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Bladerunner1122

  • 3 years ago

Let R be the region in the first quadrant bounded by the graph y=3-√x the horizontal line y=1, and the y-axis as shown in the figure to the right.

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  1. Bladerunner1122
    • 3 years ago
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    http://goo.gl/AhgJX 1. Write but do not evaluate, an integral expression that gives the volume of the solid generated when R is rotated about the horizontal line y=-1. 2. Region R is the base of a solid. For each y, where 1≤y≤3, the cross section of the solid taken perpendicular to the y-axis is a rectangle whose height is half the length of its base. Write, but do not evaluate, an integral expression that gives the volume of the solid.

  2. Bladerunner1122
    • 3 years ago
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    ? Anybody out there? xD

  3. phi
    • 3 years ago
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    |dw:1362167243909:dw|

  4. phi
    • 3 years ago
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    you need an expression for the radius of the discs, as a function of x

  5. Bladerunner1122
    • 3 years ago
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    ??

  6. Bladerunner1122
    • 3 years ago
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    Isn't the volume \[V=\pi \int\limits_{0}^{4}((-1-(3-√x))^2 -(-1-1)^2) dx\]

  7. phi
    • 3 years ago
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    |dw:1362167608670:dw|

  8. phi
    • 3 years ago
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    yes, your expression looks good.

  9. Bladerunner1122
    • 3 years ago
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    Great! So what do I do for 2? I don't have any idea what to do for it.

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