Calculating Line Integral: $\int_{dA} (x^2 dx + y^2 dy)$

Question

Normally we are given points, but this time it's different:

$$\int_{dA} (x^2 dx + y^2 dy)$$

$A$ $=$ ${(x, y)}$ in $R^2;$ $-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$ , $ -1 \leq y \leq \cos(x)$

Calculating it with Green's theorem was doable but we should do it as a line integral too.

Answer 1

Split the integral into 4 parts.

1. $-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}$, $y(x) = \cos(x)$. Here, $dy=-\sin(x)dx$, $y^2=\cos^2(x)$ and $$-\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^2-\sin(x)\cos^2(x))dx = -\frac{\pi^3}{12}$$ (the minus sign before the integral comes from ccw orientation).
2. $-\frac{\pi}{2}\leq x\leq \frac{\pi}{2}$, $y(x) = -1$. Here, $dy=0$, and the integral becomes $$\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x^2dx = \frac{\pi^3}{12}$$
3. $-1 \leq y \leq 0$, $x(y) = -1$. Here, $dx = 0$, and $$-\int\limits_{-1}^0y^2dy = -{1\over 3}$$
4. $-1 \leq y \leq 0$, $x(y) = 1$. Here, again, $dx = 0$, and $$\int\limits_{-1}^0y^2dy = {1\over 3}$$

Summing the parts, we see that the integral is zero.

Now, using Green's theorem: $d(x^2dx+y^2dy) = 2x \cdot dx\land dx + 2y \cdot dy\land dy = 0$, so the integral is zero, as in our previous calculation.