# Thread: how do i setup this integral for the area of the surface rotating about the y-axis?

1. ## how do i setup this integral for the area of the surface rotating about the y-axis?

y=x^6 with the intervals 0≤x≤1

how do i setup this integral for the area of the surface obtained by rotating the curve about the y-axis??

i dont get it at all.. i understood how to rotate it about the x-axis which i got integral from [0,1] of int(2πx^6sqrt[1+(6x^5)^2])

but i dont know how ot set it up for the y-axis?

2. Originally Posted by Reefer
y=x^6 with the intervals 0≤x≤1

how do i setup this integral for the area of the surface obtained by rotating the curve about the y-axis??

i dont get it at all.. i understood how to rotate it about the x-axis which i got integral from [0,1] of int(2πx^6sqrt[1+(6x^5)^2])

but i dont know how ot set it up for the y-axis?
$y = x^6$

$x = y^{\frac{1}{6}}$

$\displaystyle \frac{dx}{dy} = \frac{1}{6x^{\frac{5}{6}}}$

$\displaystyle A = 2\pi \int_0^1 y^{\frac{1}{6}} \sqrt{1 + \frac{1}{36y^{\frac{5}{3}}}} \, dy$

3. this is with respect to dx though.. as in the thickness is dx, not dy.

if the thickness was dy i would have never asked this question.. but since my professor says the thickness is dx im confused as to how to do the problem..

4. Originally Posted by Reefer
this is with respect to dx though.. as in the thickness is dx, not dy.

if the thickness was dy i would have never asked this question.. but since my professor says the thickness is dx im confused as to how to do the problem..
"thickness" is relative. see for yourself ...

PlanetMath: area of surface of revolution