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Question 1181729: The diameter of a sphere coincides with the axis of a right circular cone, and the surface of the cone intersects the surface of the sphere in in a great circle. Find: (a) the vertical angle of the cone: (b) the volume of the cone; (c) the volume common to the two solids: (d) the volume of that portion of the cone which lies outside the sphere: (e) the volume of that portion of the sphere which lies outside the cone. Denote radius of sphere by R.
Solve Problem if (a) R=2: (b) R=3.72.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's the solution, broken down step by step:
**1. Visualize the Setup:** Imagine a sphere perfectly nestled inside a cone, with the sphere's diameter matching the cone's axis. The sphere touches the cone along a great circle.
**2. Key Geometric Relationships:**
* Let *R* be the radius of the sphere.
* The cone's height (h) is equal to the sphere's diameter, so h = 2R.
* The slant height (s) of the cone is the distance from the apex to any point on the edge of the circular base.
* The radius of the cone's base (r) can be found using the Pythagorean theorem.
**3. (a) Vertical Angle of the Cone:**
* Consider a right triangle formed by the cone's height (2R), the cone's radius (r), and the cone's slant height (s). Half of the cone's vertical angle (θ/2) is the angle between the slant height and the axis (height).
* sin(θ/2) = r / s
* Since the sphere touches the cone along a great circle, r = R.
* s = sqrt(r² + h²) = sqrt(R² + (2R)²) = sqrt(5R²) = R√5
* sin(θ/2) = R / (R√5) = 1/√5
* θ/2 = arcsin(1/√5)
* θ = 2 * arcsin(1/√5) ≈ 51.83°
**4. (b) Volume of the Cone:**
* Volume of a cone = (1/3)πr²h
* V_cone = (1/3)πR²(2R) = (2/3)πR³
**5. (c) Volume Common to the Two Solids:**
* The volume common to both is simply the volume of the sphere.
* V_common = (4/3)πR³
**6. (d) Volume of the Cone Outside the Sphere:**
* V_outside_sphere = V_cone - V_common
* V_outside_sphere = (2/3)πR³ - (4/3)πR³ = (-2/3)πR³
**Important Correction:** The volume of the cone *outside* the sphere should be a positive value. We made an error in the calculation. It should be:
V_outside_sphere = V_cone - V_common
V_outside_sphere = (2/3)πR³- (4/3)πR³
Answer by ikleyn(52784)  (Show Source):
You can put this solution on YOUR website! .
The diameter of a sphere coincides with the axis of a right circular cone, and the surface
of the cone intersects the surface of the sphere in in a great circle. Find:
(a) the vertical angle of the cone:
(b) the volume of the cone;
(c) the volume common to the two solids:
(d) the volume of that portion of the cone which lies outside the sphere:
(e) the volume of that portion of the sphere which lies outside the cone. Denote radius of sphere by R.
Solve Problem if (a) R=2: (b) R=3.72.
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The solution in the post by @CPhill is INCORRECT.
It is incorrect, since @CPhill incorrectly interprets the problem.
He solves the problem in assumption that the cone TOUCHES the sphere
along a great circle.
But, firstly, it is  , and, secondly, the problem  something  .
It says that the cone' surface  the sphere surface at the great circle,
which is totally different condition.
Thus, the value of @CPhill' solution is ZERO.
He is just deceiving you.
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Regarding the post by @CPhill . . .
Keep in mind that @KPhill is a pseudonym for the Google artificial intelligence.
The artificial intelligence is like a baby now. It is in the experimental stage
of development and can make mistakes and produce nonsense without any embarrassment.
It has no feeling of shame - it is shameless.
This time, again, it made an error.
Although the @CPhill' solution are copy-paste Google AI solutions, there is one essential difference.
Every time, Google AI makes a note at the end of its solutions that Google AI is experimental
and can make errors/mistakes.
All @CPhill' solutions are copy-paste of Google AI solutions, with one difference:
@PChill never makes this notice and never says that his solutions are copy-past that of Google.
So, in my view, doing this way, this guy, @CPhill, simply makes dishonest business at this forum.
Every time, @CPhill embarrasses to tell the truth.
But I am not embarrassing to tell the truth, as it is my duty at this forum.
And the last my comment.
When you obtain such posts from @CPhill, remember, that NOBODY is responsible for their correctness,
until the specialists and experts will check and confirm their correctness.
Without it, their reliability is ZERO and their creadability is ZERO, too.
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Special note for a visitor who posted this problem.
It borders on madness to pack 5 (five) questions into one post.
The rules of this forum allow one question per post.
Even if assume that the rules are made of rubber, then 3 questions per post is the maximum.
Every time, as you violate this rule, you EITHER will wait several years, OR will get a mess.
Regarding the rules of this forum, see this page
https://www.algebra.com/tutors/students/ask.mpl?action=ask_question&topic=Equations&return_url=http://www.algebra.com/algebra/homework/equations/
from which you post your problems.
It is assumed that you read these rules before posting.
It is also assumed that you do understand what is written in that page and follow the rules.
Those who violate them, work against their own interests.
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