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richyw Group Title

Volume by DOUBLE integral. Still struggling with the same double integral problem! Test is in two hours! Can someone check my attempt?

  • one year ago
  • one year ago

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  1. richyw Group Title
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    set up the double integral to evaluate the volume of the solid bounded by \(y=x^2\), \(z=3y\) and \(z=2+y\)

    • one year ago
  2. richyw Group Title
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    I have set it up like this. does it look correct? \[V=\int^1_0\int^{2+y}_{3y}x^2\,dzdx\]

    • one year ago
  3. amistre64 Group Title
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    how did you determine the solid that is bounded?

    • one year ago
  4. richyw Group Title
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    I plugged \(y=x^2\) into \(z=3y\) and \(z=2+y\). And found \(x=\pm 1\).

    • one year ago
  5. richyw Group Title
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    I am very confused, all of the text book questions I breezed through and this one has me second guessing...

    • one year ago
  6. amistre64 Group Title
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    |dw:1352407804481:dw|

    • one year ago
  7. amistre64 Group Title
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    im thinking the solid is something like this; a y=x^2 cylindar that hits the z=3y plane and the z=y+2 plane; giving us 2 seperate volumes to calculate

    • one year ago
  8. amistre64 Group Title
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    |dw:1352408012819:dw|

    • one year ago
  9. amistre64 Group Title
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    its been a while so i gotta wonder if im interpreting the shape correctly

    • one year ago
  10. amistre64 Group Title
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    |dw:1352408252457:dw| or does it look more like a wedge of an orange

    • one year ago
  11. amistre64 Group Title
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    the wedge looks more "bounded" by the 3 surfaces to me

    • one year ago
  12. amistre64 Group Title
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    so, if we use half of this, it being symmetric in shape; x = 0 to 1 y = x^2 to 1 z = 3y to 2+y

    • one year ago
  13. amistre64 Group Title
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    \[\huge \int_{x=0}^{x=1}~\int_{y=x^2}^{y=1}~\int_{z=3y}^{z=2+y}~~dz~dy~dx\]

    • one year ago
  14. richyw Group Title
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    yeah it has to be a double integral though. that's my huge problem. I'm missing something...

    • one year ago
  15. amistre64 Group Title
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    solve the inner integral and it turns into a double :/

    • one year ago
  16. richyw Group Title
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    I have posted it here before. furthermore I haven't gotten a DI to equal the result of that TI.

    • one year ago
  17. richyw Group Title
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    yup that's what I might do on the exam haha.

    • one year ago
  18. amistre64 Group Title
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    if you wanna move about the limits of integrations to form different forms of the same TI into to create different forms of the DI ... that would prolly help work out some issues i think the problem is that is is not bounded by a usual plane

    • one year ago
  19. amistre64 Group Title
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    do you have an answer key to check with?

    • one year ago
  20. richyw Group Title
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    no

    • one year ago
  21. richyw Group Title
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    the cylinder has vertical sides, so I think this is correct.

    • one year ago
  22. richyw Group Title
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    anyways thanks a lot!

    • one year ago
  23. amistre64 Group Title
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    \[\int_{0}^{1}~\int_{x^2}^{1}~\int_{3y}^{2+y}~~dz~dy~dx\] \[\int_{0}^{1}~\int_{x^2}^{1}(2+y-3y)~dy~dx\] \[\int_{0}^{1}~\int_{x^2}^{1}2-2y~dy~dx\] \[2\int_{0}^{1}~\int_{x^2}^{1}1-y~dy~dx\] \[2\int_{0}^{1}~(1-\frac12)-(x^2-\frac12(x^2)^2)~dx\] \[2\int_{0}^{1}~\frac12-x^2+\frac12x^4~dx\]

    • one year ago
  24. amistre64 Group Title
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    and then double that result since thats only half of it :) otherwise; use each z as a cap over the y=x^2 and subtract the lower cap volume from the upper cap volume

    • one year ago
  25. amistre64 Group Title
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    \[2\left(2\int_{0}^{1}~\frac12-x^2+\frac12x^4~dx\right)\] \[2\left(\int_{0}^{1}~1-2x^2+x^4~dx\right)\] \[\int_{0}^{1}~2-4x^2+2x^4~dx\] \[2-\frac43+\frac25=\frac{16}{15}\] maybe :)

    • one year ago
  26. richyw Group Title
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    thanks a lot. sorry if I wasn't too active. trying to cram! I'll take a good look at it tonight though and figure it out for sure.

    • one year ago
  27. amistre64 Group Title
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    good luck :)

    • one year ago
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