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richyw

  • 2 years ago

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

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

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

  3. amistre64
    • 2 years ago
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    how did you determine the solid that is bounded?

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

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

  6. amistre64
    • 2 years ago
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    |dw:1352407804481:dw|

  7. amistre64
    • 2 years ago
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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

  8. amistre64
    • 2 years ago
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    |dw:1352408012819:dw|

  9. amistre64
    • 2 years ago
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    its been a while so i gotta wonder if im interpreting the shape correctly

  10. amistre64
    • 2 years ago
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    |dw:1352408252457:dw| or does it look more like a wedge of an orange

  11. amistre64
    • 2 years ago
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    the wedge looks more "bounded" by the 3 surfaces to me

  12. amistre64
    • 2 years ago
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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

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

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

  15. amistre64
    • 2 years ago
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    solve the inner integral and it turns into a double :/

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

  17. richyw
    • 2 years ago
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    yup that's what I might do on the exam haha.

  18. amistre64
    • 2 years ago
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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

  19. amistre64
    • 2 years ago
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    do you have an answer key to check with?

  20. richyw
    • 2 years ago
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    no

  21. richyw
    • 2 years ago
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    the cylinder has vertical sides, so I think this is correct.

  22. richyw
    • 2 years ago
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    anyways thanks a lot!

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

  24. amistre64
    • 2 years ago
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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

  25. amistre64
    • 2 years ago
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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 :)

  26. richyw
    • 2 years ago
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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.

  27. amistre64
    • 2 years ago
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    good luck :)

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