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Question 1181731: A sphere is inscribed in a right circular cone of altitude h and radius of base r. Write a formula in terms of r and h for the volume of the sphere.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to derive the formula for the volume of the inscribed sphere in terms of *r* and *h*:
**1. Diagram and Key Relationships:**
Draw a cross-section of the cone and sphere. You'll see a circle (representing the sphere) inscribed in a triangle (representing the cone).
* Let *R* be the radius of the inscribed sphere.
* The radius of the cone's base is *r*.
* The height (altitude) of the cone is *h*.
* The slant height of the cone (the hypotenuse of the triangle) is *s* = sqrt(r² + h²).
**2. Similar Triangles:**
There are two similar right triangles in the cross-section:
* The large triangle representing the cone, with sides *r*, *h*, and *s*.
* A smaller triangle formed by the radius of the sphere (*R*), the difference between the cone's height and the sphere's radius (*h - R*), and a portion of the slant height.
The ratio of corresponding sides in similar triangles is equal:
R / r = (h - R) / s
**3. Solve for R:**
R / r = (h - R) / sqrt(r² + h²)
R * sqrt(r² + h²) = r(h - R)
R * sqrt(r² + h²) = rh - rR
R * sqrt(r² + h²) + rR = rh
R(sqrt(r² + h²) + r) = rh
R = rh / (sqrt(r² + h²) + r)
**4. Rationalize the Denominator (Optional but often preferred):**
Multiply the numerator and denominator by the conjugate of the denominator:
R = rh(sqrt(r² + h²) - r) / ((sqrt(r² + h²) + r)(sqrt(r² + h²) - r))
R = rh(sqrt(r² + h²) - r) / (r² + h² - r²)
R = rh(sqrt(r² + h²) - r) / h
R = r(sqrt(r² + h²) - r)
**5. Volume of the Sphere:**
The volume *V* of a sphere is given by:
V = (4/3)πR³
Substitute the expression for *R* we derived:
V = (4/3)π[r(sqrt(r² + h²) - r)]³
Therefore, the volume of the inscribed sphere in terms of *r* and *h* is:
V = (4/3)πr³(sqrt(r² + h²) - r)³
Answer by ikleyn(52803)  (Show Source):
You can put this solution on YOUR website! .
A sphere is inscribed in a right circular cone of altitude h and radius of base r.
Write a formula in terms of r and h for the volume of the sphere.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
After attentive reading the solution by @CPhill, I see the errors in it,
that require me to fix them.
So, I copy the solution by @CPhill and make my editing right there.
Here's how to derive the formula for the volume of the inscribed sphere in terms of *r* and *h*:
**1. Diagram and Key Relationships:**
Draw a cross-section of the cone and sphere. You'll see a circle (representing the sphere) inscribed in a triangle (representing the cone).
* Let *R* be the radius of the inscribed sphere.
* The radius of the cone's base is *r*.
* The height (altitude) of the cone is *h*.
* The slant height of the cone (the hypotenuse of the triangle) is *s* = sqrt(r² + h²).
**2. Similar Triangles:**
There are two similar right triangles in the cross-section:
* The large triangle representing the cone, with sides *r*, *h*, and *s*.
* A smaller triangle formed by the radius of the sphere (*R*), the difference between the cone's height and the sphere's radius (*h - R*), and a portion of the slant height.
The ratio of corresponding sides in similar triangles is equal:
R / r = (h - R) / s
**3. Solve for R:**
R / r = (h - R) / sqrt(r² + h²)
R * sqrt(r² + h²) = r(h - R)
R * sqrt(r² + h²) = rh - rR
R * sqrt(r² + h²) + rR = rh
R(sqrt(r² + h²) + r) = rh
R = rh / (sqrt(r² + h²) + r)
**4. Rationalize the Denominator (Optional but often preferred):**
Multiply the numerator and denominator by the conjugate of the denominator:
R = rh(sqrt(r² + h²) - r) / ((sqrt(r² + h²) + r)(sqrt(r² + h²) - r))
R = rh(sqrt(r² + h²) - r) / (r² + h² - r²)
R = rh(sqrt(r² + h²) - r) / h <<<---===  : should be R = rh(sqrt(r² + h²) - r) / h²
R = r(sqrt(r² + h²) - r) <<<---=== as a consequence, should be R = r(sqrt(r² + h²) - r)/h
otherwise, the 'R's dimension is m^2, instead of 'm'.
**5. Volume of the Sphere:**
The volume *V* of a sphere is given by:
V = (4/3)πR³
Substitute the expression for *R* we derived:
V = (4/3)π[r(sqrt(r² + h²) - r)/h]³ <<<---=== associated correction here
Therefore, the volume of the inscribed sphere in terms of *r* and *h* is:
V = (4/3)πr³(sqrt((r² + h²) - r)/h)³ <<<---=== associated correction here
Now everything is correct.
In previous version by @CPhill, even the dimensions of the expressions were wrong.
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Regarding the post by @CPhill . . .
Keep in mind that @CPhill 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-paste that of Google.
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.
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
Dear @CPhill, I will include such comment to every your post, where I will find an error,
UNTIL you will place a correct acknowledgment in every your post.
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