so our volume is equal to \[V= \int\limits_{0}^{1} ( 4x ) dx - \int\limits_{0}^{1} ( 2x^2)^2 dx\]

  5. anonymous
    • 5 years ago
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    \[ V= \int\limits_{0}^{1} 4x - 4x^4 dx \]

  6. anonymous
    • 5 years ago
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    Whoops, there should be a actor of pi outside the the integrals

  7. anonymous
    • 5 years ago
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    so answer = pi [ 2 - 4/5] = 6pi/ 5 = \[\frac{6\pi}{5}\]

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