Help me find Find S (Integration problem)

y = √x , y = 1/4x , x=25


So with that information do I integrate with r espect to the y axis?

I know they intercept when x = 0 and 16

so what do I do with the x = 25? i don't understand that part =( help me please! I don't want the answer, just help me understand. I have a calc exam tomorrow.

i suspect that you want to find the area S between y= sqrtx and y= (1/4) x ie, y=x/4 between x=0 to x=25 using a simple integral .-
The curves intercept themselves when y=y sqrtx= (x/4) , x= x^2/16

x(x-16)=0
x=0 and x=16

dS = height dx
From 0 From 16 They are asking you the area S beyond of the curves intercept until x=25

S1= INT (sqrtx -x/4 ) dx (0 S2 = INT ( (x/4) -sqrtx ) dx 16
S1= (2/3) x^(3/2) - x^2/8 = (2/3) 16^(3/2) - 16^2/8 = (2/3) 64 -32 = 128/3 -32 = 32/3

S2 = x^2/8 - (2/3)x^(3/2) = 25^2/8 - (2/3) 25^(3/2) -( -32/3 )
S2= 625/8 - 250/3 +32/3 = 625/8- 218/3 = (1875- 1744 )/24 =131/24

S= S1+S2 = 32/3 +131/24 = 387/24 =129/8

First of all, they don't "intercept"; they "intersect" at x = 0 and x = 16.

Is the x = 25 an axis of rotation? Are you trying to find the volume of revolution?

If so, you would integrate w.r.t. x, since both functions are functions of x.

dV = π[(√x - 25)^2 - (x/4 - 25)^2] dx

This is called the Washer Method.