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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
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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dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.1to what are you referring? area under curve ?

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.1so the function f(x) = 2x^2 is bounded by the x_axis

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.1that doesn't make sense ...the area would be infinite

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0uhh 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
 4 years ago
Best ResponseYou've already chosen the best response.1now its spinning, so im guessing volume? can you post the original problem, then maybe i can help you set it up

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haha 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
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1328607842409:dw what axis is it being revolved around?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh and i neglected to say that i am first revolving it around the y axis

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.1ok does it say you have to use a particular method...like disc or washer or shell

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok 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 yaxis dont you could use the disk method right? v=pi\[\int\limits_{a}^{b}\][R(y)]^2 dy ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0*if it goes across the yaxis you could use the disk method right? and all of the equation was supposed to be connected lol

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.1no 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
 4 years ago
Best ResponseYou've already chosen the best response.0yes i am. ok so shell method it is...now back to the question of bounds...what are they?

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.1revolving around vertical axis so integrate respect to x x goes from 0 to 2 right?

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.1radius = x_value height = y_value \[V = 2\pi \int\limits_{0}^{2}x*(2x^{2}) dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohhhh ok i see i gotcha. omygoodness thank you so so much!!
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