find volume of solid obtained by rotating region bounded by curves y=x and y=sq.rt x, about line x=2
To find the volume of the solid obtained by rotating the region bounded by the curves y = x and y = √x about the line x = 2, we can use the method of cylindrical shells.
First, let's find the points of intersection of the two curves by solving the equation y = x and y = √x simultaneously.
Setting y = x and y = √x equal, we have:
x = √x.
Squaring both sides, we get:
x^2 = x.
Rearranging the equation, we have:
x^2 - x = 0.
Factoring out an x, we have:
x(x - 1) = 0.
Setting each factor equal to zero, we find:
x = 0 and x = 1.
So the region is bounded by the curves y = x and y = √x from x = 0 to x = 1.
Now, let's consider a vertical strip of width ∆x located at an arbitrary x-value within this interval. The height of this strip will be the difference between the two functions: y = x - √x.
The radius of the cylindrical shell at this x-value is the distance from the line x = 2 to the x-value, which is 2 - x.
The differential volume of this thin cylindrical shell is given by:
∆V = 2π(radius) * (height) * ∆x.
Substituting the values for the radius and height, we have:
∆V = 2π(2 - x)(x - √x)∆x.
To find the total volume, we need to integrate this expression over the interval from x = 0 to x = 1:
V = ∫[0,1] 2π(2 - x)(x - √x) dx.
Now, we can proceed to evaluate the integral using standard techniques of integration:
V = 2π ∫[0,1] (2x - x^2 - 2√x + x√x) dx.
V = 2π [x^2 - (1/3)x^3 - (4/3)x^(3/2) + (2/5)x^(5/2)] evaluated from x = 0 to x = 1.
V = 2π [(1 - (1/3) - (4/3) + (2/5)) - (0 - 0 - 0 + 0)].
Simplifying further, we get:
V = 2π [2/15].
Therefore, the volume of the solid obtained by rotating the region bounded by the curves y = x and y = √x about the line x = 2 is (4π/15) cubic units.