# Homework Help: Area of a vertically sliced circle

1. May 30, 2010

### bur7ama1989

1. The problem statement, all variables and given/known data

A circle of radius 7 inches is sliced vertically, parallel to the y-axis, into three pieces. Each piece has an equal area. What is the width, x-axis, of each piece?

2. Relevant equations

f(x)= +/- sqrt((r^2)-(x^2)) where "r" is the radius.

3. The attempt at a solution

My sign for the integral is $Using the equation for a circle above I took advantage of the obvious symmetry about the x-axis to isolate the part of the circle above the x-axis relevant to the given radius. g(x)= sqrt((7^2)-(x^2)). My attempt was to first define the area as an integral from the origin until x=7. (I chose x=7 because of the symmetry about the y-axis.) I(x)= 0->7$sqrt((7^2)-(x^2))dx

Here is where I hit a wall. The prof tried explaining to the class about Pythagoras and I thought I understood but apparently I didn't. Any assistance or pushes in the right direction would be greatly appreciated.

2. May 30, 2010

### HallsofIvy

First the area of the circle is $\pi(7^2)= 49\pi$. If it is divided into 3 pieces of equal area, each piece must have area $49\pi/3$.

Taking X to be the x coordinate of the left most slice, its area is given by
$$2\int_{-7}^X \sqrt{49- x^2} dx$$
and that must be equal to $49\pi/3$ .

Go ahead and do the integral (the substitution $u= 7sin(t)$ will work), set it equal to $49\pi/3$, and solve for X. The symmetry tells you that the other slice is at 7+ X.

3. May 30, 2010

### ƒ(x)

Do you even need calculus? If you consider just the part that is above the y-axis, then each slice has to be equal to $49\pi/6$. and, the two end pieces are quarter ellipses.

so, if b is the x coordinate of the right slice, then

49pi/6 = (pi/4)(7-b)(49-b^2)^(1/2)

Last edited: May 30, 2010
4. May 30, 2010

### bur7ama1989

I do need to use calculus. I need to use trigonometric substitutions and integration. The method provided by HallsofIvy is the one i must use.

I followed through with the integral but now I am stuck at trying to solve for X.

What I have is:

sin^-1(x/7)+sin(2(sin^-1(x/7))) = pi/6

How do i break this down?

5. May 30, 2010

### tiny-tim

No, a cap of a circle is a sector minus a triangle, and you can easily find the area of each.