A little math help for the exam!

[MeKC]

Hello dear hive members.

The clock is now so many, and I really need to go too sleep. I have been up the whole night working on a exams project that needs to be finish in 8 hours.

I got stuck one place, because Im not a math genius, so I hoped some one here can explain me, or just write down the line that will calculate
the red area on the picture below.

I need to know how big the red area you see on the picture are, so please
help if you know how to. I will be a GREAT help for me.

Thank you for your time

 

[The Reborn Devil]

The answer is 63.617
Muahaha


Edit: I believe it is 33.3333 instead

 

[Deleted member 157129]

Ok, you know the radius of the bigger circle is 15 (the R15 on the picture is pointing at it, so I believe it is safe to assume that's the radius). You also know four straight lines, and if you look closely, you'll see the difference between them is 6, and that this difference is from the centre of the larger circle and out, in both horizontal and vertical direction (indicating that it is a circle and not an eclipse).

The length 6 we have found now is the cathetus (catheti) of a triangle, what we want is the hypotenuse. In order to find that in this kind of triangle (90 degrees corner) we add cathetus² with cathetus², the result is hypotenuse². Take the square root of that result and you've got the distance from the centre of the big circle to the centre (corner) of the little circle. This means that the hypotenuse is the excess part of the radius 15 we already know, so subtract it from it, and the result you got left is the radius of the little, red circle. Now, as the red circle is not full, you only want one fourth of it, you need to divide the full areal of the red circle (PI*R²) with 4. There's your answer.

 

[uberfoop]

That would be a fantastic solution, except that the red area is not a quarter circle.


I found ~37.6 as my answer, potentially not in the most efficient way possible though, seeing as how I had to use some whacky vector stuff thanks to my forgetting how to deal with chords efficiently.

Note that all values in here are truncated sketchily and so my answer and any values in here are possibly a tad rough and iffy, but should still provide a close answer.

Anyway, I first found the coordinates of the points where the red area terminates along the boundary of the circle by setting up the circle equation y=sqrt(15^2-x^2). I got -13.7477,6 and -6,13.7477. Extending from the origin, these form vectors of length 15. I used the arccos(a dot b/lallbl) formula to determine that the angle between the vectors was .74776 radians. Then I used the (r^2)*theta/2 area formula to calculate the area of the sector to be 84.123. This is where my chord forgetfulness comes in and I used the areas of two triangles to get my answer. I figured that the area of the sector minus the area of a triangle formed by the two vectors would give the area between the outside of the circle and a chord line drawn from the endpoints of the two vectors. The remaining area of the red zone would be a triangle with points -6,6 and the endpoints of the two vectors. Anyhow, the area of the vector triangle is just lalxlbl*sin(theta)/2, in this case (15^2)*sin(.74776)/2, which is about 76.5. 84.123-76.5 gives a chord slice area of 7.623. The second triangle has two sides of length of 7.7477 and is a right triangle, so its area is ~30.014. Add this to the chord slice area and you get an answer of about 37.6.


An alternative way to do this would be to integrate with the circle formula minus the line y=6 from -13.7477 to -6 (ie, integral from -13.7477 to -6 of sqrt(15^2-x^2)-6); on a graphing calculator it would give you a fast and accurate approximation/answer (depending on the calc). By hand it would involve insane trig sub madness. My TI-83 I have here says the area is 37.637 based on this method.

 

[HINDYhat]

Exact answer is:
-225*pi/4 - 54 + 225*arctan(5/2)

I did it using polar coordinate integration. uberfoop, when you've got to do trig substitution, it's often nicer to switch to polar coordinates. I did, and it was rather simple by hand.

I doubt it's anything that the op could understand though. To conclude, fuck precalculus.

 

[The Reborn Devil]

So it's 37.1006519141?
I wasn't that far off then

 

[uberfoop]

Exact answer is:
-225*pi/4 - 54 + 225*arctan(5/2)

I did it using polar coordinate integration. uberfoop, when you've got to do trig substitution, it's often nicer to switch to polar coordinates. I did, and it was rather simple by hand.

I'm kind of skeptical of this answer, given the cleanliness of it despite the irrationality of so many of the input values that I would have expected and the fact that it contradicts what I got several times through using multiple methods of calculation.



Whatever, though. The ultimate QFTage here is:

To conclude, fuck precalculus

 

[HINDYhat]

The irrationality of input values? The only irrationality comes from arctan(5/2) and pi, which happen to both be present at the beginning and end of my calculations. Do you want me to explain exactly how I did it?

EDIT: Heh, I think I made a little mistake. Nonetheless, the answer can still be expressed in simple terms. I'll update soon.

EDIT 2: Here we go:
225*pi/4 + 36 - 225*arcsin(2/5) - 18*sqrt(21) ~= 37.63693390

 

[Mulgrim]

math is only good when you're winning ..