Homework Help

Determine the diameter of a hole that is drilled vertically through the center of the...

user profile pic

david3 | Student | (Level 3) eNoter

Posted March 25, 2013 at 10:40 PM via web

dislike 1 like

Determine the diameter of a hole that is drilled vertically through the center of the solid bounded by the graphs of the equations

z = 20e^−(x^2 + y^2) / 4,

z = 0, and 

x^2 + y^2 = 16

if one-tenth of the volume of the solid is removed. (Round your answer to four decimal places)

1 Answer | Add Yours

user profile pic

mathsworkmusic | (Level 1) Educator

Posted March 31, 2013 at 11:19 AM (Answer #1)

dislike 1 like

The solid takes the form of a scaled (by ` ``80pi`) bivariate bell (Normal/Gaussian) curve where the variance of the two variables `x` and `y` is `sigma^2 =2` and the correlation `rho=0` .

Ignoring the scale factor of `80pi` which only affects the height of the solid and not the radius (which we are interested in), a radius of 4 (the radius of the function `x^2+y^2 = 16`) on this graph is equivalent to a radius of `4/sigma= 4/sqrt(2)` on a standard bivariate bell curve.

The quantile `4/sqrt(2)` gives a one-sided proportion of a standard Normal/Gaussian curve of

`0.5- Phi(-4/sqrt(2)) = 0.498`

where `Phi` is the distribution function of a standard Normal/Gaussian variable.

A twentieth (2 x tenth) of this is 0.0249. The required radius is then at the 100(0.5 - 0.0249) = 47.51th percentile of the standard Normal curve.

Now, using lookup tables for the inverse of the standard Normal distribution function

`Phi^(-1)(0.4751) = -0.06241`

The radius we require is` ` `sigma=sqrt(2)` times this value, working on our standard deviation being `sigma` and not 1 as for the standard Normal distribution.

The required radius of the drilled section is sqrt(2)(-0.06241) = 0.0883.

The required diameter of the drilled section is then 2*0.0883 = 0.1765


` `



Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes