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anonymous

  • 4 years ago

Calculus: Rotating volume

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  1. anonymous
    • 4 years ago
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  2. anonymous
    • 4 years ago
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    # 22

  3. anonymous
    • 4 years ago
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    so?

  4. dumbcow
    • 4 years ago
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    x = y^3, x = 4y y^3 = 4y y^3 -4y = 0 y(y^2 -4) = 0 y(y-2)(y+2) = 0 they intersect at y= 0,2, -2 to graph: y = cubert(x), y = x/4 intersect at x = 0, 8, -8 |dw:1328001094873:dw|

  5. anonymous
    • 4 years ago
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    ok

  6. anonymous
    • 4 years ago
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    ummm lol How wld u set up? Sorry I need giuded help :D

  7. anonymous
    • 4 years ago
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    * set up the integral?

  8. dumbcow
    • 4 years ago
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    \[V =\pi \int\limits_{-8}^{8}R^{2} -r^{2} dx\] so you are integrating over the circular areas of each cross-section R is outer radius from axis r is inner radius from axis

  9. anonymous
    • 4 years ago
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    ohhhh i see Thanks :D

  10. anonymous
    • 4 years ago
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    Thanks Dumbcow

  11. dumbcow
    • 4 years ago
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    welcome make sure you get R and r in terms of x, also this set up only works if you are revolving around a horizontal axis

  12. anonymous
    • 4 years ago
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    well it says the y axis

  13. dumbcow
    • 4 years ago
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    i just noticed something, because of the shape of the region, the outer and inner radius flip when x is negative. so just integrate from 0 to 8 but multiply by 2. due to symmetry the two regions will have identical volume

  14. anonymous
    • 4 years ago
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    ok I see Thanks :D

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