# Homework Help: Minimizing area of a semicircle bounded by two lines

1. Feb 8, 2014

### californicate

1. The problem statement, all variables and given/known data
The figure shows a semicircle with radius 1, horizontal diameter , and tangent lines at
and . At what height above the diameter should the horizontal line be placed so as to minimize
http://imgur.com/grrCqWF

2. Relevant equations
The equation of a circle, x2 + y2 = a2

3. The attempt at a solution
I labeled the horizontal line y=a, and chose to express the area in terms of y, with the origin at the midpoint between P and Q. So A = 0a √(1-y2)dy + ∫?. My problem is currently the second half of the integral as I'm evaluating the area above the function and below some other line, not above the x-axis. Should I be integrating in terms of x? Any help is appreciated, thanks.

2. Feb 8, 2014

3. Feb 8, 2014

### LCKurtz

You can work the area in the right half of the picture and double it for the total area. Where did you get the $\sqrt{1-y^2}$ from. The radius of the circle is $a$. If you use a $dy$ element your integrand would be $x_{right} - x_{left}$ and if you use a $dx$ integral the integrand would be $y_{upper}-y_{lower}$, each with appropriate limits. You can use either for this problem, but you need to try again.

4. Feb 8, 2014

### californicate

It states in the question the radius of the circle is 1, not a, and so x^2 + y^2 =1 and so x=sqrt(1-y^2)

5. Feb 8, 2014

### haruspex

For the region above the line, what is the range of y? For any given y in that range, what is the value of x on the left half of the circle?

6. Feb 8, 2014

### californicate

So the range of y above the line is from a(the line) to 1

The value of x is -√1-y2, if it's to the left of the origin at the midpoint.

If I integrate in terms of x, say, I have the integral from 0 to some point of (equation of circle - a) + the integral from that same point to 1 of (a-equation). However, I don't know where that point of intersection is. Does this matter? It's really from where to where I should be integrating that's confusing me.

7. Feb 8, 2014

### LCKurtz

It is confusing for you to state in the "relevant equations" that the equation of the circle is $x^2+y^2= a^2$ when it is $x^2+y^2= 1$ and then to use $y=a$ for the equation of the horizontal line.

Now that we have that straight, let's call the regions "left of circle", "above circle", and "right of circle" so we know which one we are talking about.

Pick one of the regions, tell us which region and whether you plan to use a $dy$ or $dx$ integral. Then we can discuss the integrand and limits. In any case you need to know the $(x,y)$ coordinates where the line $y=a$ intersects the circle. So tell us that too.

8. Feb 8, 2014

### haruspex

It looked like you were trying to use y as the variable of integration, so if you don't mind I'll stick with that.
Yes, you have the range for y. If you take a horizontal slice dy thick at height y, the left hand end is at -√1-y2 (relative to the midpoint). Where is the right hand end? How long is the slice? What is its area?

9. Feb 9, 2014

### californicate

I'm thinking I should only consider the integral of one half and then double it, as it is symmetrical.

I know the total range of x is from 0 to 1. I'll have to integrate this in two parts as the upper and lower function alternate.

The first would be the integral from 0 to the meeting point of the circle and the line a (sqrt(1-x^2)-a). The second would be the integral from the meeting point to 1 of (a-sqrt(1-x^2)).

Since this meeting point is where a= sqrt(1-x^2) so a=1-x^2. So it's coordinates are (sqrt(1-a^2)), a. Is this how I should integrate it?

10. Feb 9, 2014

### haruspex

You're talking about a range of y, yes? That should not have an x in the formula. Why not just 0 to a?