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anonymous

  • 5 years ago

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating rectangle and label its height and width. 6x+y^2=13, x=2y find area of the region.

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  1. amistre64
    • 5 years ago
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    x = -(1/6)y^2 + 13/6 ; x = 2y

  2. amistre64
    • 5 years ago
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    id integrate with respect to y in the end; maybe :)

  3. anonymous
    • 5 years ago
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    when you integrate with respect to y, do you have the it in terms of x or y ? im a little confused D:

  4. amistre64
    • 5 years ago
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    terms of y

  5. amistre64
    • 5 years ago
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    we need to equate the 2 equations i wrote and find the 'boundarys'

  6. amistre64
    • 5 years ago
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    (-1/6)y^2 -2y +13/6 = 0 will give us the interval between the ys

  7. anonymous
    • 5 years ago
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  8. amistre64
    • 5 years ago
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    multiply by -6 to normalize this thing right? y^2 +12y -13 = 0

  9. anonymous
    • 5 years ago
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    i think it should be integrated wrt x.check the curve please...which i have given...

  10. amistre64
    • 5 years ago
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    y = -13, and y = 1

  11. amistre64
    • 5 years ago
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    coulda tried to save it as a jpeg ya know :) loads faster

  12. amistre64
    • 5 years ago
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    wrt x is fine; but trickier; wrt y is easier and more straight forward

  13. anonymous
    • 5 years ago
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    for the region left where x=2y intersects the parabola. the y is x/2, and for the right the y=sqrt(13-6x).

  14. amistre64
    • 5 years ago
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    when we integrate this area, we need to subtract one function from the other to get the total area; do we do absolute values? or let the areas lie where they are either pos or neg?

  15. anonymous
    • 5 years ago
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    so the integral would look something like this.. \[\int\limits_{-13}^{1} (-1/6)y^2 +13/6-2y)dy\]

  16. amistre64
    • 5 years ago
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    i believe so :)

  17. anonymous
    • 5 years ago
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    than you both so much :)

  18. anonymous
    • 5 years ago
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    thank*

  19. amistre64
    • 5 years ago
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    if it turns out negative; just take the absolute value; it simply means you subtracted in the wrong order for example: 5-3 = 2 3-5 = -2; but |-2| = 2

  20. anonymous
    • 5 years ago
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    gotcha. thanks a bunch :)

  21. amistre64
    • 5 years ago
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    youre welcome :)

  22. anonymous
    • 5 years ago
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    it all worked out. :D you rock :)

  23. amistre64
    • 5 years ago
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    :) thnx

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