Question

y = sqrt(x - 1), y = 0, x = 5      Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer. (about the x-axis)

Accepted Answer

The volume of the solid obtained by rotating the region bounded by the curves y=sqrt(x-1), y=0, x= 5 , about x axis, can be evaluated using the washer method, such that:
V = int_a^b pi*(f^2(x) - g^2(x))dx
Since the problem provides you the endpoint x = 5, you need to find the other endpoint of interval, hence, you need to solve for x the following equation, such that:
sqrt (x-1) = 0 => x - 1 = 0 => x = 1
V = int_1^5 pi*(sqrt(x-1) - 0)^2 dx
V = pi*int_1^5 (x - 1)dx
V = pi*int_1^5 (x)dx - pi*int_1^5 dx
V = (pi*x^2/2 - pi*x)|_1^5
V = (pi*5^2/2 - pi*5 - pi*1^2/2 + pi*1)
V = (25pi)/2 - pi/2 - 4pi
V = 12pi - 4pi
V = 8pi 
Hence, evaluating the volume of the solid obtained by rotating the region bounded by the curves y=sqrt(x-1), y=0, x= 5 , about x axis , using the washer method, yields V = 8pi.

Answer

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_1^5[ pi* (x-1)] dx `

`= pi [( x^2)/2 -x ] | _1^5`

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

`= pi[ 12 -5 + 1] `
` = 8pi `

Math

`y = sqrt(x - 1), y = 0, x = 5` Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk...

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