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Andresfon12 Group Title

find the volume of the given solid: under the surface z=2x+y^2 and above the triangle vertices (1,1),(4,1) and (1,2).

  • one year ago
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  1. Andresfon12 Group Title
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    |dw:1360708771803:dw|

    • one year ago
  2. TuringTest Group Title
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    what is the equation of this part of the area?|dw:1360709404014:dw|

    • one year ago
  3. Andresfon12 Group Title
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    y=2x?

    • one year ago
  4. TuringTest Group Title
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    no,you can see that the slope from x=1 to x=4 is down for that part of the graph, so the coefficient of x will be negative you should use point-slope form way back from algebra I to do this|dw:1360709566120:dw|

    • one year ago
  5. Andresfon12 Group Title
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    @TuringTest yes i can see that

    • one year ago
  6. Andresfon12 Group Title
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    so -1/3?

    • one year ago
  7. TuringTest Group Title
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    yes, so now use point-slope form on either point \((x_0,y_0)\) to get the equation of the line \[y-y_0=-\frac13(x-x_0)\]

    • one year ago
  8. Andresfon12 Group Title
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    y-1 = -1/3 (x-1)?

    • one year ago
  9. TuringTest Group Title
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    no, because (1,1) is not one of the points on the diagonal line

    • one year ago
  10. Andresfon12 Group Title
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    ok

    • one year ago
  11. Andresfon12 Group Title
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    the only one that i can use the (1,2) and (1,4)

    • one year ago
  12. TuringTest Group Title
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    correct, we are using point-slope form on the diagonal line, so we can only get the equation from points that lie \(on\) the line.

    • one year ago
  13. Andresfon12 Group Title
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    y-2= -1/3 (x-1)

    • one year ago
  14. TuringTest Group Title
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    yes, now solve for y...

    • one year ago
  15. Andresfon12 Group Title
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    y= -1/3x+1/3 (+2) y=-1/3 x+7/3

    • one year ago
  16. TuringTest Group Title
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    yes, and now you need the line that bounds the region below:|dw:1360710827125:dw|

    • one year ago
  17. Andresfon12 Group Title
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    same using slope

    • one year ago
  18. TuringTest Group Title
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    er, yeah you could, but you should really be able to eyeball this one. Note here that all we have been doing so far is basic algebra to find the equations of the bounds of the region. Just because you are in calc2 does not mean you can forget the basics!! On the contrary, here is where you actually need them most.

    • one year ago
  19. Andresfon12 Group Title
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    really is calc 3

    • one year ago
  20. TuringTest Group Title
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    Well, what some people call cal3 others call calc2. MIT for instance only has 2 basic calc classes: single and multivariable. But enough semantics, what's the equation of the bottom line?

    • one year ago
  21. Andresfon12 Group Title
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    @TuringTest is 1?

    • one year ago
  22. TuringTest Group Title
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    y=1, yes so y is bound by the two equations we found, so they will be the bounds for the inner integral, which will be with respect to x. what are the bounds on x, which we will use for the outer integral.

    • one year ago
  23. Andresfon12 Group Title
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    x=1, 4 and y= 1 , 1/3x+7/3?

    • one year ago
  24. TuringTest Group Title
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    yes :)

    • one year ago
  25. Andresfon12 Group Title
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    \[\int\limits_{1}^{4} \int\limits_{1}^{1/3x+7/3} xy \] dy dx

    • one year ago
  26. TuringTest Group Title
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    the integrand is the function for z, so use that, not xy gotta go, happy integrating!

    • one year ago
  27. Andresfon12 Group Title
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    @TuringTest thank you so much

    • one year ago
  28. Andresfon12 Group Title
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    sorry is not z=2x+y^2 i just mistype the book is really saying xy

    • one year ago
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