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monroe17

  • 2 years ago

Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y=3sqrt(x) and y=3x^2 about the y-axis

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  1. NoelGreco
    • 2 years ago
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    What are the points of intersection?

  2. yakeyglee
    • 2 years ago
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    For two functions \(f\) and \(g\) where \(f \ge g\), the volume between \(x=a\) and \(x=b\) by the cylindrical shells method is given by the following formula. You find those bounds by finding the points of intersection of \(f\) and \(g\), presumably. \[V=2\pi \int_a^b x (f-g) dx\]

  3. monroe17
    • 2 years ago
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    0 and 1 are the points of intersection.. but I can't seem to figure out the function to evaluate the integral at ;/

  4. NoelGreco
    • 2 years ago
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    |dw:1361312140525:dw| Each shell is going to have the shape of a rectangle when it is unfolded from the circular shell shape. You are really looking for the volume of that rectangle. V = length X height X thickness The thickness is going to be dx or dy, the height is the difference between the two functions, and the length is the circumference of each shell.

  5. yakeyglee
    • 2 years ago
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    Using the formula, we have \(f=3\sqrt x\) and \(g=3x^2\). So\[V=2\pi \int_a^b x (3 \sqrt x + 3x^2) dx.\]You can simplify and integrate that, yes?

  6. yakeyglee
    • 2 years ago
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    I should have put \(\displaystyle \int_0^1\), but you get the idea.

  7. monroe17
    • 2 years ago
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    how did you know what f and g were..? do you just assume?

  8. yakeyglee
    • 2 years ago
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    They were the equations you were given (\(f\) is the upper bound and \(g\) is the lower bound). Does that make sense?

  9. monroe17
    • 2 years ago
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    what defines an upper and lower bound?

  10. NoelGreco
    • 2 years ago
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    I think you want the DIFFERENCE between the two functions in the integrand.

  11. yakeyglee
    • 2 years ago
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    @NoelGreco, I do. Look at the integral again. @monroe17, the upper bound is the function that defines the top part of the bounded area, and the lower bound defines the lower part of the bounded area (look at their positions in the picture you drew).

  12. NoelGreco
    • 2 years ago
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    I looked at both. You have the sum, not the difference.

  13. NoelGreco
    • 2 years ago
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    Just make sure you get \[\frac{ 9\Pi }{ 10 }\]

  14. yakeyglee
    • 2 years ago
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    That is what I meant. \[V=2\pi \int_\color{red}{0}^\color{red}{1} x (3 \sqrt x \color{red}{-} 3x^2) dx.\]

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