f773aa…: “Inside of a frowning trapezoid”

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Let $C$ be the circle of radius $3$ centered at the origin, i.e., $C = \{ (x,y) \in \R ^2 : x^2 + y^2 = 3^2 \}$ and assume $C$ is oriented in the counterclockwise direction.

Further let $H$ be the “northern semicircle” meaning $H = \{ (x,y) \in C : y \geq 0 \}$ oriented compatibly with $C$ .

Compute $\displaystyle \oint _C \left ( 2x\d x - 2x \d y\right ) = \answer {-18\pi }$

Let $D$ be the disk of radius $3$ centered at the origin, i.e., $D = \{ (x,y) \in \R ^2 : x^2 + y^2 \leq 3^2 \}.$

Then apply Green’s theorem to deduce $\oint _C \left ( 2x\d x - 2x \d y\right ) = \iint _D - 2 \d x \d y$ which is $-2$ times the area of $D$ .

The area of $D$ is $\pi \cdot 3^2$ .

So the integral equals $-2 \cdot \pi \cdot 3^2 = -18\pi$ .

Compute $\displaystyle \oint _C \left ( (1+2y)\d x + (2x-2y)\d y\right ) = \answer {0}$ .

Note that the field is $\vector {1+2y,2x-2y}$ .

In other words, if $F : \R ^2 \to \R$ is the function $F(x,y) = 2xy + x - y^2$ , then $\grad F = \vector {1+2y, 2x-2y}$ .

So by the fundamental theorem of line integrals, this integral vanishes.

$\displaystyle \int _H \left ( (1+2y)\d x + (2x-2y)\d y\right ) = \answer {-6}$

Note that the field is $\vector {1+2y,2x-2y}$ .

In other words, if $F : \R ^2 \to \R$ is the function $F(x,y) = 2xy + x - y^2$ , then $\grad F = \vector {1+2y, 2x-2y}$ .

So by the fundamental theorem of line integrals, this integral equals the difference of the potential function $F$ at the endpoints.

The endpoints $\partial H$ consist of the points $(3,0)$ and $(-3,0)$ .

So the integral equals $F(-3,0) - F(3,0) = -6$ .

$\displaystyle \int _H \left ( 2x\d x - 2x \d y\right ) = \answer {-9\pi }$

Since $H$ doesn’t bound a region, we can’t apply Green’s theorem directly.

One method to set $\vec {p}(t) = \left ( 3 \cos t, 3 \sin t \right )$ which, for $0 \leq t \leq \pi$ , traces out $H$ . Then the given integral can be appropriately transformed, e.g., $dx = -3 \sin t \d t$ and so forth.

There are other methods, though: consider the curve $\ell$ which is the straight line segment from $(3,0)$ to $(-3,0)$ . Then $H - \ell$ bounds the upper half-disk, with area $\frac {9}{2}\pi$ .

Note that $\int _\ell \left ( 2x\d x - 2x \d y\right ) = \int _{x=3}^{-3} 2 x \d x = 0$ .

Consequently the given integral equals the circulation around $H - \ell$ , which, bounding the upper half-disk, is, by Green’s theorem, equal to $-2 \cdot \frac {9}{2}\pi$ .