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Andresfon12

  • one year ago

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).

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

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

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

  4. TuringTest
    • one year ago
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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|

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

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

  7. TuringTest
    • one year ago
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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 \[y-y_0=-\frac13(x-x_0)\]

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

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

  10. Andresfon12
    • one year ago
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    ok

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

  12. TuringTest
    • one year ago
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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.

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

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

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

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

  17. Andresfon12
    • one year ago
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    same using slope

  18. TuringTest
    • one year ago
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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.

  19. Andresfon12
    • one year ago
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    really is calc 3

  20. TuringTest
    • one year ago
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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?

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

  22. TuringTest
    • one year ago
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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.

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

  24. TuringTest
    • one year ago
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    yes :)

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

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

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

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

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