Proof of average height of half circle

[Jurgen M]

I need proof how find average height of half circle?

Lets say pressure distribution is half circle with Pmax = radius,I must find average/resultant pressure..

 

[Office_Shredder]

I have no idea what the average value of half a circle means conceptually, can you elaborate? Are you trying to compute an integral or something?

 

[Jurgen M]

I have no idea what the average value of half a circle means conceptually, can you elaborate? Are you trying to compute an integral or something?

I edit my post, is now clear?

 

[Mark44]

I edit my post, is now clear?

Slightly clearer, if I understand what you're trying to do. Assume for the moment that $P(x) = \sqrt{1 - x^2}$ gives the pressure. This is the upper half of a circle of radius 1, centered at the origin. The average value will be this:
$$\frac 1 {b - a}\int_a^b P(x) dx= \frac 1 2\int_{-1}^1 \sqrt{1 - x^2}~dx $$
My simple example function is a semicircle of radius 1 that is centered at the origin. You need a formula for your semicircle that you'll plug into the integral I wrote.

 

[Jurgen M]

Slightly clearer, if I understand what you're trying to do. Assume for the moment that $P(x) = \sqrt{1 - x^2}$ gives the pressure. This is the upper half of a circle of radius 1, centered at the origin. The average value will be this:
$$\frac 1 {b - a}\int_a^b P(x) dx= \frac 1 2\int_{-1}^1 \sqrt{1 - x^2}~dx $$
My simple example function is a semicircle of radius 1 that is centered at the origin. You need a formula for your semicircle that you'll plug into the integral I wrote.

I find average pressure will be area of semi circle divide by diameter . (1/2 x r2 x 3.14) / 2r
But how they find this, is your calculation proof for this?

 

[PeroK]

Slightly clearer, if I understand what you're trying to do. Assume for the moment that $P(x) = \sqrt{1 - x^2}$ gives the pressure. This is the upper half of a circle of radius 1, centered at the origin. The average value will be this:
$$\frac 1 {b - a}\int_a^b P(x) dx= \frac 1 2\int_{-1}^1 \sqrt{1 - x^2}~dx $$
My simple example function is a semicircle of radius 1 that is centered at the origin. You need a formula for your semicircle that you'll plug into the integral I wrote.

That integral gives the area of the semicircle, which perhaps we already know!

 

[jedishrfu]

It looks like you’re trying to rectangularize a half circle with one side being the diameter of the circle and the vertical size being the average value of the vertical vectors in your diagram.

Is that right?

if so it doesn’t really need calculus to figure it out. One needs to know only the area of the half circle and its diameter.

 

[Mark44]

That integral gives the area of the semicircle, which perhaps we already know!

The factor out front, $\frac 1 {b - a}$, together with the integral, gives the average value of the function inside the integral.

 

[Jurgen M]

It looks like you’re trying to rectangularize a half circle with one side being the diameter of the circle and the vertical size being the average value of the vertical vectors in your diagram.

Is that right?

if so it doesn’t really need calculus to figure it out. One needs to know only the area of the half circle and its diameter.

If find from pure logic that if I do this I will find average value...but don't know how to proof this with calculus..

 

[mathman]

Radius r. Width =$2r$, area=$\frac{\pi r^2}{2}$, average height =$\frac{\pi r}{4}$.

 

[Jurgen M]

Radius r. Width =$2r$, area=$\frac{\pi r^2}{2}$, average height =$\frac{\pi r}{4}$.

$\frac{\pi r}{4}$ is not proof

 

[Mark44]

Here's what you wrote in post #1:

I need proof how find average height of half circle?

Lets say pressure distribution is half circle with Pmax = radius,I must find average/resultant pressure..

It's not clear to me what it is you're trying to do. Are you supposed to
(1) prove/derive the formula for the average value of a function over an interval [a, b]? That formula is $\bar f = \frac 1 {b - a}\int_a^b f(x)dx$, which is what I wrote earlier.
Or (2), calculate the average pressure of a pressure distribution in the shape of a circle or radius Pmax?
If it's #2, the average pressure is $\frac{\pi P_{max}}4$.

 

[jim mcnamara]

I'm not sure what you want. But this question asking for proof is often asked lots of places.
Here is an example:
https://math.stackexchange.com/ques...rive-area-of-circle-using-n-triangles/1257663

Plus to me it seems like homework. Generally we want homework posted in homework forums.
Since @Mark44 has tried to help, let's get his opinion.

 

[Jurgen M]

Here's what you wrote in post #1:

It's not clear to me what it is you're trying to do. Are you supposed to
(1) prove/derive the formula for the average value of a function over an interval [a, b]? That formula is $\bar f = \frac 1 {b - a}\int_a^b f(x)dx$, which is what I wrote earlier.
Or (2), calculate the average pressure of a pressure distribution in the shape of a circle or radius Pmax?
If it's #2, the average pressure is $\frac{\pi P_{max}}4$.

Yes prove/derived how we find that average height of semi circle is 3.14x r /4
Can you explain your formula?
What is a to b, -1 to 1 ?

 

[Mark44]

Yes prove/derived how we find that average height of semi circle is 3.14x r /4

Your question is still not clear. Which one of the following two questions are you asking?

1. Do you want to derive the formula for the average value of an arbitrary function?
2. Do you want to calculate the average value of the upper half-circle of radius R, whose center is at (0, 0)?

If it's #1, the average value of f, $\bar f$, on an interval [a, b], is the area between the graph of f and the horizontal axis, divided by the length of the interval. To derive this formula, set up a Riemann sum with area increments and take the limit as the number of subintervals increases to infinity.
If it's #2, just use the formula that I wrote in post #12.

Can you explain your formula?
What is a to b, -1 to 1 ?

Yes, if the radius of the semicircle is 1, and the semicircle is centered at (0, 0).

 

[Jurgen M]

Your question is still not clear. Which one of the following two questions are you asking?

1. Do you want to derive the formula for the average value of an arbitrary function?
2. Do you want to calculate the average value of the upper half-circle of radius R, whose center is at (0, 0)?

2.

 

[Jurgen M]

That integral gives the area of the semicircle, which perhaps we already know!

So who has right, you or Mark44?

 

[Mark44]

2.

If it's #2, just use the formula that I wrote in post #12.

Assuming the semicircle is centered at (0, 0), with a radius of $P_{max}$, then the average pressure is $P_{ave} = \frac 1 {2P_{max}}\int_{-P_{max}}^{P_{max}} \sqrt{P_{max}^2 - x^2}~dx$.

So who has right, you or Mark44?

Both of us are right.
PeroK's comment was about an easy way to calculate the integral above. The integral divided by the length of the interval gives the average value.

Most calculus textbooks have a general formula for the average value of a function on some interval [a, b]; namely, $\bar f = \frac 1 {b - a}\int_a^b f(x)~dx$. This is what I used in the equation I wrote in the first paragraph.

 

[Jurgen M]

Assuming the semicircle is centered at (0, 0), with a radius of $P_{max}$, then the average pressure is $P_{ave} = \frac 1 {2P_{max}}\int_{-P_{max}}^{P_{max}} \sqrt{P_{max}^2 - x^2}~dx$.

Both of us are right.
PeroK's comment was about an easy way to calculate the integral above. The integral divided by the length of the interval gives the average value.

Most calculus textbooks have a general formula for the average value of a function on some interval [a, b]; namely, $\bar f = \frac 1 {b - a}\int_a^b f(x)~dx$. This is what I used in the equation I wrote in the first paragraph.

Thanks for answer.

 

[mathman]

$\frac{\pi r}{4}$ is not proof

Why not? Area=height x width.