Answer:
Volume for any [tex]m[/tex] in [tex][0,1)[/tex] is [tex]3 \pi+3m\pi[/tex].
Volume for [tex]m=0[/tex] is [tex]3 \pi[/tex].
We get the same thing using the formula (volume of cylinder formula) for [tex]m=0[/tex]. (See below.)
Step-by-step explanation:
Introduction:
[tex]V=\int_{\text{given x-interval of the functions given}} \pi (\text{radius})^2 dx[/tex] (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:
[tex]V=\int_{\text{given x-interval of the functions given}} \pi (R^2-r^2) dx[/tex]
The radius, [tex]R[/tex] is for bigger circle.
The radius, [tex]r[/tex] is for smaller circle.
What are the radi?:
[tex]R=2+mx[/tex]
[tex]r=1-mx[/tex]
What are the radi squared?:
I will use the identity, [tex](a+b)^2=a^2+2ab+b^2[/tex] to find the square of each radius.
[tex]R^2=(2+mx)^2=2^2+2(2mx)+(mx)^2[/tex]
[tex]R^2=(2+mx)^2=4+4mx+m^2x^2[/tex]
[tex]r^2=(1-mx)^2=1^2+2(1(-mx))+(-mx)^2[/tex]
[tex]r^2=(1-mx)^2=1-2mx+m^2x^2[/tex]
What is the positive difference of the radi squared?:
Let's find [tex]R^2-r^2[/tex] .
[tex]R^2-r^2=[4+4mx+m^2x^2]-[1-2mx+m^2x^2][/tex]
[tex]R^2-r^2=[4-1]+[4mx-(-2mx)]+[m^2x^2-m^2x^2][/tex]
[tex]R^2-r^2=3+6mx+0[/tex]
[tex]R^2-r^2=3+6mx[/tex]
Finding the volume for any [tex]m[/tex] in [tex][0,1)[/tex]:
[tex]V=\int_{\text{given x-interval of the functions given}} \pi (R^2-r^2) dx[/tex]
[tex]V=\int_0^1 \pi (3+6mx)dx[/tex]
[tex]V= \pi (3x+\frac{6mx^2}{2})|_0^1[/tex]
[tex]V=\pi \{[(3(1)+\frac{6m(1)^2}{2}]-[3(0)+\frac{6m(0)^2}{2}]\}[/tex]
[tex]V=\pi \{[3+3m]-[0+0] \}[/tex]
[tex]V=\pi (3+3m)[/tex]
[tex]V=3 \pi+3m \pi[/tex]
Finding the volume for [tex]m=0[/tex]:
At [tex]m=0[/tex], we have [tex]V=3 \pi+3(0) \pi=3 \pi[/tex].
Confirmation using the volume of cylinder for [tex]m=0[/tex]:
If [tex]m=0[/tex], then we have horizontal lines [tex]f(x)=2[/tex] and [tex]f(x)=1[/tex].
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 [tex]V=\pi (2)^2(1)=\pi(4)(1)=4\pi[/tex].
The smaller cylinder has a radius of 1 units. So the volume of it is [tex]V=\pi(1)^2(1)=\pi(1)^2(1)=\pi[/tex].
The difference of these cylinder’s volume will give us desired volume of the resulting 3d-figure which is [tex]4\pi-1\pi=3\pi[/tex] units cubed.
Conclusions:
Volume for any [tex]m[/tex] in [tex][0,1)[/tex] is [tex]3 \pi+3m\pi[/tex].
Volume for [tex]m=0[/tex] is [tex]3 \pi[/tex].
We get the same thing using the formula (volume of cylinder formula) for [tex]m=0[/tex]. (See above.)
