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anonymous

  • one year ago

The region between the graph f(x)=x^(1/2) and the x-axis for x∈[0,16] is revolved about the line y=8. Which integral will determine the volume of the solid that is generated? I got π(x-64)dx, x∈[0,16] using the yA=1/2[(f(x))^2 - (g(x))^2]dx x∈[a,b] equation, but the answer is ∫π[64−(8−x^(1/2))^2]dx x∈[0,16]. No idea where I went wrong, but maybe this isn't the right equation to use?

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  1. amistre64
    • one year ago
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    are we using shells or discs?

  2. amistre64
    • one year ago
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    shells add the area of rectangles: 2pi r discs add the area of circles: pi r^2

  3. amistre64
    • one year ago
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    it looks like disc method to me

  4. amistre64
    • one year ago
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    |dw:1442770213833:dw|

  5. anonymous
    • one year ago
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    This is a problem semi-related to centroids. I assumed it was Disc Method, but either I did something wrong or I'm overthinking everything.

  6. amistre64
    • one year ago
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    we have a circle of radius 8, that have a center removed from them of a radius of 8-sqrt(x)

  7. amistre64
    • one year ago
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    pi (8^2 - (8-sqrt(x))^2) is the area of a disc for a given value of x

  8. amistre64
    • one year ago
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    it looks like you have an error in your application. can you write it out in more detail? what were your thoughts?

  9. amistre64
    • one year ago
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    washer method is another name for removing a disc out of a disc ...

  10. anonymous
    • one year ago
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    I was looking at this in terms of centroids. Most of the equations are the same as the ones for volume. I actually have the Washer Method equation down (I just realized PFFT). The Disc Method equation is just V=πr^2. The way your wrote it as π[8^2 - (8-x^(1/2))^2] is because of the rotation around the y=8, right?

  11. amistre64
    • one year ago
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    yes we have a cylindar, or radius 8 (center at y=8, and edge at y=0) we are gutting out the middle of it, so at an given x value, we are 8, minus sqrt(x) |dw:1442770823342:dw| since sqrt(x) + r = 8

  12. amistre64
    • one year ago
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    pi(R^2 - r^2) gives us the portion that we are concerned with.

  13. anonymous
    • one year ago
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    Thank you very much for you answers~!

  14. amistre64
    • one year ago
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    youre welcome

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