Mean Value of Ordinates of Semicircle

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In summary, the function f(x)=\sqrt{a^2-x^2} (-a<= x <=a) is a semicircle with a center at the origin and a bounding diameter on the x-axis between points A(-a,0) and B(a,0). The mean value of the ordinates of this semicircle, or the average value of y, can be found by taking the integral of the function over the interval divided by the length of the interval. This results in a value of a\pi/4, which is also the area of half of a circle with radius a. This problem can be solved without using calculus by recognizing that the area of the semicircle is half the area of a
  • #1
John O' Meara
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The graph of the function f(x)=[tex]\sqrt{a^2-x^2}[/tex] (-a<= x <=a) is a semicircle, centre the origin and bounding diameter the portion of the x-axis between the points A(-a,0) and B(a,0). Show that the mean value with respect to x of the ordinates of the semicircle is [tex]\pi[/tex]a/4. ( This question can be done without using calculus.)
Does ordinates mean the x-coordinates only of the points?
My attempt, mean value =[tex]\frac{1}{b-a}\int_a^{b}f(x)dx = \frac{1}{2a}\int_{-a}^{a}(\sqrt{a^2-x^2})dx = \frac{1}{4a}\int_{-a}^{a}\ln({a^2-x^2})dx[/tex].
I'm interested to know how it can be done without using calculus? Thanks for the help.
 
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  • #2
In fact, it's the average value of y. So your first two integrals are just fine. Can't say the same about the third. Where did the wacky ln come from? Hmm. How to do this without calculus?? Don't know.
 
  • #3
The "average value" of a function, over some interval, is the integral of the function over that interval divided by the length. As Dick said, the "ordinal" is the y component, not the x.

As for doing it without calculus, isn't the integral of a function the area under the curve? What is the area here?
 
  • #4
To solve the problem try [tex] x=asin\theta[/tex], then [tex]\sqrt{a^2-x^2} =\sqrt{a^2-a^2sin^2\theta} = \sqrt{a^2(1-sin^2\theta)} = acos\theta[/tex]. Therefore [tex]\\ \frac{1}{2a}\int_{x=-a}^{x=a}\sqrt{a^2-x^2}dx =
\frac{1}{2a} \int_{x=-a}^{x=a}\sqrt{a^2-x^2}acos\theta d\theta \\[/tex]
[tex]= \frac{1}{2a} \int_{\theta=-\frac{\pi}{2}}^{\theta=\frac{\pi}{2}} \sqrt{a^2(1-sin^2\theta)}acos\theta d\theta \\ =
\frac{a}{2} \int{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2}(1+cos\theta) d\theta \\ = a\frac{\pi}{4}[/tex]. So I learned how to solve the problem using calculus.
 
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  • #5
Great! Now how would you do it without calculus? This is in "pre-calculus mathematics"
 
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  • #6
The area in question is 1/2 the area of a circle = [tex] \frac{1}{2}\pi \times r^2 [/tex] multiplied by [tex] \frac{1}{2r} \mbox{ to get the mean value, which =} a\pi/4 \mbox{ where a=r.}[/tex]
 

1. What is the formula for finding the mean value of ordinates of a semicircle?

The formula for finding the mean value of ordinates of a semicircle is (4/π) x radius.

2. How is the mean value of ordinates of a semicircle calculated?

The mean value of ordinates of a semicircle is calculated by finding the average of the y-values of all points on the semicircle.

3. Why is the mean value of ordinates of a semicircle important?

The mean value of ordinates of a semicircle is important because it represents the average distance from the center of the semicircle to its circumference. It is also used in various mathematical calculations and geometric proofs.

4. Can the mean value of ordinates of a semicircle be negative?

No, the mean value of ordinates of a semicircle cannot be negative as it represents a distance and distances cannot be negative.

5. How does the mean value of ordinates of a semicircle relate to its area?

The mean value of ordinates of a semicircle is directly proportional to its area. This means that as the mean value of ordinates increases, the area of the semicircle also increases. It can be calculated using the formula (π/4) x mean value of ordinates squared.

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