Compute $I := \int_Df(x, y)$ with $f(x, y) := 2y \sin (x)$ and $D$ rectangle with vertices $A := (0, 0)$, $B := (π, 0)$, $C := (π, 1)$ and $D := (0, 1)$.
Compute $I := \int_D \sqrt{x^2 + y^2}$ with $D :=$ {$(x, y) ∈ \mathbb R^2: x^2 + y^2 − 2x ≤ 0$}. Hint: Use polar coordinates.
Use the Green theorem to compute $\int_Γ(x^2 + y^2) dx + 2xy \ dy$ with $Γ$ being the boundary of the bounded domain delimiting the curve $y = x^4$ and $x = y^4$ traversed counterclockwise.
My attempts: 1. $$ \begin{align} I &:= \int_D 2y \sin (x) = \int_0^1 \int_0^\pi 2y \sin (x) dxdy \\ &= \int_0^1 [-2y \cos(x)]_0^\pi dy = \int_0^1 -2y \cos(\pi) + 2y \cos(0) dy \\ &= \int_0^1 -2y + 2y dy = \int_0^1 0 dy = c, \end{align} $$ where $c$ is a constant. Here I took $\pi = 3.1415$.... And not $\pi$ in radian. Is my answer correct?
- $\sqrt{x^2 + y^2} = r$ in polar coordinates. And $x^2 + y^2 - 2x = (x-1)^2 - y^2 - 1$ and so $D$ is the disc of center $(1,0)$ and radius less than or equal to $1$, i.e. $(x-1)^2 - y^2 - 1 \leq 0$ implies that $(x-1)^2 - y^2 \leq 1$. Hence, $I := \int_D \sqrt{x^2 + y^2} = \int_0^{2\pi} \int_0^{2 \cos(\theta)} r×r drd\theta$, where $r^2 - 2r \cos(\theta) \leq 0$, so $0 \leq r \leq 2 \cos(\theta)$
Then I compute the integral over these boundaries and I get $0$ as an answer. Are my boundaries for $\theta$ and $r$ correct? Or should $-2 \cos(\theta) \leq r \leq 2 \cos(\theta)$
- $$\begin{align} \int_Γ(x^2 + y^2) dx + 2xy dy &= \int_0^1 \int_{y^4} ^{y^{1/4}} \frac{\partial (2xy) }{\partial x} - \frac{\partial (x^2 + y^2) }{\partial y} dxdy \\ &= \int_0^1 \int_{y^4} ^{y^{1/4}} 2y - 2y dxdy \\ &= \int_0^1 \int_{y^4} ^{y^{1/4}} 0 dxdy = \int_0^1 c dy \end{align}$$ where $c$ is a constant. Finally, $\int_0^1 c dy = [cy] _0^1 = c$.
Is this correct?
