선적분

Solution 1:

We are being asked to evaluate the line integral $\int_C\mathbf{F}\cdot\mathrm{d}\mathbf{r}$ with $C$ being the specified semi-circle and $\mathbf{F}=xy\,\hat{\boldsymbol{\jmath}}$. As ${ \boldsymbol{\nabla} }\times\mathbf{F}\ne\mathbf{0}$, the vector field $\mathbf{F}$ is not conservative. So we'll evaluate the integral directly. First, using the figure,

we parametrize $C$ by

$$\mathbf{r}(\theta) = \big(x(\theta)\,,\,y(\theta)\big) =(1-\cos\theta)\,\hat{\boldsymbol{\imath}} +\sin\theta\,\hat{\boldsymbol{\jmath}}\qquad 0\le\theta\le\pi$$

So the integral is

$$\begin{aligned} \int_C xy\,\mathrm{d}y =\int_0^\pi x(\theta)\,y(\theta)\, y'(\theta)\ \mathrm{d}\theta =\int_0^\pi (1-\cos\theta)\,\sin\theta\,\cos\theta \ \mathrm{d}\theta \end{aligned}$$

Making the substitution $u=\cos\theta$, $\mathrm{d}u =-\sin\theta\,\mathrm{d}\theta$, $u(0)=1$, $u(\pi)=-1$,

$$\begin{aligned} \int_C xy\,\mathrm{d}y =\int_1^{-1} (1-u)\, u\, (-\mathrm{d}u) =\int_{-1}^1 (u-u^2)\,\mathrm{d}u =-2 \int_0^1 u^2\,\mathrm{d}u =-2 \frac{1^3}{3}=-\frac{2}{3} \end{aligned}$$

Solution 2:

We can write $x$ in terms of $y$ over $C$ in two pieces:

• Let $C_1$ be the quarter-circle $x=1-\sqrt{1-y^2}$ as $y$ goes from 0 to 1, and
• Let $C_2$ be the quarter-circle $x=1+\sqrt{1-y^2}$ as $y$ goes from 1 to 0.

Then:

$$\begin{aligned} \int_C xy\mathrm{d}y &=\int_{C_1} xy\mathrm{d}y+\int_{c_2}xy\mathrm{d}y\\ &=\int_0^1\left(1-\sqrt{1-y^2}\right)y\,\mathrm{d}y+\int_1^0\left(1+\sqrt{1-y^2}\right)y\,\mathrm{d}y\\ &=\int_0^1 y\,\mathrm{d}y-\int_0^1y\sqrt{1-y^2}\,\mathrm{d}y+\int_1^0y\,\mathrm{d}y+\int_1^0 y\sqrt{1-y^2}\,\mathrm{d}y\\ &=-2\int_0^1 y\sqrt{1-y^2}\,\mathrm{d}y \\ \text{Using the substitution $u=1-y^2$, $\mathrm{d}u=-2y\,\mathrm{d}y$:} \\ &=\int_1^0u^{1/2}\mathrm{d}u=-\frac23 \end{aligned}$$