Verifying Volume of Region about x=1

[Lo.Lee.Ta.]

Find the volume between y=-2x + 4, x-axis, x=1, about the line x=1. Check my work? :)

Hi, everyone.

1. Find the volume of the region between y= -2x + 4, x-axis, x=1, about the line x=1.

I tried to post this before, but I don't think it went through!

2. Alright, so I first drew it out, and the shape is a cone.
Since it revolves around a y-axis, the limits should be also in terms of y.
Every cross-section is a circle.

So this is how I wrote it out:

∫0 to 4 of ∏[((-y/2) + 1)^2]dy

∫0 to 4 of ∏[(y^2/4) -y + 1]

= 1/4 * (y^3)/3) - (y^2)/2) + x |0 to 4

= ((y^3)/12) - 1/2(y^2) + x |0 to 4

= ((4)^3)/3 - 1/2(4)^2 + 4 -(0)

= 64/12 - 8 + 4

= 1.33 or 4/3

So... Is that right?
Thank you so much for checking my work! :)

 

[Mark44]

Hi, everyone.

1. Find the volume of the region between y= -2x + 4, x-axis, x=1, about the line x=1.

I tried to post this before, but I don't think it went through!

2. Alright, so I first drew it out, and the shape is a cone.
Since it revolves around a y-axis, the limits should be also in terms of y.
Every cross-section is a circle.

What is the exact problem statement? What you have is confusing. Is the region bounded by y = -2x + 4, the x-axis, and the line x = 1 revolved about some line? You don't mention the word "revolved" in your problem statement.

Is the region revolved about the line x = 1 or about the y-axis (the line x = 0)?

So this is how I wrote it out:

∫0 to 4 of ∏[((-y/2) + 1)^2]dy

∫0 to 4 of ∏[(y^2/4) -y + 1]

= 1/4 * (y^3)/3) - (y^2)/2) + x |0 to 4

= ((y^3)/12) - 1/2(y^2) + x |0 to 4

= ((4)^3)/3 - 1/2(4)^2 + 4 -(0)

= 64/12 - 8 + 4

= 1.33 or 4/3

So... Is that right?
Thank you so much for checking my work! :)

 

[Lo.Lee.Ta.]

Yes, it is revolved about the line x=1.

Sorry about the confusion.

 

[HallsofIvy]

Hi, everyone.

1. Find the volume of the region between y= -2x + 4, x-axis, x=1, about the line x=1.

y= -2x+ 4= 0 (crossing the x-axis) when x= 2 so x ranges from 1 to 2.
It is NOT a cone, it is "frustrum" of a cone, not including the "point".

I tried to post this before, but I don't think it went through!

2. Alright, so I first drew it out, and the shape is a cone.
Since it revolves around a y-axis, the limits should be also in terms of y.

You mean around a line parallel to the x-axis.

Every cross-section is a circle.

Yes, having radius x- 1. Since y= -2x+ 4, 2x= 4- y, x= 2- y/2 and x- 1= 2- y/2- 1= 1- y/2. The area of such a circle is [itex]\pi(1- y/2)^2[itex] and taking its thickness to be "dy", its volume is [tex]\pi(1- y/2)^2dy= \pi (1- y+ y^2/4)dy[tex].

So this is how I wrote it out:

∫0 to 4 of ∏[((-y/2) + 1)^2]dy

∫0 to 4 of ∏[(y^2/4) -y + 1]

You forgot the "dy" but no matter.

= 1/4 * (y^3)/3) - (y^2)/2) + x |0 to 4

= ((y^3)/12) - 1/2(y^2) + x |0 to 4

and you really mean "y", not "x" there. But again, you put y= 4 into that so, no matter!

= ((4)^3)/3 - 1/2(4)^2 + 4 -(0)

And here, you should have "(4)^3/12" not "(4)^3/3"

= 64/12 - 8 + 4

64/12= 16/3 so this is 16/3- 24/3+ 12/3= (26- 24)/3= 4/3.

= 1.33 or 4/3

So... Is that right?
Thank you so much for checking my work! :)

Yes, that looks good to me. The fact that it is only a part of a cone is not relevant.

 

[Lo.Lee.Ta.]

Thanks for checking it! :)

I made a lot of mistakes in my work, but I managed to get the right answer! I'll have to fix those!