- #1
fahraynk
- 186
- 6
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.