## 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

1. anonymous

to what are you referring? area under curve ?

2. anonymous

yes

3. anonymous

so the function f(x) = 2x^2 is bounded by the x_axis

4. anonymous

that doesn't make sense ...the area would be infinite

5. 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...

6. anonymous

now its spinning, so im guessing volume? can you post the original problem, then maybe i can help you set it up

7. 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

8. anonymous

|dw:1328607842409:dw| what axis is it being revolved around?

9. anonymous

oh and i neglected to say that i am first revolving it around the y axis

10. anonymous

ok does it say you have to use a particular method...like disc or washer or shell

11. 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 ??

12. 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

13. anonymous

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

yes i am. ok so shell method it is...now back to the question of bounds...what are they?

15. anonymous

revolving around vertical axis so integrate respect to x x goes from 0 to 2 right?

16. anonymous

radius = x_value height = y_value $V = 2\pi \int\limits_{0}^{2}x*(2x^{2}) dx$

17. anonymous

ohhhh ok i see i gotcha. omygoodness thank you so so much!!

18. anonymous

yw