If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution.
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
024. ``q Query 24
Question: `q Query 9.5.12 vol of sphere diam 14.8 **** What is the volume of the sphere and how did you obtain it?
Your solution:
Confidence Assessment:
Given Solution:
`a I use the formula for finding the volume of a sphere which is 4/3(3.14)(r^3).
Since the diameter is 14.8, the radius is half that which is 7.4.
V = 4/3 * 3.14 * 7.4^3
V = 4/3 * 3.14 * 405.224
V = 1696.54
The volume is 1696.54
Self-critique (if necessary):
Self-critique Rating:
Question: `q Query 9.5.18 pyramid 12 x 4 altitude 10 **** What is the volume of the pyramid and how did you find it?
Your solution:
Confidence Assessment:
Given Solution:
`a I used the formula : V = 1/3Bh
The base = 12 * 4 = 48
V = 1/3 * 48 * 10
V = 1/3 * 480
V = 160
The volume is 160ft.^3
Self-critique (if necessary):
Self-critique Rating:
Question: `q Query 9.5.24 bottle 3 cm alt 4.3 cm **** What is the volume of the bottle and how did you find it?
Your solution:
Confidence Assessment:
Given Solution:
`a ** The figure is a right circular cylinder with V = 3.14 * r^2 * h
Since the diameter is 3, then the radius is 1.5
V = 3.14 * 1.5^2 * 4.3
V = 3.14 * 2.25 * 4.3
V = 30.38 cm^3 **
Self-critique (if necessary):
Self-critique Rating:
Question: `q Query 9.5.36 sphere area 144 `pi m^2 **** What are the radius, diameter and volume of the sphere and how did you find them?
Your solution:
Confidence Assessment:
Given Solution:
`a ** Sphere area is 4 pi r^2, so we have
4 pi r^2 = 144 pi m^2. Dividing by 4 pi we get
r^2 = 36 m^2. Taking the square root of both sides we get
r = 6 m.
From this we find that the diameter is 2 * 6 m = 12 m and the volume is 4/3 pi * (6 m)^3 = 288 pi m^3. **
Self-critique (if necessary):
Self-critique Rating:
Question: `q Query 9.5.48 cone alt 15 rad x vol 245 `pi **** What is the value of x and how did you find your result?
Your solution:
Confidence Assessment:
Given Solution:
`a ** We have V = 1/3 pi r^2 h. To solve for r we multiply both sides by 3 / (pi * h) to get
3 V / (pi * h) = r^2 then take the square root to get
r = sqrt(3 V / ( pi * h) ). Substituting we get
r = sqrt( 3 * 245 pi / (pi * 15) ) = sqrt(49) = 7.
Self-critique (if necessary):
Self-critique Rating:
Question: `q Query 9.5.51 plane intersects sphere passing 7 in from center forming circle with area 576 `pi **** What is the volume of the sphere and how did you obtain it?
Your solution:
Confidence Assessment:
Given Solution:
`a ** The circle does have radius sqrt(576 in^2) = 24 in. However that is not the radius of the sphere since the plane containing the circle passes 7 in from the center of the sphere. So the center of the circle is not the center of the sphere.
The center of the circle is 7 in from the center of the sphere. Note also that a line from the center of the sphere to the center of the circle will be perpendicular to the plane of the circle.
Thus if you start at the center of the sphere and move the 7 in straight to the center of the circle, then move along the plane of intersection (in any direction) for 24 in (at which point you encounter the rim of the circle, whose radius you recall is 24 in), then back to the center of the sphere you will have traced out a right triangle with legs 7 in and 24 in. The hypotenuse of the triangle is the radius R of the sphere.
So we have
R^2 = 7^2 + 24^2 = 625 and
R = 25.
The radius of the sphere is 25 in.
Its volume will therefore be 4/3 pi r^3:
V = 4/3 pi r^3
= 4/3 pi * (25 in)^3
= 4/3 pi * 16000 in^3 (approx.)
= 67000 in^3, approx. **
STUDENT COMMENT: I don't
understand how you drew a triangle inside of the circle. How did you know to do
that instead of adding the 7 to the radius of the plane?
INSTRUCTOR RESPONSE: The triangle wasn't inside the circle. It was,
however, inside the sphere.
Its 7-inch leg runs from the center of the sphere to the center of the circle.
Its 24 in leg runs from the center of the circle to any point on the circle.
That point is also on the sphere.
A radius of the sphere runs from the center of the sphere to any point on the
sphere. The hypotenuse of the right triangle runs from the center of the circle
to a point on the sphere, so the hypotenuse is a radius of the sphere.
The key part of the explanation is quoted in the paragraph below. You can
probably get by OK without completely understanding this (nothing this
challenging on the test), but if you still have trouble with the given solution
and want to get to the bottom of it, then tell me, phrase by phrase, what you do
and do not understand in the above explanation and in this paragraph:
'Thus if you start at the center of the sphere and move the 7 in straight to the
center of the circle, then move along the
plane of intersection (in any direction) for 24 in (at which point you encounter
the rim of the circle, whose radius you recall
is 24 in), then back to the center of the sphere you will have traced out a
right triangle with legs 7 in and 24 in. The
hypotenuse of the triangle is the radius R of the sphere.'
Self-critique (if necessary):
Self-critique Rating:
Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment.
I don't think that my last answer is correct. Also, in the first problem, the answer should be 1696.54 cm^3, instead of just 1696.54
vvvv