integrationline-integralspolar coordinates
I am trying to calculate the line integral $\int_C (x^2+y^2) \,ds$ on
$r=e^\theta$ over the interval $\theta=[0,\pi]$.
$\int_C (x^2+y^2) \,ds = \int_C r^2 \,ds$ leads me to $\int_0^\pi e^{3t} \,dt = \frac{1}{3} e^{3t} \Big|_0^\pi = \frac{1}{3}\big( e^{3\pi}-e^0 \big)\ldots$
The correct answer is $\dfrac{\sqrt{2} \cdot e^{3t}}{3}$. What am I doing wrong?
Not a bad problem on computer-assisted multivariable integration.
Since $f(-x,y)=f(x,y)$, we can ignore parts of the domain that are symmetric about the $y$-axis. For example, the integral could just as well be $$\int_{-3}^{3} \int_{x/3 + 2}^{3 } f(x,y)\,\mathrm{d}y\,\mathrm{d}x $$ without $+\sqrt{9-x^2}$ in the upper limit. This does not change it fundamentally, though: we still need a numeric integration routine to arrive at $\approx -2.01$ (which is correct). And for the polar coordinates it's probably better to keep the semicircular shape as is.
The polar integral that you set up evaluates to $$\int_{\pi/4}^{3\pi/4} \int_{r_1}^{r_2} r \cos\theta \, \mathrm{d}r \, \mathrm{d}\theta \approx -0.81$$ which is not the number you gave.
But the discrepancy between 1 and 2 remains, why? Because the polar integral does not include a triangular bit of $R$ in the bottom left corner. This is the triangle between $x=-3$, $y=x/2+3$, and $\theta=3\pi/4$. In polar terms, this piece contributes $$\int_{3\pi/4}^{\pi-\arctan (1/3)} \int_{r_1}^{-3/\cos\theta} r \cos\theta \, \mathrm{d}r \, \mathrm{d}\theta \approx -1.20$$ which reconciles the computations.
From the iterate integral, we know that the region is $x=\sqrt{2y-y^2}$ with some rearranging one can get $x^2+y^2-2y+1-1=0$ which is $x^2+(y-1)^2=1$. Equivalently in polar it translates to $r^2=2r\sin(\theta) \iff r=2\sin\theta$. The iterated integral also has a constriction on y: $0<y<1$. From the figure below, its obvious that this doesn't cover the whole circle. Therefore, we have to find the angle:
We know the circle is of origin (0,1)
$\tan(\theta)=1 \implies \theta=\frac{\pi}{4}$
So, $$\int_0^{\pi/4} \int_0^{2\sin\theta}r(1-r^2)drd\theta$$
Best Answer
\begin{align} (x,y) & = (e^\theta \cos \theta, e^\theta \sin \theta)\\[8pt] (x',y') & = e^\theta (\cos \theta - \sin \theta, \cos\theta + \sin\theta) \ d\theta\\[8pt] ds & = \|(x',y')\| = e^\theta \sqrt 2 \ d\theta \end{align}
Or, staying in polar
arc length in polar coordinates
$$ds = \sqrt {r^2 + r'^2} \, d\theta$$