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

Answer:

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 $[0,1)$:

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

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

$V= \pi (3x+\frac{6mx^2}{2})|_0^1$

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

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

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

$V=3 \pi+3m \pi$

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.)

Answer 2

Answer:

(m³/3 + 5m/2 + 3)pi

Step-by-step 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