Question

What am I doing wrong here?

Answer 1

how r u going about it?

Answer 2

wait a bit

Answer 3

k

Answer 4

i really don't understand from ur working what it is that u r trying to do.

Answer 5

generally when a region of graph is revolved around x-axis, its volume is given by \[\pi \int\limits_{a}^{b}y^2dx\]

Answer 6

problem here is that the region here is not adjacent to x-axis but to y-axis

Answer 7

so then it needs to be rewritten in terms of y?

Answer 8

i would think of getting around it, by finding inverses of all the curves and line involved

Answer 9

yea, writing in terms of y then replacing the x with y and y with x

Answer 10

@legomyego180 made sense?

Answer 11

yup, thanks

Answer 12

how do I rewrite y=1 in terms of x?

Answer 13

no problem. any time

Answer 14

there is no x term here, so skip the writing in terms of y step, and replace y with x

Answer 15

alrighty

Answer 16

@legomyego180 did u get the right answer by this method?

Answer 17

I took a little break. Ill try it out

Answer 18

nah i got \[\frac{ 3\pi }{ 5 }\]

Answer 19

can u post ur working?

Answer 20

@legomyego180

Answer 21

\[\pi \int\limits_{0}^{1}(y ^{\frac{ 1 }{ 3 }})^2\]

Answer 22

you want me to integrate it out as well? I thought maybe finding the integral was where my mistake was

Answer 23

I plugged it into wolfram and got the same answer

Answer 24

can u post a image of ur working like before, that way i have better chances of seeing what might be going wrong

Answer 25

If its being flipped across the x-axis I think it needs to be written with respect to x, I think we had it right the first time.

Answer 26

I was using the formula incorrectly, should have been:\[\int\limits_{0}^{1}\pi(1)^2-\pi(x^3)^2=\frac{ 6\pi }{ 7 }\]

Answer 27

Thanks for the help and working through it with me

Answer 28

my pleasure. U did fine work