Question

volume generated by the two curves
\[x^m,x^n\]
where \[m>n\]

Answer 1

\[x^m, x^n\]

Answer 2

Graphing ?

Answer 3

integration

Answer 4

@mathslover definite integration hai...

Answer 5

VOLUME

Answer 6

Are you sure its volume ?

I have heard of area under the curve using definite integration.... but volume ??

Are you referring to something 3 dimensions

Answer 7

@Jonask

Answer 8

its volume generated by revolving the surfaces of x^m and x\^n i am not sure about my solution
\[V=\pi(+\frac{1}{2n+1}-\frac{1}{2m+1})\]

Answer 9

this is in the first quadrant

Answer 10

\[m,n \in R\]

Answer 11

what's the axis of revolution?

Answer 12

y axis

Answer 13

does it matter whether m or n is even or odd

Answer 14

since\[ m>n\] implies \[x^n>x^m\] ,for(0,1)

Answer 15

Sorry Jonask, can't help, its over my head :(

Answer 16

\[V=\pi \int\limits _a^by^2dx\]\[\huge V=\pi \int\limits _0^1(x^{2m}-x^{2n}) dx\]

Answer 17

thanks @mathslover

Answer 18

you got it right alr :), use the washer method

Answer 19

oh wait

Answer 20

hence
oh i shud start with x^2n-x^2m
xince m>n

Answer 21

the axis of revolution is y-axis, so it should be dy if you use the washer method or dx if you use the shell method, check against your integrand

Answer 22

\[\huge V=\pi \int\limits _0^1(x^{2n}-x^{2m})\]

Answer 23

if y=x^m , can you write x in terms of y? Similarly for y=x^n?

Answer 24

well i dont like radicals what about by shells

Answer 25

so you are saying that the set up mmade above is for revovling around x-axis

Answer 26

yes

Answer 27

let me think about it, plz wait

Answer 28

2*Pi*x*(x^n-x^m), integrate this wrt x from x=0 to x=1

Answer 29

yes it works but in my exam i made a mistake i used the latter method
so its wrong
but yours is correct
makes sense