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anonymous

  • 5 years ago

1) A region is bounded by the line y = x and the parabola y = x2 - 6x + 10. What is the volume of the solid generated by revolving the region about the x-axis?

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  1. amistre64
    • 5 years ago
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    first we should determine our bounds right?

  2. anonymous
    • 5 years ago
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    first draw a rough of a parabola (x^2-6x+10) and dissect it with a diagonal line (y=x). then set x=x^2-6x+10. Solve this it would tell you where parabola and the line intersect.

  3. amistre64
    • 5 years ago
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    yep.... or x^2 = -7x +10 (x-5)(x-2)......[2,5]

  4. amistre64
    • 5 years ago
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    well it looked right in me head lol

  5. amistre64
    • 5 years ago
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    0 = x^2 -7x +10...better

  6. amistre64
    • 5 years ago
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    pi [S] [x^2 -7x +10]^2 dx [2,5] right?

  7. amistre64
    • 5 years ago
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    \[F(x) = \pi \int\limits_{2}^{5}[x^2 -7x +10]dx\]

  8. amistre64
    • 5 years ago
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    F(5) - F(2) = answer of the volume of the solid formed....

  9. amistre64
    • 5 years ago
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    forgot the ^2 in that equation..... there should be an edit button lol

  10. amistre64
    • 5 years ago
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    or should we integrate each 'y=' seperately on the interval and then just subtract the higher from the lower?

  11. anonymous
    • 5 years ago
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    I think you integrate the given parabola? Must check on this.

  12. anonymous
    • 5 years ago
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    Yes you integrate the given function. The other created function is helpful only to create the boundaries.

  13. amistre64
    • 5 years ago
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    we tend to get an area shaped like this right? if we find the volume of the solid formed by the y=x in the interval and then subtract the volume of the solid formed by y=x^2 -6x +10...we will end up with the volume of the solid of this area..... and from the last time i messed this up, I recall we had to use pi(y^2) as the intgrands :)

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  14. amistre64
    • 5 years ago
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    piR^2 - pir^2 = pi(R^2 - r^2) so we should be able to use the (y=x)^2 - (y=x^2....)^2 as the integrands

  15. amistre64
    • 5 years ago
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    but I would feel safer just getting one solid and subtracting the other from it ;)

  16. anonymous
    • 5 years ago
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    Sure, If you are a bright student you can do several methods to check, but a non-math major: to find the volume of a solid integrate from a to b, pi*(f(x))^2. (f(x) is given function. Find a, b boundaries as we did above.

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