calculating the line integral

Question

It is a simple line integral, but I haven't really grasped the idea yet.

$$ \mbox{Calculate the integral}\quad\int_{L}x^{2}\,y\,d\ell \quad\mbox{where}\quad L \equiv \left\{\,x^{2} + y^{2} = R^{2}\ \mid\ y\geq 0\,\right\} $$ My attempt:

Form what is known above we get $y=\sqrt{\,R^{2} - x^{2}\,}$, $y \geq 0$, so $0 \leq x \leq R$.

So the integral we have to calculate is $\displaystyle\int_{0}^{R}x^{2}\,\sqrt{\,R^{2} - x^{2}\,}\,\,\mathrm{d}x$.

But I got stuck here. I also tried polar coordinates but got a wrong result. Any ideas ?.

Answer 1

Polar coordinates facilitates the evaluation of the line integral. We can write $x=R\cos(\phi)$, $y=R\sin(\phi)$, and $d\ell =R\,d\phi$.

Therefore, we have

$$\begin{align} \int_L x^2y\,d\ell &= \int_0^{\pi} R^4 \cos^2(\phi)\sin(\phi)\,d\phi\\\\ &=R^4\left.\left(-\frac13 \cos^3(\phi)\right)\right|_{\phi=0}^{\phi=\pi}\\\\ &=\frac23R^4 \end{align}$$

Answer 2

Hint (from where you are stuck):

$$\int x^2 \sqrt{R^2 - x^2}dx = \int \left((x^2 - R^2) + R^2\right)\sqrt{R^2 - x^2}dx $$

$$ = \int \left(x^2 - R^2\right) \sqrt{R^2 - x^2}dx +\int R^2 \sqrt{R^2 - x^2}dx =$$

$$ = -\int (R^2 - x^2)^{3/2}dx + \int R^2 \sqrt{R^2 - x^2}dx $$

and these integrals can be solved using the substitution $x = \sin(t), dx = \cos(t)dt$

Note that Mark Viola's answer is much easier.

Answer 3

You can of course apply the line integral's definition: the half circle can be parametrized as $\gamma(t)=(R \cos(t), R \sin(t))$, with $t \in [0,\pi]$, so the derived vector is $\gamma'(t)=(-R \sin(t),R \cos(t))$. Then $f(x,y)=x^2y$ $$ \int_{\gamma} f(x(t),y(t)) ||\gamma' (t)|| dt= $$ $$ \int_{0}^{\pi} R^3 \cos^2(t) \sin (t) \sqrt{R^2 \sin^2(t) + R^2 \cos^2(t)} dt = $$ $$ \int_{0}^{\pi} R^4 \cos^2 (t) \sin (t) dt $$ from now on the computation are the same as in Mark Viola's answer.