# Find the volume of the solid formed by rotating the region enclosed by y=x^2 and y=2x about the x-axis.

lemjay | Certified Educator

calendarEducator since 2012

starTop subjects are Math and Science

Let,

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.

`y=2x`

`x^2=2x`

Set the equation in quadratic form `ax^2+bx+c=0` .

`x^2-2x=0`

Then, factor.

`x(x-2)=0`

Set each factor to zero and solve for x.

`x=0`               and              `x-2=0`

`x=2`

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` .

`r_o^2= (2x)^2=4x^2`

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.

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## Related Questions

justaguide | Certified Educator

calendarEducator since 2010

starTop subjects are Math, Science, and Business

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)`

=> `(64*pi)/15`

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.

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