Area of a vertically sliced circle

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Homework Help Overview

The problem involves a circle with a radius of 7 inches that is sliced vertically into three pieces of equal area. Participants are exploring how to determine the width of each piece along the x-axis using calculus and geometric reasoning.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of integrals to find the area of the circle and how to set up the equations for the slices. There is a mention of using trigonometric substitutions and the symmetry of the circle. Some question whether calculus is necessary for the solution.

Discussion Status

The discussion is active, with participants providing different approaches and questioning the necessity of calculus. Some guidance has been offered regarding setting up integrals and solving for the variable X, but no consensus has been reached on the method to proceed.

Contextual Notes

There is a mention of needing to use calculus and trigonometric substitutions, as well as constraints related to the area calculations and the symmetry of the circle. Participants are also considering the implications of dividing the area into equal parts.

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Homework Statement



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?

Homework Equations



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

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.
 
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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.
 
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:
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?
 
bur7ama1989 said:
I do need to use calculus.

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

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