Question

A pie with a 4 inch radius is to be cut with 2 parallel lines into 3 pieces with equal area. Where should the cuts be made?

Answer 1

well u want to cut a pie in 3 so pi/3 units from the center to the right, and pi/3 units from the center to the left

Answer 2

Is this a calculus class problem?

Answer 3

Solve anyway you know how.

Answer 4

The equation of a circle is: \[
(x-h)^2+(y-k)^2=r^2
\]Where the center is \((h, k)\) and the radius is \(5\).
Let's center it at the origin and set the radius to to \(4\): \[
\begin{array}{rcl}
(x-(0))^2+(y-(0))^2&=&(4)^2 \\
x^2+y^2&=&16\\
y^2 &=& 16 - x^2 \\
\sqrt{y^2} &=& \sqrt{16 - x^2} \\
|y| &=& \sqrt{16 - x^2}
\end{array}
\]We'll assume \(y\) is positive.
That will give us the equation of the top semicircle:\[
y = \sqrt{16 - x^2}
\]

Answer 5

Notice the symmetry.
We want to find \(c\) such that: \[ \Large
\frac{1}{3}\int_{-4}^{4} ydx = \int_{x-c}^{x+c}ydx
\]Given \(y = \sqrt{16-x^2}\)

Answer 6

Fortunately, we know one of those integrals is going to be easy, since it's just half a circle. \[
\frac{1}{3}\int_{-4}^{4}ydx = \frac{1}{3}\left[ \frac{1}{2}\pi 4^2 \right]
\]

Answer 7

@Argos There, it's pretty much solved. You just need to do the calc.

Answer 8

Also since \(y\) is an even function, we can make one of the integrals easier: \[
\Large
\int_{x-c}^{x+c}ydx = 2\int_{0}^{x+c}ydx
\]

Answer 9

Putting it all together, we get: \[ \Large
2\int_{0}^{x+c}ydx = \frac{1}{3}\left[ \frac{1}{2}\pi 4^2 \right]
\]Solve for \(c\). The lines will be \(x = c\), \(x=-c\) where the origin is the center of the pie.

Answer 10

i mean just integrate

just integrate sqrt(16 - x^2) - pi/3 from -4 to 4 and you will obtain an area of 16 pi/3, which is the area of the shaded region, which is 1/3 of the total area of the circle.

so the cut line is pi/3