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

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marizi eNotes educator| Certified Educator

The region bounded by `y=sqrt(x)` , `y =-x/2+4` , `x=0` ,and `x=8` revolved about the x-axis is shown on the attached image. We may apply Disk Method wherein we use a rectangular strip representation such that it is perpendicular to the axis of rotation.  In this case, we need to sets of rectangular strip since the upper bound of the rectangular strip before and after `x=4` differs.

The vertical orientation of the rectangular strip shows the thickness of strip =dx.

 That will be the basis to use the formula of the Disc method in a form of:

`V = int_a^b A(x) dx ` where `A(x) = pir^2` and `r =y_(above)-y_(below)` .

The `r` is radius of the disc which is the same as the length of the rectangular strip.

As shown on the attached file, the `r= sqrt(x)-0 = sqrt(x)` from the boundary values of `x=0 ` to `x=4` .

For the boundary values from `x=4` to `x=8` , we have `r=-x/2+4-0 =-x/2+4` .

Then the integral set-up will be:

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

`V = int_0^4 pixdx+ int_4^8 pi(-x/2+4)^2dx`

We may apply the basic integration property: `int c f(x) dx - c int f(x) dx`

`V = pi int_0^4x dx+ pi int_4^8 (-x/2+4)^2dx`

For the integration of `piint_0^4x dx` , we apply the Power rule for integration: `int x^n dx = x^(n+1)/(n+1)` .

`pi int_0^4xdx =pi*x^((1+1))/((1+1)) |_0^4`

                   ` = pi*x^(2)/(2) |_0^4`

                           or `(pix^2)/2|_0^4`

 Using the definite integral formula: `int_a^b f(x) dx = F(b) - F(a)` , we get:

`(pix^2)/2|_0^4=(pi(4)^2)/2-(pi(0)^2)/2`

            `= 8pi - 0`

            ` =8pi`

 

For the integration of `piint_4^8 (-x/2+4)^2dx` ,we apply FOIL method to expand.

`(-x/2+4)^2 = (-x/2+4)(-x/2+4) = x^2/4-4x+16 ` .

 The integral becomes:

`piint_4^8 (x^2/4-4x+16)dx`

Apply basic integration property: `int (u+-v+-w) dx = int (u) dx +- int (v) dx+- int (w) dx.`

`piint_4^8 (-x/2+4)^2dx=pi [int_4^8 (x^2/4) dx -int_4^8(4x)^2dx+int_4^8 16 dx]`

Apply Power rule for integration: `int x^n dx = x^(n+1)/(n+1)` and basic integration property: `int c dx = cx` .

`int_4^8 (x^2/4) dx=1/4 int_4^8 x^2 dx`

                     ` =1/4x^((2+1))/((2+1))|_4^8`

                   ` =1/4*x^3/3|_4^8`

                    ` =x^3/12|_4^8`

`int_4^8 (4x) dx=4 int_4^8 x dx`

                     ` =4x^(1+1)/(1+1)|_4^8`

                      ` =4*x^2/2|_4^8`

                       `=2x^2|_4^8`

`int_4^8 16 dx = 16x|_4^8`

Then,

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

 

 Using the definite integral formula:` int_a^b f(x) dx = F(b) - F(a)` , we get:

`piint_4^8 (-x/2+4)^2dx=pi[(8)^3/12-2(8)^2+16(8)]-pi[(4)^3/12-2(4)^2+16(4)]`

`= pi[128/3-128+128] -pi[16/3-32+64]`

`=(128pi)/3-(112pi)/3`

`= (16pi)/3`

 

or use u-subtitution by letting `u =-x/2+4 ` then` du =-1/2dx` or `-2 du =dx`

`pi int (-x/2+4)^2dx =pi int(u)^2* -2 du`

                              `= -2pi* u^3/3`

Plug-in `u= -x/2 +4` or `u=(-x+8)/2` on  `-2pi* u^3/3` , we get:

`pi int (-x/2+4)^2dx =-2pi((-x+8)/2)^3/3|_4^8`

              `=-2pi((-x+8)^3/8) *1/3|_4^8`

              `= ((x-8)^3pi)/12|_4^8`

 Using the definite integral formula: `int_a^b f(x) dx = F(b) - F(a)` , we get:

`piint_4^8 (-x/2+4)^2dx = ((8-8)^3pi)/12 -((4-8)^3pi)/12`

            ` = 0 - (-16pi)/3`

            ` =(16pi)/3`

          

 Then combining the result of the integrals, we get:

`V = pi int_0^4xdx+ pi int_4^8 (-x/2+4)^2dx`

`V=8 pi +(16pi)/3`

`V =(24pi)/3+(16pi)/3`

`V=(40pi)/3 ` or `41.89` (approximated value)

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