Question 1181731
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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.
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        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.



<pre>
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     <<<---=== {{{highlight(highlight(error))}}}: 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
</pre>

Now everything is correct.


In previous version by @CPhill, even the dimensions of the expressions were wrong.



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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;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.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;It has no feeling of shame - it is shameless.



This time, again, &nbsp;it made an error.



Although the @CPhill' solution are copy-paste &nbsp;Google &nbsp;AI solutions, &nbsp;there is one essential difference.


Every time, &nbsp;Google &nbsp;AI &nbsp;makes a note at the end of its solutions that &nbsp;Google &nbsp;AI &nbsp;is experimental
and can make errors/mistakes.


All @CPhill' solutions are copy-paste of &nbsp;Google &nbsp;AI &nbsp;solutions, with one difference:
@PChill never makes this notice and never says that his solutions are copy-paste that of Google.


Every time, &nbsp;@CPhill embarrasses to tell the truth.

But I am not embarrassing to tell the truth, &nbsp;as it is my duty at this forum.



And the last my comment.


When you obtain such posts from @CPhill, &nbsp;remember, &nbsp;that &nbsp;NOBODY &nbsp;is responsible for their correctness, 
until the specialists and experts will check and confirm their correctness.


Without it, &nbsp;their reliability is &nbsp;ZERO and their creadability is &nbsp;ZERO, &nbsp;too.



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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.