Mathematics, 13.04.2020 00:12 hjeffrey168

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.

Suppose a solid is formed by revolving the function f(x)=2+mx around the x-axis where 0≤m<1 and 0

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Answer from: coralaguilar1702

(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

Answer from: isiss18

Volume for any 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, is for bigger circle.

The radius, 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 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 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.)

Suppose a solid is formed by revolving the function f(x)=2+mx around the x-axis where 0≤m<1 and 0

Answer from: Quest

answer:

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Answer from: Quest

step-by-step explanation:

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