Surface area of a curve about the y axis

the equation is https://fastres01.qut.edu.au/webwork...0c96109db1.png. I found that x=sqrt(1-y) and that x'=-1/(2*sqrt(y-1)).

I've been using integral(2*pi*x*sqrt(1+x'^2)) but i keep just getting mixed up with my calculations.

help would be appreciated.

Re: Surface area of a curve about the y axis

If you rotate this function about the y-axis, each cross-section parallel to the x-axis will be a circle. The circumference of each circle is $\displaystyle \displaystyle \begin{align*} 2\pi r = 2\pi x = 2\pi \, \sqrt{ 1 - y } \end{align*}$, and if you add up all these circumferences over $\displaystyle \displaystyle \begin{align*} 0 \leq y \leq 1 \end{align*}$ then you will get the total surface area. So

$\displaystyle \displaystyle \begin{align*} SA &= \int_0^1{2\pi \, \sqrt{ 1 - y} \, dy} \end{align*}$

Re: Surface area of a curve about the y axis

so i used that equation and my end result was (4pi)/3, but its apparently incorrect. did i do something wrong?

Re: Surface area of a curve about the y axis

Can you show us your working please?

Re: Surface area of a curve about the y axis

I put 2pi out the front and integrated sqrt(1-y) and got -2/3 (1-y)^3/2. when i subbed in 1 I got 0 and when i subbed in 1 i got -2/3, and 0 - -2/3= 2/3. and then multiplied by 2pi

Re: Surface area of a curve about the y axis

I agree with your answer and solution, so I'm not sure what's gone wrong.

Re: Surface area of a curve about the y axis

when you wrote y between 0 and 1, did you actually mean x perhaps cause that is what the question asks. its the only thing i can think of for it not too work