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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 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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    I have no idea how to solve this one, so any and all help is appreciated.

  3. phi
    • 3 years ago
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    |dw:1362168263188:dw|

  4. Bladerunner1122
    • 3 years ago
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    How is this supposed to work math wise? I'm having trouble visualizing how that becomes Calc.

  5. Bladerunner1122
    • 3 years ago
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    finally it's back up for me now.

  6. Bladerunner1122
    • 3 years ago
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    ... so is that last part there the "actual" answer? I'm trying to wrap my head around your post.

  7. phi
    • 3 years ago
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    does this make sense each plane is a sheet of volume x*x/2 * dy

  8. Bladerunner1122
    • 3 years ago
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    Yeah, I think so. Thanks.

  9. phi
    • 3 years ago
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    each plane is a sheet of volume x*x/2 * dy now put everything in terms of all x or all y if we stick with x: y=3-√x and \[dy = -\frac{1}{2} x^{-\frac{1}{2}} \ dx\] the limits of integration will be x=4 to 0 \[ -\frac{1}{4} \int_{4}^{0} x^{\frac{3}{2}}\ dx\] or flipping the limits, and negating \[ \frac{1}{4} \int_{0}^{4} x^{\frac{3}{2}}\ dx\] if we go with y: x= (3-y)^2 \[ \frac{1}{2} \int_{1}^{3} (3-y)^4 \ dy\]

  10. phi
    • 3 years ago
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    fixed a typo on finding how dy equates to dx. the first way changes dy to a function of x and dx and then integrates over x the second way changes x to a function of y and integrates the volume x^2/2 (in terms of y) over y

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