anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
are we using shells or discs?
amistre64
  • amistre64
shells add the area of rectangles: 2pi r discs add the area of circles: pi r^2
amistre64
  • amistre64
it looks like disc method to me

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amistre64
  • amistre64
|dw:1442770213833:dw|
anonymous
  • anonymous
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.
amistre64
  • amistre64
we have a circle of radius 8, that have a center removed from them of a radius of 8-sqrt(x)
amistre64
  • amistre64
pi (8^2 - (8-sqrt(x))^2) is the area of a disc for a given value of x
amistre64
  • amistre64
it looks like you have an error in your application. can you write it out in more detail? what were your thoughts?
amistre64
  • amistre64
washer method is another name for removing a disc out of a disc ...
anonymous
  • anonymous
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?
amistre64
  • amistre64
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
amistre64
  • amistre64
pi(R^2 - r^2) gives us the portion that we are concerned with.
anonymous
  • anonymous
Thank you very much for you answers~!
amistre64
  • amistre64
youre welcome

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