Calculating Surface Area

[Vlatko]

How do you calculate the surface area of a body which forms when you rotate the body on the picture around side a.

[Guv]

It looks to me as though formulae for cones and truncated cones are applicable here.

I have a vague memory about a general algorithm for calculating the area of a surface of revolution. Perhaps it uses an integral involving the length of the plane curve (or line segmnents) being rotated. The memory is vague, so it could easliy be a faulty interpretation of something else.

[paulw]

Integration through a rotation of Pi => -Pi is definitely the way to do it. I can't remember the details though

P.

[Starman]

Needs more information:

The points between c-b and b-d are not fixed.
We need an angle or at least one height to calculate the area.

Curved surface area of right circular cone =Pi*r*l where r is the raduis of the base and l is the slant height.

From the information given it is impossible to have a fixed value for r.

[Vlatko]

Well, there is no angle or height given. Only the four sides.

[Guv]

Starman: You are correct. Nice catch! I did not notice that the shape was indeterminate.

I was about to make nasty remarks about your intelligence, ancestry, and various facets of your personality. I remembered the formula for area of a cone as having a square root, and thought you had made a mistake.

When I looked it up, there was a formula in terms of the height and radius of the base, which requires a square root. It just so happens to be equivalent to your formula, since the slant height expressed in terms of radius and height requires a square root.

I have deleted the file with scathing comments about you.

[Starman]

Thank Guv (I think)

Here's one with a square root in it though

Maybe there's a general formula for the area of all the possible shapes that it could be ?


General formula for surface area of rotation:

S(area) = 2* Pi * integral [y sqrt (1 + (dy/dx)^2 ].dx

the notation is horrid as I can't do a long S.

Unfortunately I don’t understand it so maybe someone else could help from here.

[Guv]

Starman: Your integral formula for the area of a surface of rotation is understandable. Without a fancy Equation editor, it’s about the best we can do here.

Your formula looks tractable for some shapes and might not be much help for others. In general, there are potential problems if the Integrand is not a function of the variable of integration only. For your formula both y and dy/dx would have to be functions of x, except for cases where there is some trickery applicable.

For a shape like the one we are considering, your formula could be applied piecemeal to each of the line segments. For each line segment, y is a simple linear function of x, and the derivative is a constant (equal to the slope of the line). Looks like cake & pie if we had a bit more information about the figure.

My intuition is only right about 70-90% of the time for math and physics. This time it was wrong. For a surface of rotation, I thought there might be an integral which used the perimeter being rotated as the function being integrated. When you pointed out that the figure in question is indeterminate, this intuition was shown to be pretty bad. The fattest and thinnest possible shapes for the figure would have the same perimeter, but different surface areas.

BTW: Integrand is flagged as an error by my WordPerfect, with no alternative spellings. Does anybody know if this is a legitimate word? It could be that the WP spell checker does not have a good math vocabulary.

[Vlatko]

So guys, does anyone have a specific number for the surface area.

[Guv]

Vlatko: The area of the surface of revolution cannot be determined because the figure as described can have many different shapes.

This was mentioned in a post by Starman.

Look at the figure closely. Imagine keeping side a fixed and rotating side d. to make a bigger a-d angle. You could make sides b & c become a straight line, or you could make the figure convex at the b-c intersection. If you rotate side d to make the a-d angle smaller, the b-c angle also becomes smaller.

Suppose the four sides were equal in length. The figure could be a square or could be an almost flat parallelogram. The surface of rotation generated by the square would have more area that the surface generated by rotating an almost flat parallelogram.

[Vlatko]

the figure as described can have many different shapes.

I don't fully understand. The angles are fixed. They do not change. When i rotate the figure i get only one kind of a shape.

BTW, it has to be a solution. This is a task from a math highschool exam (17 year olds should sove this). I will ask someone very good in math to solve it and i will post the answer. It can't be as complex as you guys are saying, cos as i said it is for 17 year olds.

[Guv]

Vlatko: You are not paying attention. At least one angle must be known to specify a particular shape.

Go to the source of the problem and recheck to see if something has been omitted from the diagram you provided to us.

Study the figure carefully and reread my previous post.

There are theorems in geometry which prove that triangles are rigid figures, and that other simple closed polygon shapes are not rigid figures.

• If you know the length of three sides, you have specified a particular triangle.
  
• Knowing the length of all four sides of a quadrilateral does not specify a particular shape.
  
• Consider a 4-sided figure with all sides 10 inches long. It can be a square or a parallelogram. If you specify one angle as a right angle, you have a 10X10 square. If one angle is less than 90 degrees, you have a parallelogram.

I do not know what more we can say about this problem

There is not enough information to solve the problem. The problem is not difficult if you have some more information.

Given the way the figure looks, I believe that the formula for the surface area of a cone can be used to determine the area you want.

As drawn, sides c & d will form cones when rotated. You will need the heights of the b-c and b-d vertices (radii of the cones). Side b will form a truncated cone when rotated. You will need to extend side b until it intersects side a, and determine the length of that extension. Then you can subtract the surface area of two cones get the third surface area.

If you can work out the information described in the previous paragraph, you can solve the problem. The formula posted by Starman is applicable.

[Vlatko]

Thanks guys for all the effort.
It is not that i didn't pay attention, just i didn't understand some of the math terms you were using cos i haven't learned math in English. Anyway i now see what you mean. I doublechecked the body and that is all info, the 4 sides. I'll have to check if the test is correct.

Thanks again.

[parksie]

As an aside, look at the POV-Ray documentation. It gives full mathematical explanations for how it works with things like this.

[Vlatko]

What is POV-Ray documentation??

[parksie]

The documentation for POV-Ray

[Vlatko]

Argh! Let's try again. What is POV-Ray

[parksie]

Free raytracer: www.povray.org

[Vlatko]

Ok, thanks.