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anonymous

  • 4 years ago

ok i completely forgot...im setting bounds and so i set y=2x^2 to y=0... so 2x^2=0 what are my bounds? i cant think its too late lol. is it x=0 and x=?? idk. helppp

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  1. dumbcow
    • 4 years ago
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    to what are you referring? area under curve ?

  2. anonymous
    • 4 years ago
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    yes

  3. dumbcow
    • 4 years ago
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    so the function f(x) = 2x^2 is bounded by the x_axis

  4. dumbcow
    • 4 years ago
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    that doesn't make sense ...the area would be infinite

  5. anonymous
    • 4 years ago
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    uhh now im just confused lol. im trying to find the bounds for a spinning object thing. so i suppose that would make it infinite? i dont know im so confused...

  6. dumbcow
    • 4 years ago
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    now its spinning, so im guessing volume? can you post the original problem, then maybe i can help you set it up

  7. anonymous
    • 4 years ago
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    haha sorry. this section is kicking my butt. but yes that would help me a ton. the original problem: find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the given lines... y=2x^2, y=0, x=2

  8. dumbcow
    • 4 years ago
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    |dw:1328607842409:dw| what axis is it being revolved around?

  9. anonymous
    • 4 years ago
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    oh and i neglected to say that i am first revolving it around the y axis

  10. dumbcow
    • 4 years ago
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    ok does it say you have to use a particular method...like disc or washer or shell

  11. anonymous
    • 4 years ago
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    ok i apologize i was trying to respond but my connection was lost for the longest time and no it doesnt. if it goes across the y-axis dont you could use the disk method right? v=pi\[\int\limits_{a}^{b}\][R(y)]^2 dy ??

  12. anonymous
    • 4 years ago
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    *if it goes across the y-axis you could use the disk method right? and all of the equation was supposed to be connected lol

  13. dumbcow
    • 4 years ago
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    no problem yes you can use either method here if you use the disc method though there will be 2 radius shell method would have easier setup are you familiar with it? -->\[2\pi \int\limits_{?}^{?}radius*height\]

  14. anonymous
    • 4 years ago
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    yes i am. ok so shell method it is...now back to the question of bounds...what are they?

  15. dumbcow
    • 4 years ago
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    revolving around vertical axis so integrate respect to x x goes from 0 to 2 right?

  16. dumbcow
    • 4 years ago
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    radius = x_value height = y_value \[V = 2\pi \int\limits_{0}^{2}x*(2x^{2}) dx\]

  17. anonymous
    • 4 years ago
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    ohhhh ok i see i gotcha. omygoodness thank you so so much!!

  18. dumbcow
    • 4 years ago
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    yw

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