What is the Solution for Finding c in the Integral of ArcCos Problem?

[XJellieBX]

Homework Statement

The region between y=cos x and the x-axis for x \in [0, \pi/2] is divided into two subregions of equal area by a line y=c. Find c.

2. The attempt at a solution
First I drew a graph of the region bounded between the function and the x-axis in [0, \pi/2]. Next I found the total area of the region:
A_{1,2} = \int^{\pi/2}_{0} cos x dx = sin x |^{\pi/2}_{0} = 1

Dividing the total area by 2, gives the area of each portion divided by y=c.

A_{1} = 1/2 = \int^{c}_{0} cos^{-1} (y) dy = ycos^{-1} (y) - \sqrt{1-y^{2}} |^{c}_{0}
= c (cos^{-1} (c)) - \sqrt{1-c^{2}} + 1

--> c (cos^{-1} (c)) - \sqrt{1-c^{2}} = -1/2

I'm not sure how to solve for c from the equation above. Any input will be appreciated.

 

[gabbagabbahey]

--> c (cos^{-1} (c)) - \sqrt{1-c^{2}} = -1/2

I'm not sure how to solve for c from the equation above. Any input will be appreciated.

I don't think think an exact solution is possible. Try estimating the solution using Newtons Method or any other numerical technique you have learned.

 

[Mark44]

You're not going to be able to solve for an exact value of c using algebraic techniques. The best you can do is to get an approximate value. A simple way to do this is to start with an educated guess, like, say c = 0.4. (The value has to be less than 1/2 because there is more area in the lower part of the region than in the upper part.)

You want to find a value c so that c*cos^(-1)(c) = sqrt(1 - c^2) + 1/2 = 0. c = .4 probably isn't right, but it will give you an idea of what to use for your next try. Keep doing this, with better and better approximations until your successive approximations are in agreement in two, three, or four decimal places, or however precise you want to be.

 

[XJellieBX]

I had thought it would be just approximating. Thank you =)