Ok, this is all the information that is given:

y= integral 1->x(((t)^1/2 - 1)^1/2)dt
1</= x </= 16

Find the area of the surface obtained by rotating the curve about the y-axis

Please Help

According to my understanding
this is an integration question, with respect to t.

When we want to find volumes or surface areas and such through this method,
the process is more or less the same.
Only that a particular scope is defined and you have to compute the answer by inserting the parameters into the devised formula.

Geez it has been a while since delving into differentiation,
but will go ahead and scratch mathemetical molecules up my brain.



Here I would like to remind all that Integration is the reverse process of differentiation.
Illustrated as follows:-

d [f(x)]/dx = f '(x) <==> ∫f '(x)dx = f (x)

Here goes nothing:

Q. y= ∫[1 - (((t)^1/2 - 1)^1/2)dt]
= ∫1 - ∫ (((t)^1/2 - 1)^1/2)dt

= t - ∫under root of t raised to (1/2 - 1) .dt

For simplification

(((t)^1/2 - 1)^1/2)dt = under root of t^(-1/2) .dt
= t^(-1/2)*(1/2).dt
= t^(-1/4) . dt

Hence we arrive at

y= [t - ∫t^(-1/4) . dt] --- observe parameters
y= [t - {1/(1 - 1/4) x t^ (1 - 1/4)}] --- observe parameters
y= [t - (4/3)t^(3/4)] --- observe parameters


As x = (((t)^1/2 - 1)^1/2)dt
compute a formula for t first.

Now insert the values for x, thus finding the two significant t limits.
All that is left to be done is to rotate the curve i.e. solve for the equation.

Do post for any further queries,
Pin~Jinx / anarchist