# Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. `y = sqrt(x), y = -1/2x + 4, x = 0, x = 8`

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The volume of the solid generated by revolving the region bounded by the given graphs about the x-axis can be found using the "disks" method. According to this method, the volume is found by the sum (integral) of the disks with the height dx and radius y(x):

`int_0 ^ 8 pi (y(x))^2 dx`

As can be seen from the graph on the attached image, the given region needs to be broken into two parts: one under the graph of the square root function, and another under the graph of the linear function. These two functions intersect at the point (4, 2), which can also be seen on the graph.

So the integral will break up into two:

`int_0 ^4 pi(sqrt(x))^2 dx + int_4 ^8 pi(-1/2x + 4)^2 dx`

The first integral yields

`int_0 ^ 4 pi xdx = pi x^2/2 |_0 ^4 = 8pi`

The second integral yields

`int_4 ^8 pi(1/4x^2 - 4x + 16) dx = pi(1/12 x^3 - 2x^2 + 16x) |_4 ^8`

`pi (512/12 - 128 + 128) - pi(64/12 - 32+64) = pi(448/12 - 32) = (16pi)/3`

**The sum of the two results is `(40pi)/3` **

**which is the volume of the given solid.**