Surface Area and Volumes of Spheres

kayla121497:

point A is the center of the sphere. point C is on the surface of the sphere. point B is the center of the circle that lies in plane P and includes point C. the radius of the circle is 12mm. AB= 5 mm. What is the volume of the sphere to the nearest cubic mm?

I am wondering if there is something missing - like an angle maybe?

Perhaps you could supply a picture??? It's difficult to answer without knowing exactly where B lies..

kayla121497:

point A is the center of the sphere. point C is on the surface of the sphere. point B is the center of the circle that lies in plane P and includes point C. the radius of the circle is 12mm. AB= 5 mm. What is the volume of the sphere to the nearest cubic mm?

An intuitive and quick method to solve: Treat the circle on the plane as a spherical segment (analogous to a chord) situated 5 mm below the center of the sphere.

Find the angle at common point C by the arc-Tan of (5mm/12mm). This returns an angle of 22.62 degrees.
Divide 5mm by the Sine of this angle (5mm)/Sin(22.62) =(5mm/ 0.384)=13mm. This value is the radius of the sphere.

Then use the sphere volume formula (4/3)Pi (r ³). (4/3)Pi (13 ³])=

Hi DavidQD
I thought of this but I wasn't convinced that it would always be true. I'll have to think some more.

kayla121497:

point A is the center of the sphere. point C is on the surface of the sphere. point B is the center of the circle that lies in plane P and includes point C. the radius of the circle is 12mm. AB= 5 mm. What is the volume of the sphere to the nearest cubic mm?

AB (5mm) and BC (12mm) are two sides of a right-angled triangle of which the radius of the sphere is the hypotenuse. The triangle is the well -known 5, 12, 13 triangle, so the radius of the sphere is 13mm. The volume follows from V = (4/3)*pi*r ³ as in DavidQD's reply above.

See the picture below.
Sphere.PNG

kayla121497 wrote:point A is the center of the sphere. point C is on the surface of the sphere. point B is the center of the circle that lies in plane P and includes point C. the radius of the circle is 12mm. AB= 5 mm. What is the volume of the sphere to the nearest cubic mm?

Mmmmm....I don't know about this.....here's a cross-section of a sphere where A is the center, C lies on the sphere's surface and AB =5, BC = 12 and BC lies in Plane "P"
(I think all of these meet the requirements of the problem).

Circle.JPG

Note that the radius of the sphere is ≈ 10 (At least as close to 10 as I could make it!!)

Maybe I'm reading too much into this problem, but I claim that we don't actually know that BC is at a right angle to AB

I guess my point is that we could have a right triangle, but as you see, it might not have to be. Note that ∠BAC is greater than a right angle

In hindsight, the problem only "works" if we assume that AB is perpendicular to BC.......and since we're not given any other information, then DavidQD and Alan would certainly be correct !!!

CPhill:

kayla121497 wrote:point A is the center of the sphere. point C is on the surface of the sphere. point B is the center of the circle that lies in plane P and includes point C. the radius of the circle is 12mm. AB= 5 mm. What is the volume of the sphere to the nearest cubic mm?

Mmmmm....I don't know about this.....here's a cross-section of a sphere where A is the center, C lies on the sphere's surface and AB =5, BC = 12 and BC lies in Plane "P"
(I think all of these meet the requirements of the problem).

Circle.JPG

Note that the radius of the sphere is ≈ 10 (At least as close to 10 as I could make it!!)

Maybe I'm reading too much into this problem, but I claim that we don't actually know that BC is at a right angle to AB

I guess my point is that we could have a right triangle, but as you see, it might not have to be. Note that ∠BAC is greater than a right angle

In hindsight, the problem only "works" if we assume that AB is perpendicular to BC.......and since we're not given any other information, then DavidQD and Alan would certainly be correct !!!

You need to complete the circle centred on B. You can see from the picture below (drawn to scale in Geogebra) that a circle based on BC as radius (as constructed by CPhill above) would not be wholly on the sphere. (BE is the extension of BC such that CE is the diameter of the circle)
sphere2.PNG

Ok Chris, you were thinking the same as me
but why do you say

Quote:

the problem only "works" if we assume that AB is perpendicular to BC.

If AB is not perpendicular to BC don't we just need more information? That puts us straight back to the answer you and I gave in the first place. - or am I missing something?

Melody:

...

. - or am I missing something?

I think so! See my reply immediately above yours.

A closer reading of the original shows that the circle is not specified to be wholly on the sphere, in which case Melody and CPhill are correct that the problem is indeterminate without further information. However, I doubt that the problem would have been set without the specification that the circle lies completely on the sphere.

Quote:

Alan: A closer reading of the original shows that the circle is not specified to be wholly on the sphere, in which case Melody and CPhill are correct that the problem is indeterminate without further information.

That is exactly what I was about to say. Thanks Alan.

Quote:

However, I doubt that the problem would have been set without the specification that the circle lies completely on the sphere.

Yes that sounds right too.

Not to wear this problem out, but one other thought occurred to me last night........consider the following:

Circle2.JPG

Note that AB = 5 and BC = 12. Thus, Plane "P" (containing B and C, with B being the center of a circle with radius 12 and with C lying on this circle ) would "cut" the sphere in half. What's more, we don't have any triangle at all !!!

Also, note that the radius of this sphere is just 7.

Thus......unless we know the orientation of BC, several interpretations are possible. (As Melody, Alan and DavidQD have pointed out ).

Oh well......I'll shut up now......I know when I've over-stayed my welcome.......

CPhill:

Maybe I'm reading too much into this problem, but I claim that we don't actually know that BC is at a right angle to AB
I guess my point is that we could have a right triangle, but as you see, it might not have to be. Note that ∠BAC is greater than a right angle
In hindsight, the problem only "works" if we assume that AB is perpendicular to BC.......and since we're not given any other information, then DavidQD and Alan would certainly be correct !!!

[center] Definition and Clarification[/center]

By definition a spherical segment is the “cutting” of a sphere with a pair of parallel planes. In the question, one plane is specified and passes through point (B) and the other is implied and passes through point (A).

On a sphere, the “Z” axis divides the center of the segment(s) from point (P) to the antipodal point (-P), where the end points for both lines lie on the surface of the sphere.

For clarity, define (P ₀) and (-P ₀) as the implied plane passing through Point (A), the center of the sphere. Define (P ₁) and (-P ₁) as the explicitly described plane passing through point (B), the center of the circle.

To restate: Points (A) and (B) are on the common Z-axis, which is the center of the two parallel planes, segmenting the sphere. Because the planes (lines) are parallel, the AB segment of the Z-axis is perpendicular to both lines. This explicitly defines a 90-degree angle as one of the angles to the triangle.

CPhill’s graphic depicts the intersecting point on the Z-axis, and not the angle at which the plane intersects the Z-axis.

One method to help visualize this: look at the sphere’s presents as “creating” the circle(s) in the plane(s). The concentric circles are always created in planes parallel to the plane that intersects the center of the sphere.
~~D~~

DavidQD:

...
By definition a spherical segment is the “cutting” of a sphere with a pair of parallel planes. In the question, one plane is specified and passes through point (B) and the other is implied and passes through point (A).
...
~~D~~

Unfortunately, the original question doesn't contain the words "spherical segment", nor any other words that ensure the circle is entirely on the surface of the sphere, so it is possible to interpret it as an incomplete problem with an infinite number of solutions, depending on the assumptions made! However, it seems clear to me that the intention is, indeed, for the assumption to be that the circle lies completely on the surface of the sphere, if only because we get the very nice 5, 12, 13 right-angled triangle that way!

Alan:

DavidQD:

...
By definition a spherical segment is the “cutting” of a sphere with a pair of parallel planes. In the question, one plane is specified and passes through point (B) and the other is implied and passes through point (A).

Unfortunately, the original question doesn't contain the words "spherical segment", nor any other words that ensure the circle is entirely on the surface of the sphere, so it is possible to interpret it as an incomplete problem with an infinite number of solutions, depending on the assumptions made! ...

Alan, you are correct. It does not. They were assumptions.

Kudos to Melody and Cphill for thinking outside the box (sphere in this case).

Geometry is a math branch full of postulates --that is where claims or assumptions are treated as true as a basis for reasoning or arguing. The postulates are justified when a solution is demonstrated, and the counterpoints are often dismissed when they do not (produce a solution).

Rigorous proof is a staple of mathematics, and one should never approach it lightly. Consider the question (often asked with juvenile humor in this forum): What does 1+1 equal?

“Everyone” knows the answer is 2, and most would answer accordingly, without thought for the reasons why this is true or considering the proof. (Most “normal” persons wouldn’t).

Whitehead and Russell, two (not so normal) mathematicians, considered the reasons to be of paramount importance, and transcribed the proof for “1+1=2” in minute detail in over 360 pages of equations.

(Reproduced here: The final page for the proof –pg. 362 in this edition of “Principia Mathematica”).



If an apparent trivial equation is that important then all equations are important and require more than casual thought. So, thank you for reminding me to always think outside of any container I might be in at the moment, whether it is a box, sphere, or the limitations of my own brain.

Note to CPhil: Never “shut up” about science or math, else some of us might still think there are 4 elements -- instead of the 5 that really exist. 

Again, Kudos to Melody and Cphill.

~~D~~