Thread: Find diameter of hole with quarter volume of whole shape.

1. Find diameter of hole with quarter volume of whole shape.

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

b^2 = (256 + 16sqrt(12)) and 256 - 16sqrt(12)

Take three square root and multiply by 2, yielding

Diameter = 35.295 or 28.325

2. Originally Posted by Seed5813
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!

3. Originally Posted by stapel
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.

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4. Originally Posted by Seed5813
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...?

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

. . . . .$\mbox{b. }\, x\, =\, 0$

. . . . .$\mbox{c. }\, y\, =\, 8$

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

Thank you!

5. Originally Posted by stapel
"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...?

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

. . . . .$\mbox{b. }\, x\, =\, 0$

. . . . .$\mbox{c. }\, y\, =\, 8$

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

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6. Originally Posted by Seed5813
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!

7. Originally Posted by stapel
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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