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Find diameter of hole with quarter volume of whole shape.

Seed5813

New member
Joined
Jan 29, 2018
Messages
24
The region bound by
Code:
y=(1/8)x^2
and the y axis is rotated about the line
Code:
y=8
A hole is drilled that takes up 1/4th the volume of the original volume, find the diameter to 3 decimal places.

2pi Int_0^8(8x - (x^3)/8)dx
... =256pi

Dividing by 4 yields 64pi

Code:
64pi = 2pi Int_0^z(8x - (x^3)/8)dx
32 = Int_0^z(8x - (x^3)/8)dx
32 = 4b^2 - (b^4)/32

-(b^4)/32 + 4b^2 - 32 = 0

By quadratic formula:
b^2 = (256 + 16sqrt(12)) and 256 - 16sqrt(12)

Take three square root and multiply by 2, yielding

Diameter = 35.295 or 28.325
 

stapel

Super Moderator
Staff member
Joined
Feb 4, 2004
Messages
15,937
The region bound by
Code:
y=(1/8)x^2
and the y axis is rotated about the line
Code:
y=8
A hole is drilled that takes up 1/4th the volume of the original volume, find the diameter to 3 decimal places.
I'm not understanding the "region" here. The first curve is a parabola whose axis in the y-axis. So the y-axis is "inside" the curve. How is the y-axis then a bound on the region? Also, the parabola opens upward, including through y = 8. The figure would be infinite!

Is there maybe a picture that goes with this? Or other information? Thank you! ;)
 

Seed5813

New member
Joined
Jan 29, 2018
Messages
24
I'm not understanding the "region" here. The first curve is a parabola whose axis in the y-axis. So the y-axis is "inside" the curve. How is the y-axis then a bound on the region? Also, the parabola opens upward, including through y = 8. The figure would be infinite!

Is there maybe a picture that goes with this? Or other information? Thank you! ;)
Its a bullet shape. The Y axis is the flat side of the bullet and you stop the parabola at y=8 and rotate around that.

Sent from my LGLS755 using Tapatalk
 

stapel

Super Moderator
Staff member
Joined
Feb 4, 2004
Messages
15,937
Its a bullet shape. The Y axis is the flat side of the bullet and you stop the parabola at y=8 and rotate around that.
"The flat side of the bullet" is the base. The parabola and the line y = 8 could approximate this, but then the y-axis would be irrelevant. Do you perhaps mean that the "bullet" shape is cut in half, from its tip to its base? So the bounds are as follows...?

. . . . .\(\displaystyle \mbox{a. }\, y\, =\, \frac{1}{8}x^2,\, x\, \geq\, 0\)

. . . . .\(\displaystyle \mbox{b. }\, x\, =\, 0\)

. . . . .\(\displaystyle \mbox{c. }\, y\, =\, 8\)

And the resulting region is rotated about the horizontal line y = 8...?

Thank you! ;)
 

Seed5813

New member
Joined
Jan 29, 2018
Messages
24
"The flat side of the bullet" is the base. The parabola and the line y = 8 could approximate this, but then the y-axis would be irrelevant. Do you perhaps mean that the "bullet" shape is cut in half, from its tip to its base? So the bounds are as follows...?

. . . . .\(\displaystyle \mbox{a. }\, y\, =\, \frac{1}{8}x^2,\, x\, \geq\, 0\)

. . . . .\(\displaystyle \mbox{b. }\, x\, =\, 0\)

. . . . .\(\displaystyle \mbox{c. }\, y\, =\, 8\)

And the resulting region is rotated about the horizontal line y = 8...?

Thank you! ;)
Yes. Its cut in half but after being rotated it makes a bullet shape

Sent from my LGLS755 using Tapatalk
 

stapel

Super Moderator
Staff member
Joined
Feb 4, 2004
Messages
15,937
Yes. Its cut in half but after being rotated it makes a bullet shape
How? You have half of a parabola:

Code:
shape:

     ^y
     | 
---------------- y=8   
     |####'
     |###'
     |###'
     |###'
     |##'
     |#'
     |'
-----+------------>x
     |
You're twirling it up, over, and around the given horizontal line. How is that process recreating the left-hand half of the bullet, while not creating anything above the given axis?

Code:
flipped:

     ^y
     | 
     |
     |'
     |#'
     |##'
     |###'
     |###'
     |###'
     |####'
---------------- y=8   
     |####'
     |###'
     |###'
     |###'
     |##'
     |#'
     |'
-----+------------>x
     |
Please show what you're doing. Thank you! ;)
 

Seed5813

New member
Joined
Jan 29, 2018
Messages
24
How? You have half of a parabola:

Code:
shape:

     ^y
     | 
---------------- y=8   
     |####'
     |###'
     |###'
     |###'
     |##'
     |#'
     |'
-----+------------>x
     |
You're twirling it up, over, and around the given horizontal line. How is that process recreating the left-hand half of the bullet, while not creating anything above the given axis?

Code:
flipped:

     ^y
     | 
     |
     |'
     |#'
     |##'
     |###'
     |###'
     |###'
     |####'
---------------- y=8   
     |####'
     |###'
     |###'
     |###'
     |##'
     |#'
     |'
-----+------------>x
     |
Please show what you're doing. Thank you! ;)
Its a parabola bounded by x=0 and y=8.


Sent from my LGLS755 using Tapatalk
 
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