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anonymous

  • 4 years ago

Find the volume of the solid generated by revolving the described region about the given axis: The region bounded above by the line y=4 , below by the curve y=x^2 , rotated along the following lines: y=4

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  1. anonymous
    • 4 years ago
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    GT as in Georgia Tech?

  2. anonymous
    • 4 years ago
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    \[\Pi \int\limits\limits\limits_{-2}^{2}(x^2-4)^2 \]

  3. Rogue
    • 4 years ago
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    If you think about it for it a bit, its going to be the volume of the solid generated by revolving \[y = x ^{2} - 4\] about the x - axis. The equation crosses the x-axis positive and negative 2, so those are your limits of integration. This is a disk, which has circular partitions. If we add up the area of all the circular partitions, we can get the volume. \[A = \pi r ^{2}\] The integral of area is volume. Your radius is x^2 - 4. So now just do the integration. \[V = \pi \int\limits_{-2}^{2} (x ^{2} - 4)^{2} dx\]

  4. TuringTest
    • 4 years ago
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    nash and rogue are right just to clean this up a little we can note that the integrand is even, so\[\pi\int_{-2}^{2}(x^2-4)^2dx=2\pi\int_{0}^{2}(x^4-8x^2+16)dx\]which is a very straightforward integration.

  5. anonymous
    • 4 years ago
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    Thank you guys, my final answer was 512/15pi, which was right!

  6. anonymous
    • 4 years ago
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    So what about with rotation y=8, what would the limits be?

  7. TuringTest
    • 4 years ago
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    same area bounded by y=x^2 and y=4 about y=8 ?

  8. TuringTest
    • 4 years ago
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    ...or is it now the area bounded between y=x^2 and y=8 about y=8 ?

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