EQ1: `y=x^2` and EQ2: `y=2x`
Determine the intersection points of these two equations. Use substitution method of system of equations.
So, substitute EQ1 to EQ2.
Set the equation in quadratic form `ax^2+bx+c=0` .
Set each factor to zero and solve for x.
`x=0` and `x-2=0`
Substitute values of x to EQ2 and solve for y.
`x=0` , `y=2x=2*0=0`
`x=2` , `y=2x=2*2=4`
Hence, the two equations intersect at (0,0) and (2,4).
So the bounded region of `y=x^2` and `y=2x` is:
(Note: Red-Graph of `y=x^2` and Blue-Graph of `y=2x` )
Apply the disk method of volume of solids of revolution. The formula is:
`V=pi int_a^b (r_o^2-r_i^2)dx`
The limits of the integral are the x-coordinates of the intersection points.
So, a=0 and b=2.
Base on the graph, the inner radius is the distance between the axis of rotation and red curve. Note that the axis of rotation is the x-axis which is y=0.
`r_i = y_(red)-y_(ax is)=x^2-0 = x^2`
Take the square of `r_i` .
`r_i^2 = (x^2)^2=x^4`
And the outer radius is the distance between the axis of rotation and the blue line. So,
`r_o=y_(blue)-y_(ax is) = 2x - 0 = 2x`
Take the square of `r_o` .
Substitute expression of `r_i^2` and `r_o^2` to the formula of volume.
`V=pi int_0^2 (4x^2-x^4)dx = pi((4x^3)/3 - x^5/5)` `|_0^2`
`V= pi[(4*2^3)/3 - 2^5/5] = pi(32/3 - 32/5) = 64/15pi =13.4`
Hence, volume of the solid obtained by rotating the bounded region of the given equations about the x-axis is 13.4 cubic units.
The volume of the solid formed by rotating the region enclosed by `y=x^2` and `y = 2x` about the x-axis has to be determined.
From the graph given below the area that is being rotated about the x-axis is given.
The volume can be divided into rings with infinitesimally small width. Each ring has an internal radius of x^2 and an external radius of 2x. The area of each ring is `pi*((2x)^2 - (x^2)^2)` and the volume is `pi*(4*x^2 - x^4) dx`
Adding the volume of all the rings gives the volume of the solid. This is given by `int_(0)^2 pi*(4*x^2 - x^4) dx`
=> `pi*int_(0)^2 4*x^2 - x^4 dx`
=> `pi*(4*x^3/3 - x^5/5)_(0)^2`
=> `pi*((4/3)*(2^3 - 0) - (1/5)*(2^5 - 0))`
=> `pi*(32/3 - 32/5)`
The volume of the solid formed by rotating the region enclosed by `y=x^2` and `y = 2x` about the x-axis is `(64*pi)/15` cube units.