anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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dumbcow
  • dumbcow
to what are you referring? area under curve ?
anonymous
  • anonymous
yes
dumbcow
  • dumbcow
so the function f(x) = 2x^2 is bounded by the x_axis

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dumbcow
  • dumbcow
that doesn't make sense ...the area would be infinite
anonymous
  • anonymous
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...
dumbcow
  • dumbcow
now its spinning, so im guessing volume? can you post the original problem, then maybe i can help you set it up
anonymous
  • anonymous
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
dumbcow
  • dumbcow
|dw:1328607842409:dw| what axis is it being revolved around?
anonymous
  • anonymous
oh and i neglected to say that i am first revolving it around the y axis
dumbcow
  • dumbcow
ok does it say you have to use a particular method...like disc or washer or shell
anonymous
  • anonymous
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 ??
anonymous
  • anonymous
*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
dumbcow
  • dumbcow
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? -->\[2\pi \int\limits_{?}^{?}radius*height\]
anonymous
  • anonymous
yes i am. ok so shell method it is...now back to the question of bounds...what are they?
dumbcow
  • dumbcow
revolving around vertical axis so integrate respect to x x goes from 0 to 2 right?
dumbcow
  • dumbcow
radius = x_value height = y_value \[V = 2\pi \int\limits_{0}^{2}x*(2x^{2}) dx\]
anonymous
  • anonymous
ohhhh ok i see i gotcha. omygoodness thank you so so much!!
dumbcow
  • dumbcow
yw

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