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anonymous

  • 4 years ago

*CALC2 HELP PLEASE* The region enclosed by the curve x=sqrt(y), and by the lines x= -y and y=5 rotated around the x-axis.

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  1. anonymous
    • 4 years ago
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    Thanks for the medal, but do you have any tips?

  2. anonymous
    • 4 years ago
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    well, what you've got yourself there is an integral with some pretty odd bounds, give me a second so I can draw a picture to see how to approach it.

  3. anonymous
    • 4 years ago
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    Thanks man.. The shape is pretty funky..

  4. anonymous
    • 4 years ago
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    is it 3d?

  5. anonymous
    • 4 years ago
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    |dw:1327642062679:dw|

  6. anonymous
    • 4 years ago
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    yeah it's rotating

  7. anonymous
    • 4 years ago
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    another, question, which area contained by those lines?

  8. anonymous
    • 4 years ago
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    It doesn't specify really, so I'm confused there as well..

  9. anonymous
    • 4 years ago
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    The whole area I shaded in

  10. anonymous
    • 4 years ago
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    Because graph is this

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  11. anonymous
    • 4 years ago
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    hmm.. that is really weird..

  12. anonymous
    • 4 years ago
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    aww, I thought dumbcow had it figured out.. haha

  13. anonymous
    • 4 years ago
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    so depending on what the area is its just a matter of doing the volume of revolution, but without knowing which specific area is... I want to assume that the area is the bit off to the left from y=-x up to y=5 and from y=-x over to y=x^2

  14. anonymous
    • 4 years ago
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    did they specify x=+sqrt(y)?

  15. dumbcow
    • 4 years ago
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    ok i think you can approach this using the washer method where you have an outer area - inner area each cross section is a circle and you integrate wrt x from -5 to sqrt5 the outer radius would be 5 since that is the max y-value. the inner radius would be the y-value of the appropriate function f(x) |dw:1327642598764:dw| \[\large V = \pi \int\limits_{-5}^{0}(25-x^{2}) dx + \pi \int\limits_{0}^{\sqrt{5}}(25-x^{4}) dx\]

  16. anonymous
    • 4 years ago
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    dumbcow: after integration I got a negative volume?

  17. dumbcow
    • 4 years ago
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    hmm no it should be positive make sure when you evaluate you subtract upper limit - lower limit

  18. anonymous
    • 4 years ago
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    ok let me try again

  19. anonymous
    • 4 years ago
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    the integration would be.. pi(25x-x^3/3) and pi(25x-x^5/5), right?

  20. dumbcow
    • 4 years ago
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    yes

  21. anonymous
    • 4 years ago
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    I got 402.296 on wolfram, and it says incorrect on my homework.. :/

  22. anonymous
    • 4 years ago
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    Nahhh I'm just kidding.. Worked perfect. Thanks guys!!

  23. dumbcow
    • 4 years ago
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    isn't wolfram great. ok well do they want in terms of pi? i'll look at it again i may have made a mistake somewhere

  24. dumbcow
    • 4 years ago
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    never mind :)

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