We can use a rectangular strips to represent the region bounded by `x+y=4,` `y=0` , and `y=x` revolved about the x-axis. As shown on the attached graph, we consider *two sets of rectangular strip perpendicular to the x-axis (axis of revolution)* to be able to use the **Disk Method** ....

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We can use a rectangular strips to represent the region bounded by `x+y=4,` `y=0` , and `y=x` revolved about the x-axis. As shown on the attached graph, we consider *two sets of rectangular strip perpendicular to the x-axis (axis of revolution)* to be able to use the **Disk Method**. This is the case since the upper bound of the rectangular strip differs before and after` x=2` .

In this method, we follow the formula:` V = int_a^b A(x) dx` since we are using a vertical orientation of each rectangular strip with a **thickness =dy**.

Note: `A = pir^2` where **r= length of the rectangular strip**.

We may apply `r = y_(above) - y_(below)` .

For the region within the boundary values of x: `[ 0,2]` , we follow `r = x-0=x`

For the region within the boundary values of x: `[ 2,4]` , we follow `r = 4-x-0=4-x`

Note: `x+y=4` can be rearranged as `y=4-x.`

Then the integral set-up will be:

`V = int_0^2 pi*(x)^2dx+int_2^4 pi*(4-x)^2dx`

For the first integral: `int_0^2 pi*(x)^2dx` , we may apply Power rule of integration: `int x^n dx = x^(n+1)/(n+1)` .

`int_0^2 pi*(x)^2dx= pi* x^((2+1))/((2+1))|_0^2`

` =(pix^3)/3|_0^2`

Apply definite integration formula: `int_a^b f(y) dy= F(b)-F(a)` .

`(pix^3)/3|_0^2 =(pi(2)^3)/3-(pi(0)^3)/3`

` =(8pi)/3- 0`

` =(8pi)/3`

For the indefinite integral of `int_2^4 pi*(4-x)^2dx` , we may u-substitution by letting `u =4-x` then `du =-dx` or `(-1)du =dx` .

The integral becomes :

`int pi*(4-x)^2dx =int pi*u^2*(-1) du`

Apply basic integration property: `intc*f(x) dx = c int f(x) dx.`

`int pi*u^2*(-1) du = -pi int u^2 du`

Apply power rule for integration: `int x^n dy= x^(n+1)/(n+1).`

`-pi int u^2 du =-pi* u^((2+1))/((2+1))`

`= (-piu^3)/3`

Plug-in` u=4-x ` on `(-piu^3)/3` we get:

`int_2^4 pi*(4-x)^2dx =(-pi(4-x)^3)/3|_2^4 or ((x-4)^3pi)/3|_2^4`

Apply the definite integral formula: `int _a^b f(x) dx = F(b) - F(a)` .

`((x-4)^3pi)/3|_2^4 =((4-4)^3pi)/3-((2-4)^3pi)/3`

` = 0 - (-8pi)/3`

` = (8pi)/3`

Combing the two definite integrals, we get:

`V = int_0^2 pi*(x)^2dx+int_2^4 pi*(4-x)^2dx`

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

`V =(16pi)/3` or `16.76 ` (approximated value).