GEOMETRY GURUS!!! Area of an Oval - SOLVE!

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Posted by jdjd on Nov-14-2003 06:19:

you shouldnt need calc for this

Posted by DJ Pudl on Nov-14-2003 06:27:

the problem is figuring out where one circle starts and the other stops. Since they need to be tangent to each other, how can you do this without calc?

Jonathan

Posted by DJ Pudl on Nov-14-2003 07:54:

all right, it's all written up, including a formula you can use to calculate the areas of any type of shape with this form, ie different radii and different width of the oval. I'll try to scan it in tomorrow.

The equation isn't pretty, however...

Jonathan

Posted by DJ Pudl on Nov-14-2003 21:29:

Solution

I've posted my solution as a pdf. Take a look at it, and double check my work to make sure you understand what's going on and maybe find any errors.

As a side note, once you find the point of intersection (ie the place where the slopes of the two circles is equal), you can split the top half of the oval into 3 sections and integrate each of the sections in turn. This works because the integral of a function is the area under the curve. You would then multiply your result by 2 to include the bottom half as well.

Solution

Good luck with everything,

Jonathan

Posted by cbxzcm on Nov-15-2003 00:19:

That problem is a piece of cake. Try working out these:

Posted by mezzir on Nov-15-2003 00:33:

Re: Solution

quote:
Originally posted by DJ Pudl
I've posted my solution as a pdf. Take a look at it, and double check my work to make sure you understand what's going on and maybe find any errors.

As a side note, once you find the point of intersection (ie the place where the slopes of the two circles is equal), you can split the top half of the oval into 3 sections and integrate each of the sections in turn. This works because the integral of a function is the area under the curve. You would then multiply your result by 2 to include the bottom half as well.

Solution

Good luck with everything,

Jonathan

shit mein!
that was like the original idea i had but it seemed like there must be a way w/o calc so i went with the sketchy basic geometry route
...i like my answer better

Posted by DJ Pudl on Nov-15-2003 00:48:

hehe, yeah, it certainly took a lot of work. There is probably a way to do it with geometry only, how did you calculate the area, by the way?

Posted by mezzir on Nov-15-2003 03:18:

found the area of the two pairs of sectors and added the inside rectangle
the one sketchy part is that i based it all on an assumption that pon further inspection i'm not sure was correct
tho that only should've altered it by like 20 max
huh

Posted by Photo_bot_2k1 on Nov-15-2003 03:24:

the answer is exactly
708.32425562

Posted by DJ Pudl on Nov-15-2003 08:23:

quote:
Originally posted by Photo_bot_2k1
the answer is exactly
708.32425562

any work or explanation to back that up?

Posted by ASOT100 on Nov-15-2003 11:14:

check the back of the book if it's an odd problem

lol jk

Posted by Smeagol on Nov-15-2003 19:56:

a geometrical solution

Think this should work:

I'm using the angle a the big circle uses to sweep from the leftmost point up to the point where the other circle takes over.

Now the width w can be determined as a function of a and known quantities:
w = 2 R1(1-cos(a)) + 2 R2(cos(a))

and thus

cos a = (w-2R1)/(2(R2-R1))

Now half of one of the large segments has an area A1:

A1 = aR1^2 / 2 - R1 R1sin(a)/2

as the sweeping area minus the triangle in the middle.
(I'm not very good at computerdrawing... would have helped )
Same for the smaler segment A2:

A2 = (pi/2-a)R2^2 / 2 - R2 R2 cos(a)/2

And the middle rectangle A3:

A3 = R1sin(a) R2cos(a)

A = 4A1 + 4A2 + 4A3 should now give the right total area.
Provided I didn't miss anything. )
Sorry for not calculating the actual value... I can not find my calculator...

Guess that this is more of a curiosity by now anyway.

**edited a factor two that i missed

Posted by Resnick on Nov-15-2003 23:31:

ok, first off, damn u for making me think on a saturday, second i didnt read the other solutions so i dunno if someone already solved it, and third im too lazy to solve it, but ill tell u how.

all u need to do is find the equations of the circles and the rest is easy. to do this, just imagine two circles, which is the top circle and the left circle in ur graph. now let the x axis be in the middle, and the y axis be in middle as well.

(x-36.251)^2 + y^2 = 1853.91 <---big circle
x^2 + (y-a)^2 = 5.029 <---little circle

note*: a is the distance from x axis to the centre of the little circle

now at the interection point, we can equate these to get two equations, since theres 3 unknown, we derive both and set them equal (cuz slopes are same at that point) and once we solve for a, we have eations and we done.

Posted by King_Mack on Nov-17-2003 17:26:

anyone confirm the actual area??

Posted by MERTON on Nov-17-2003 17:40:

who's the lady in king macks avvy?.. and this problem sucks!

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