Why is it the hard problems I can remeber how to do...but this one I can't for the life of me figure out.

Trying to figure this one out:




Now I know I can't use the area of a shaded area for an ellipse....but darn it what equation should I use.....

4/28/2009 12:00:01 PM

I found the shaded area. It's in the middle section of the semicircle

4/28/2009 12:00:42 PM

You have to integrate over Z, instead of r, in the example on this site:
http://www.vias.org/calculus/04_integration_04_09.html

^

4/28/2009 12:09:52 PM

nvm

[Edited on April 28, 2009 at 12:18 PM. Reason : V]

4/28/2009 12:10:19 PM

Or just use the area swept out by an arc (Pi*r^2*(angle/360)), and then add the area of the two triangles left over.

[Edited on April 28, 2009 at 12:14 PM. Reason : no integration necessary]

4/28/2009 12:13:47 PM

cut a piece of pizza out of the blue section, so you're left with 1 triangle on each side and a pizza slice in the middle. should be some fairly simple equations after that.
the pizza slice i think will be 1/8th of the area of the total circle (or 1/4th the area of the semi-circle) --- i haven't proven this one. just kinda guessed. (edit: ^ there's the equation right there)
the triangles left over are two right triangles with base 3 and hypotenuse 12. 1/2*bh is 18 for each triangle.

[Edited on April 28, 2009 at 12:17 PM. Reason : ]

4/28/2009 12:15:23 PM

Z is centered...so that integral would work....

4/28/2009 12:16:37 PM

1. find a perfectly circular cylinder.
2. ensure it's perfectly flat on the face
3. draw a line indicating the diameter, put a point at the center
4. use a protractor to draw the 90° angles of at the right distances, as the picture shows
5. cut the cylinder so that the face looks like the picture
6. weigh the remaining piece
7. slice into the cylinder and cut the shaded portion out and weigh it
8. use the same ratio of the weight and figure out the area

4/28/2009 12:21:20 PM

^fag

4/28/2009 12:22:19 PM

BURN!

4/28/2009 12:23:29 PM

Uhh, is it ok if I don't do it with geometry? My college brain doesn't think in terms of geometry anymore.

The equation for a circle is y=+/- sqrt(r^2-x^2)
so for this semi-circle it's y=sqrt(12^2-x^2)

you could drop an x,y axis on this biatch and integrate from -z/2 to z/2 (or just from 0 to z/2 and multiply the result by 2, because circles are symmetric about the y axis)

Now, my integration is a little shaky, so correct me if this is wrong but

2*int(sqrt(a^2-x*2),0,3) = 2*[x*sqrt(a^2 -x^2)]/2 + [((a^2)/2) *arcsin(x/a)] from 0 to 3

You can drop calculating it at 0, because that equals 0

so at 3, that corresponds to 35.6214039 * 2 is


71.2428078


...I think

4/28/2009 12:29:28 PM

The plane takes off.

4/28/2009 12:34:29 PM

lol... you do not need calculus for this

I would imagine there were two of those making a complete circle, then I'd find the area of the two segments, subtract it from the area of the circle, then divide by 2.....
Area of segment = 1/2 * r^2 * ( θ - sinθ )

4/28/2009 12:39:17 PM

why would you not use calculus to solve this.

Define a coordinate plane at the 'center' of the circle, feed the circle formula into a graphing calculator, integrate from zero to 3, multiply by two.

For reference:

Quote :

"When you want to represent a circle on a coordinate plane, you need to use the following equation: (x-h)^2 + (y-k)^2 = r^2, where h and k are the center points of the circle and r is the radius or the circle."

4/28/2009 1:00:35 PM

because 1/8*pi*r^2 + bh is easier

4/28/2009 1:10:40 PM

A = T*I - 89

4/28/2009 1:37:00 PM

^^ Why does that work?

4/28/2009 11:02:07 PM

^I get 128.55 for that and 74.55 when I integrate...

4/28/2009 11:47:36 PM

^ both of those are wrong

Astral Engine is correct

4/29/2009 12:03:12 AM

doesnt seem like you have enough info

does the radius bisect the z area?

4/29/2009 12:12:15 AM

Quote :

"Or just use the area swept out by an arc (Pi*r^2*(angle/360)), and then add the area of the two triangles left over. "

4/29/2009 12:12:42 AM

I could have done it wrong, but here's what I figure.

We know the length of Z, but not either of the sides. But we do have the trigonometric identities. And you can think of the circle as being like a trig circle, eg cos(60) = .5. So what that means is that the x axis is 0.5 units long if the angle is 60. We know that Z is 6, and since I can only assume it's in the dead middle of the circle, you really need 3. And the radius is 12, so you really need 3/12 = 0.25. Take the arccosine of that to get an angle of ~75.5. If you plug that number into sine, you get out ~0.97, which you then multiply by 12 to give you a height of ~11.6. Now, 75.5 isn't the angle that you want. You want to take the rest of that, 14.5, and multiply by 2. That gives you the angle of the shaded arc.

Then you just do what Shadowrunner said. Use that information to find the area of the arc, and add that to the area of the two triangles. I came out with roughly 71.24, which seems to agree with AstralEngine's calculations.

4/29/2009 12:18:27 AM