Average y coordinate of points on parametrized semicircle

[fahraynk]

Homework Statement

Find the average y coordinate of the points on the semicircle parametrized by C:[0,$\pi$]-->$R^3$,
$\theta$-->(0, a*sin$\theta$, a*cos$\theta$); a>0

Homework Equations

The Attempt at a Solution

I think the answer should be an integral of the circle in the y direction to get the total y path length divided by the total path length of the entire parametrized path. I am not sure if I did this right though. The question is kind of weird to me. The book gives the answer which is 2a/$\pi$ but no explanation.

$$\frac{ \int_0^\pi \sqrt{a^2-y^2} dy } { \int_0^\pi ||f'|| dt =\int_0^pi \sqrt{a^2(cos^2\theta + sin^2\theta)} dt = a\pi} $$
solving the numerator:

$$\int_0^\pi \sqrt{a^2-y^2} dy = \int_0^\pi a\sqrt{1-(y/a)^2} dy \\

cos\theta = \sqrt{1-(y/a)^2}\\
sin\theta = \frac{y}{a}\\
acos\theta d\theta = dy\\
\theta = arcsin(\frac{y}{a})\\
\int_0^\pi a\sqrt{1-(y/a)^2} dy = a^2 \int_0^\pi cos^2\theta d\theta = a^2 \int_0^\pi \frac{1+cos2\theta}{2} d\theta = a^2[\frac{\theta}{2}+\frac{sin(2\theta)}{4}]_0^\pi\\
=a^2\frac{\pi}{2}
$$
Plug the answer for numerator back into equation :
$$
\frac{a^2\frac{\pi}{2}}{a\pi}=\frac{a}{2}
$$

The book says the answer is $$\frac{2a}{\pi}$$

So... anyone have a clue what I am doing wrong? Also, I was wondering if I was right to take the integral from 0 to pi after I did a trig sub or if that was invalid and the constants of integration had to be changed. Arcsin(y/a) is kind of meaningless for the answer I got so I just plugged in the same constants of integration because it looked right.
I'm not really sure if what I am doing right or what I am doing wrong here, so any help is appretiated.

 

[Ray Vickson]

Homework Statement

Find the average y coordinate of the points on the semicircle parametrized by C:[0,$\pi$]-->$R^3$,
$\theta$-->(0, a*sin$\theta$, a*cos$\theta$); a>0

Homework Equations

The Attempt at a Solution

I think the answer should be an integral of the circle in the y direction to get the total y path length divided by the total path length of the entire parametrized path. I am not sure if I did this right though. The question is kind of weird to me. The book gives the answer which is 2a/$\pi$ but no explanation.

$$\frac{ \int_0^\pi \sqrt{a^2-y^2} dy } { \int_0^\pi ||f'|| dt =\int_0^pi \sqrt{a^2(cos^2\theta + sin^2\theta)} dt = a\pi} $$
solving the numerator:

$$\int_0^\pi \sqrt{a^2-y^2} dy = \int_0^\pi a\sqrt{1-(y/a)^2} dy \\

cos\theta = \sqrt{1-(y/a)^2}\\
sin\theta = \frac{y}{a}\\
acos\theta d\theta = dy\\
\theta = arcsin(\frac{y}{a})\\
\int_0^\pi a\sqrt{1-(y/a)^2} dy = a^2 \int_0^\pi cos^2\theta d\theta = a^2 \int_0^\pi \frac{1+cos2\theta}{2} d\theta = a^2[\frac{\theta}{2}+\frac{sin(2\theta)}{4}]_0^\pi\\
=a^2\frac{\pi}{2}
$$
Plug the answer for numerator back into equation :
$$
\frac{a^2\frac{\pi}{2}}{a\pi}=\frac{a}{2}
$$

The book says the answer is $$\frac{2a}{\pi}$$

So... anyone have a clue what I am doing wrong? Also, I was wondering if I was right to take the integral from 0 to pi after I did a trig sub or if that was invalid and the constants of integration had to be changed. Arcsin(y/a) is kind of meaningless for the answer I got so I just plugged in the same constants of integration because it looked right.
I'm not really sure if what I am doing right or what I am doing wrong here, so any help is appretiated.

For a function $f(\theta)$ on $[0,\pi]$ the average is
$$\bar{f} = \frac{1}{\pi} \int_0^{\pi} f(\theta) \, d \theta$$

 

[fahraynk]

For a function $f(\theta)$ on $[0,\pi]$ the average is
$$\bar{f} = \frac{1}{\pi} \int_0^{\pi} f(\theta) \, d \theta$$

What is $f(\theta)$ here?
The circle's equation is $Z^2+Y^2=a$, I tried F = Z = $\sqrt{a^2-Y^2}$ and got :
$\int_0^\pi \sqrt{a^2-Y^2} d\theta = \int_0^\pi a\sqrt{1-sin^2\theta} d\theta = \int_0^\pi acos\theta d\theta =asin\theta |_0^\pi = 0$

Then I tried $F=Y=\sqrt{a^2-Z^2}$ and it gave the right answer. Can you explain why $f(\theta)$ is the equation in terms of Y? Also, since the length of a circle is $2\pi a$ why isn't the formula $$\bar{f} = \frac{1}{a\pi} \int_1^\pi f(\theta) d\theta$$

 

[LCKurtz]

What is $f(\theta)$ here?

$f(\theta) = y = a\sin\theta$

The circle's equation is $Z^2+Y^2=a$, I tried F = Z = $\sqrt{a^2-Y^2}$ and got :
$\int_0^\pi \sqrt{a^2-Y^2} d\theta = \int_0^\pi a\sqrt{1-sin^2\theta} d\theta = \int_0^\pi acos\theta d\theta =asin\theta |_0^\pi = 0$

You have calculated the average of $z = a\cos\theta$. From the symmetry you would expect that to be $0$.

Then I tried $F=Y=\sqrt{a^2-Z^2}$ and it gave the right answer. Can you explain why $f(\theta)$ is the equation in terms of Y? Also, since the length of a circle is $2\pi a$ why isn't the formula $$\bar{f} = \frac{1}{a\pi} \int_1^\pi f(\theta) d\theta$$

It doesn't have anything to do with arc length.

 

[Ray Vickson]

What is $f(\theta)$ here?
The circle's equation is $Z^2+Y^2=a$, I tried F = Z = $\sqrt{a^2-Y^2}$ and got :
$\int_0^\pi \sqrt{a^2-Y^2} d\theta = \int_0^\pi a\sqrt{1-sin^2\theta} d\theta = \int_0^\pi acos\theta d\theta =asin\theta |_0^\pi = 0$

Then I tried $F=Y=\sqrt{a^2-Z^2}$ and it gave the right answer. Can you explain why $f(\theta)$ is the equation in terms of Y? Also, since the length of a circle is $2\pi a$ why isn't the formula $$\bar{f} = \frac{1}{a\pi} \int_1^\pi f(\theta) d\theta$$

You asked for the average of $y$ and you told us that $y = a \sin \theta$, so that is $f(\theta)$. You told us that $\theta \in [0, \pi]$ so that is the range over which the average is taken.

 

[fahraynk]

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