How to solve this line integral?

Question

I have this line integral $\int_{C}(x^2+y^2)ds$, where $C=\{(x,y)|x^2+y^2=ay\}$. I tried solving it by substituting the variables as $x=r\cos{t}$ and $y=r\sin{t}$. So when I differentiate, I get $dx=-r\sin{t}$ and $dy=r\cos{t}$.

So popping this into the integral we get:

$\int_{C}((\cos{t})^2+(\sin{t})^2)*\sqrt{(-r\sin{t})^2+(r\cos{t})^2}dt$

I could probably integrate this if I knew how to apply the given circle.

Answer 1

If $x=r\cos t$ and $y=r\sin t$, then $x^2+y^2=r^2\ne ay$; besides, what is $r$?

Instead, since$$x^2+y^2=ay\iff x^2+\left(y-\frac a2\right)^2=\frac{a^2}4,$$you should take $x=\frac a2\cos t$ and $y=\frac a2+\frac a2\sin t$. And then you compute$$\int_0^{2\pi}\left(\frac{a^2}4\cos^2t+\frac{a^2}4(1+\sin t)^2\right)\left\|\left(-\frac a2\sin t,\frac a2\cos t\right)\right\|\,\mathrm dt.$$

Answer 2

An alternative parametrization comes from polar coordinates centered at the origin. As you can check (you can draw a picture and draw the right triangle to see this), we have the polar equation $r=a\sin\theta$, $0\le\theta\le\pi$, so you can parametrize by $$x=r\cos\theta = a\sin\theta\cos\theta, \quad y = r\sin\theta = a\sin^2\theta, \quad 0\le\theta\le\pi.$$ Given the integrand $x^2+y^2$, this would be my choice.

Answer 3

Your value for $\mathrm{d}s$ looks off. Here is a trigonometric approach.

$$ \begin{align} \oint\left(x^2+y^2\right)\mathrm{d}s &=\int_0^{2\pi}\left(\frac{a^2}2-\frac{a^2}2\cos(\theta)\right)\,\frac a2\,\mathrm{d}\theta\\ &=\frac{a^3}{4}\,2\pi\\[3pt] &=\frac{\pi a^3}2 \end{align} $$