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$$\int_C x^2dS$$ where C: $$x^2+y^2=r^2$$ So, $$ x=r\cos(\theta) $$ $$ y=r\sin(\theta) $$ $$ 0 \leq \theta \leq \pi $$ How would arc length of this curve go? $$ dS = \sqrt{\left(\frac{dr}{d\theta}\right)^2+r^2}d\theta $$ What should I put in $\left(\frac{dr}{d\theta}\right)^2$ and $r^2$?

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1 Answer 1

up vote 1 down vote accepted

$r$ is constant and $> 0$ in a circle. Therefore:

$ \frac{dr}{d\theta}=0 $

And

$ ds = \sqrt{(0)^2+(r)^2} d\theta = r d\theta $

From there, your integral should be straightforward.


Another way to look at it:

\begin{align} ds &= \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \ d\theta \\ \frac{dx}{d\theta} &= -r\sin{\theta} \\ \frac{dy}{d\theta} &= r\cos{\theta} \\ \Rightarrow ds &= \sqrt{(-r\sin{\theta})^2 + (r\cos{\theta})^2}d\theta = r d\theta \\ \end{align}

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The solution I got is $\frac{r^3\pi}{2}$. –  aarnes Mar 4 '12 at 16:00

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