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anonymous

  • 4 years ago

how do I set up the integral to find the volume of the solid obtained by revolving the region about the x-axis when y=square root of x, y=1 and x=4

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  1. anonymous
    • 4 years ago
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    I think it turns into a disk but I am not sure. And I cant figure out the integral.

  2. dumbcow
    • 4 years ago
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    |dw:1328252349257:dw| this uses the washer method where for each circular cross-section you take large area and subtract smaller area Here the outer radius is sqrt(x), the inner radius is 1 \[V = \pi \int\limits_{?}^{?}R^{2} -r^{2} \] \[V = \pi \int\limits_{1}^{4}(\sqrt{x}^{2} - 1^{2}) dx = \pi \int\limits_{1}^{4}(x - 1)dx\]

  3. anonymous
    • 4 years ago
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    thank you! but what if I want to switch it and revolve it about the x axis or another point such as x=-1? how do I know how to change my integrals?

  4. dumbcow
    • 4 years ago
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    whatever the axis you're revolving it around, you just need to determine the radius or the distance from the axis to the curve

  5. dumbcow
    • 4 years ago
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    example: for this region say we revolve around x=-2: my curve is now 2 units further away from the axis, just add 2 to the radius outer would become sqrt(x) +2 inner would become 3

  6. dumbcow
    • 4 years ago
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    oops i meant y=-2

  7. anonymous
    • 4 years ago
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    so if I revolved it around the x=-2 my integral would be pi ((sqrt(x)-2)^2-(1-2)^2)dx

  8. dumbcow
    • 4 years ago
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    sorry typo there...it would be y=-2, in other words a horizontal axis no add 2 to the radius, not subtract 2 \[\pi \int\limits_{?}^{?} (\sqrt{x}+2)^{2} - 3^{2}\]

  9. dumbcow
    • 4 years ago
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    |dw:1328253465405:dw|

  10. anonymous
    • 4 years ago
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    but I thought since x=-2 thats why you would really subtract 2 from the square root and the 1

  11. dumbcow
    • 4 years ago
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    no x does not equal -2 instead of revolving around x-axis, we are revolving around the line y=-2 the distance the function is away from the line y=-2 is now 2 more than the x_axis

  12. anonymous
    • 4 years ago
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    oh okay that makes more sense

  13. anonymous
    • 4 years ago
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    thank you for all of your help

  14. dumbcow
    • 4 years ago
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    your welcome

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