Question

Suppose a solid is formed by revolving the function f(x)=2+mx around the x-axis where 0≤m<1 and 0≤x≤1, and a washer is created by drilling a hole in the solid that corresponds to the function g(x)=1-mx. Determine the volume of the resulting washer as a function of m, and confirm the result for m=0 using the formula for a cylinder.

Answer 1

(m³/3 + 5m/2 + 3)pi

Explanation:

pi integral [(f(x))² - (g(x))²]

Limits 0 to 1

pi × integral [(2+mx)² - (1-mx)²]

pi × integral[4 + 4mx + m²x² - 1 + 2mx - m²x²]

pi × integral [m²x² + 5mx + 3]

pi × [m²x³/3 + 5mx²/2 + 3x]

Upper limit - lower limit

pi × [m²/3 + 5m/2 + 3]

Verification:

m = 0

[pi × 2² × 1] - [pi × 1² × 1] = 3pi

[m³/3 + 5m/2 + 3]pi

m = 0

3pi

Answer 2

Volume for any
`m` in
`[0,1)` is
`3 \pi+3m\pi`.

Volume for
`m=0` is
`3 \pi`.

We get the same thing using the formula (volume of cylinder formula) for
`m=0`. (See below.)

Step-by-step explanation:

Introduction:

`V=\int_{\text{given x-interval of the functions given}} \pi (\text{radius})^2 dx` (Notice we will be filling the 3d-solid with area of a circles on the given interval.)

The problem is we have a hole in our 3d-solid we will need to subtract it out.

Formula:

The formula for calculating this volume will be:

`V=\int_{\text{given x-interval of the functions given}} \pi (R^2-r^2) dx`

The radius,
`R` is for bigger circle.

The radius,
`r` is for smaller circle.

What are the radi?:

`R=2+mx`

`r=1-mx`

What are the radi squared?:

I will use the identity,
`(a+b)^2=a^2+2ab+b^2` to find the square of each radius.

`R^2=(2+mx)^2=2^2+2(2mx)+(mx)^2`

`R^2=(2+mx)^2=4+4mx+m^2x^2`

`r^2=(1-mx)^2=1^2+2(1(-mx))+(-mx)^2`

`r^2=(1-mx)^2=1-2mx+m^2x^2`

What is the positive difference of the radi squared?:

Let's find
`R^2-r^2` .

`R^2-r^2=[4+4mx+m^2x^2]-[1-2mx+m^2x^2]`

`R^2-r^2=[4-1]+[4mx-(-2mx)]+[m^2x^2-m^2x^2]`

`R^2-r^2=3+6mx+0`

`R^2-r^2=3+6mx`

Finding the volume for any
`m` in
`[0,1)`:

`V=\int_{\text{given x-interval of the functions given}} \pi (R^2-r^2) dx`

`V=\int_0^1 \pi (3+6mx)dx`

`V= \pi (3x+(6mx^2)/(2))|_0^1`

`V=\pi \{[(3(1)+(6m(1)^2)/(2)]-[3(0)+(6m(0)^2)/(2)]\}`

`V=\pi \{[3+3m]-[0+0] \}`

`V=\pi (3+3m)`

`V=3 \pi+3m \pi`

Finding the volume for
`m=0`:

At
`m=0`, we have
`V=3 \pi+3(0) \pi=3 \pi`.

Confirmation using the volume of cylinder for
`m=0`:

If
`m=0`, then we have horizontal lines
`f(x)=2` and
`f(x)=1`.

The 3d-figure that results will be a cylinder with a hole in it (that is also in the shape of a cylinder).

The larger cylinder has a radius of 2 units. So the volume of it is
`V=\pi (2)^2(1)=\pi(4)(1)=4\pi`.

The smaller cylinder has a radius of 1 units. So the volume of it is
`V=\pi(1)^2(1)=\pi(1)^2(1)=\pi`.

The difference of these cylinder’s volume will give us desired volume of the resulting 3d-figure which is
`4\pi-1\pi=3\pi` units cubed.

Conclusions:

Volume for any
`m` in
`[0,1)` is
`3 \pi+3m\pi`.

Volume for
`m=0` is
`3 \pi`.

We get the same thing using the formula (volume of cylinder formula) for
`m=0`. (See above.)