Calculus Evaluate volumt of the curve y=sqrt(x-1), y=0,x=5 about the x-axis.?

First sketch a picture. The parabola has vertex at (1,0) on the x axis.

So you are rotating that area enclosed by the parabola, the x-axis, and the vertical line x= 5.

Rotation about the x-axis gives the volume as : INT [ pi* R^2] dx for x in [1,5]

If you rotate one point on the parabola, you should see a circle with radius = the height of the parabola.

So R= sqrt(x-1) and R^2= (x-1)

INT [ pi* (x-1)] dx on [1,5]

= pi [( x^2)/2 -x ] | (1,5)

= pi[ (25/2 -5 ) -( 1/2 -1)]

= pi[ 12 -5 + 1]

= 8pi

I hope this helps!

V = integral of pi(x-1) dx , from x=1 to x=5 , and it's easy to do

y^2=(x-1)

if y=o, 0=(x-1)

1=x

therefore intercept is (1,0)

if x=5

y= -2 or y=+2

so that the function intercepts at (5,-2) and (5,2)