(a) x ( t ) = cos t x(t)=\cos t x ( t ) = cos t y ( t ) = sin t y(t)=\sin t y ( t ) = sin t x ( t ) 2 + y ( t ) 2 = 1 x(t)^2+y(t)^2=1 x ( t ) 2 + y ( t ) 2 = 1 z ( t ) = y ( t ) 2 = sin 2 t z(t)=y(t)^2=\sin ^2 t z ( t ) = y ( t ) 2 = sin 2 t r ( t ) = ( cos t , sin t , sin 2 t ) \mathbf{r}(t)=(\cos t, \sin t, \sin^2 t) r ( t ) = ( cos t , sin t , sin 2 t ) 0 ≤ t ≤ 2 π 0\le t\le 2\pi 0 ≤ t ≤ 2 π
r ′ ( t ) = ( − sin t , cos t , 2 sin t cos t ) F ( r ( t ) ) = ( cos 2 t − sin t , sin 2 t + cos t , 1 ) F ( r ( t ) ) ⋅ r ′ ( t ) = − sin t cos 2 t + sin 2 t + sin 2 t cos t + cos 2 t + 2 sin t cos t = 1 + 1 3 d d t [ cos 3 t + sin 3 t ] + sin ( 2 t ) ∮ C F ⋅ d r = ∫ 0 2 π { 1 + 1 3 d d t [ cos 3 t + sin 3 t ] + sin ( 2 t ) } d t = [ t + 1 3 [ cos 3 t + sin 3 t ] − 1 2 cos ( 2 t ) ] 0 2 π = 2 π \begin{aligned}
\mathbf{r}'(t)&=(-\sin t\,,\, \cos t \,,\, 2\sin t\cos t)\\
\mathbf{F}\big(\mathbf{r}(t)\big)&=\big(\cos^2t-\sin t\,,\,\sin^2t+\cos t\,,\,1\big)\\
\mathbf{F}\big(\mathbf{r}(t)\big)\cdot\mathbf{r}'(t)&=-\sin t\cos^2 t+\sin^2 t+\sin^2t\cos t
+\cos^2t+2\sin t\cos t\\
&=1+\frac{1}{3}\frac{\mathrm{d}}{\mathrm{d}t}[\cos^3t+\sin^3 t]+\sin(2t)\\
\oint_C\mathbf{F}\cdot \mathrm{d}\mathbf{r} &=\int_0^{2\pi}
\left\{1+\frac{1}{3}\frac{\mathrm{d}}{\mathrm{d}t}[\cos^3t+\sin^3 t]+\sin(2t)\right\}\,\mathrm{d}t \\
&=\left[t+\frac{1}{3}[\cos^3t+\sin^3 t] -\frac{1}{2}\cos(2t) \right]_0^{2\pi}\\
&=2\pi
\end{aligned} r ′ ( t ) F ( r ( t ) ) F ( r ( t ) ) ⋅ r ′ ( t ) ∮ C F ⋅ d r = ( − sin t , cos t , 2 sin t cos t ) = ( cos 2 t − sin t , sin 2 t + cos t , 1 ) = − sin t cos 2 t + sin 2 t + sin 2 t cos t + cos 2 t + 2 sin t cos t = 1 + 3 1 d t d [ cos 3 t + sin 3 t ] + sin ( 2 t ) = ∫ 0 2 π { 1 + 3 1 d t d [ cos 3 t + sin 3 t ] + sin ( 2 t ) } d t = [ t + 3 1 [ cos 3 t + sin 3 t ] − 2 1 cos ( 2 t ) ] 0 2 π = 2 π
(b) S S S z = f ( x , y ) z=f(x,y) z = f ( x , y ) f ( x , y ) = y 2 f(x,y)=y^2 f ( x , y ) = y 2 x 2 + y 2 ≤ 1 x^2+y^2\le 1 x 2 + y 2 ≤ 1 C C C z z z S S S n ^ \hat{\mathbf{n}} n ^ z z z
n ^ d S = [ − ∂ f ∂ x ı ^ − ∂ f ∂ y ȷ ^ + k ^ ] d x d y = [ − 2 y ȷ ^ + k ^ ] d x d y ∇ × F = det [ ı ^ ȷ ^ k ^ ∂ ∂ x ∂ ∂ y ∂ ∂ z x 2 − y y 2 + x 1 ] = 2 k ^ ∇ × F ⋅ n ^ d S = 2 d x d y ∮ C F ⋅ d r = ∬ S ∇ × F ⋅ n ^ d S = 2 ∬ x 2 + y 2 ≤ 1 d x d y = 2 π \begin{aligned}
\hat{\mathbf{n}}\,dS&=\Big[-\frac{\partial f}{\partial x}\,\hat{\boldsymbol{\imath}} -\frac{\partial f}{\partial y}\,\hat{\boldsymbol{\jmath}}+\hat{\mathbf{k}}\Big]\,\mathrm{d}x\,\mathrm{d}y
=\Big[-2y\,\hat{\boldsymbol{\jmath}}+\hat{\mathbf{k}}\Big]\,\mathrm{d}x\,\mathrm{d}y\\
{ \boldsymbol{\nabla} }\times\mathbf{F}&=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in]
\frac{\partial }{\partial x} &
\frac{\partial }{\partial y} &
\frac{\partial }{\partial z} \\[0.05in]
x^2-y & y^2+x & 1 \end{matrix}\right]
=2\hat{\mathbf{k}}\\
{ \boldsymbol{\nabla} }\times\mathbf{F}\cdot\hat{\mathbf{n}}\,dS&=2\,\mathrm{d}x\,\mathrm{d}y\\
\oint_C\mathbf{F}\cdot \mathrm{d}\mathbf{r} &=\iint _S{ \boldsymbol{\nabla} }\times\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S
=2\iint _{x^2+y^2\le 1}\,\mathrm{d}x\,\mathrm{d}y\\
&=2\pi
\end{aligned} n ^ d S ∇ × F ∇ × F ⋅ n ^ d S ∮ C F ⋅ d r = [ − ∂ x ∂ f ^ − ∂ y ∂ f ^ + k ^ ] d x d y = [ − 2 y ^ + k ^ ] d x d y = det ^ ∂ x ∂ x 2 − y ^ ∂ y ∂ y 2 + x k ^ ∂ z ∂ 1 = 2 k ^ = 2 d x d y = ∬ S ∇ × F ⋅ n ^ d S = 2 ∬ x 2 + y 2 ≤ 1 d x d y = 2 π
원본 교재 풀이(영어) (a) Observe that x ( t ) = cos t x(t)=\cos t x ( t ) = cos t y ( t ) = sin t y(t)=\sin t y ( t ) = sin t x ( t ) 2 + y ( t ) 2 = 1 x(t)^2+y(t)^2=1 x ( t ) 2 + y ( t ) 2 = 1 z ( t ) = y ( t ) 2 = sin 2 t z(t)=y(t)^2=\sin ^2 t z ( t ) = y ( t ) 2 = sin 2 t r ( t ) = ( cos t , sin t , sin 2 t ) \mathbf{r}(t)=(\cos t, \sin t, \sin^2 t) r ( t ) = ( cos t , sin t , sin 2 t ) 0 ≤ t ≤ 2 π 0\le t\le 2\pi 0 ≤ t ≤ 2 π
r ′ ( t ) = ( − sin t , cos t , 2 sin t cos t ) F ( r ( t ) ) = ( cos 2 t − sin t , sin 2 t + cos t , 1 ) F ( r ( t ) ) ⋅ r ′ ( t ) = − sin t cos 2 t + sin 2 t + sin 2 t cos t + cos 2 t + 2 sin t cos t = 1 + 1 3 d d t [ cos 3 t + sin 3 t ] + sin ( 2 t ) ∮ C F ⋅ d r = ∫ 0 2 π { 1 + 1 3 d d t [ cos 3 t + sin 3 t ] + sin ( 2 t ) } d t = [ t + 1 3 [ cos 3 t + sin 3 t ] − 1 2 cos ( 2 t ) ] 0 2 π = 2 π \begin{aligned}
\mathbf{r}'(t)&=(-\sin t\,,\, \cos t \,,\, 2\sin t\cos t)\\
\mathbf{F}\big(\mathbf{r}(t)\big)&=\big(\cos^2t-\sin t\,,\,\sin^2t+\cos t\,,\,1\big)\\
\mathbf{F}\big(\mathbf{r}(t)\big)\cdot\mathbf{r}'(t)&=-\sin t\cos^2 t+\sin^2 t+\sin^2t\cos t
+\cos^2t+2\sin t\cos t\\
&=1+\frac{1}{3}\frac{\mathrm{d}}{\mathrm{d}t}[\cos^3t+\sin^3 t]+\sin(2t)\\
\oint_C\mathbf{F}\cdot \mathrm{d}\mathbf{r} &=\int_0^{2\pi}
\left\{1+\frac{1}{3}\frac{\mathrm{d}}{\mathrm{d}t}[\cos^3t+\sin^3 t]+\sin(2t)\right\}\,\mathrm{d}t \\
&=\left[t+\frac{1}{3}[\cos^3t+\sin^3 t] -\frac{1}{2}\cos(2t) \right]_0^{2\pi}\\
&=2\pi
\end{aligned} r ′ ( t ) F ( r ( t ) ) F ( r ( t ) ) ⋅ r ′ ( t ) ∮ C F ⋅ d r = ( − sin t , cos t , 2 sin t cos t ) = ( cos 2 t − sin t , sin 2 t + cos t , 1 ) = − sin t cos 2 t + sin 2 t + sin 2 t cos t + cos 2 t + 2 sin t cos t = 1 + 3 1 d t d [ cos 3 t + sin 3 t ] + sin ( 2 t ) = ∫ 0 2 π { 1 + 3 1 d t d [ cos 3 t + sin 3 t ] + sin ( 2 t ) } d t = [ t + 3 1 [ cos 3 t + sin 3 t ] − 2 1 cos ( 2 t ) ] 0 2 π = 2 π
(b) Let S S S z = f ( x , y ) z=f(x,y) z = f ( x , y ) f ( x , y ) = y 2 f(x,y)=y^2 f ( x , y ) = y 2 x 2 + y 2 ≤ 1 x^2+y^2\le 1 x 2 + y 2 ≤ 1 C C C z z z n ^ \hat{\mathbf{n}} n ^ S S S z z z
n ^ d S = [ − ∂ f ∂ x ı ^ − ∂ f ∂ y ȷ ^ + k ^ ] d x d y = [ − 2 y ȷ ^ + k ^ ] d x d y ∇ × F = det [ ı ^ ȷ ^ k ^ ∂ ∂ x ∂ ∂ y ∂ ∂ z x 2 − y y 2 + x 1 ] = 2 k ^ ∇ × F ⋅ n ^ d S = 2 d x d y ∮ C F ⋅ d r = ∬ S ∇ × F ⋅ n ^ d S = 2 ∬ x 2 + y 2 ≤ 1 d x d y = 2 π \begin{aligned}
\hat{\mathbf{n}}\,dS&=\Big[-\frac{\partial f}{\partial x}\,\hat{\boldsymbol{\imath}} -\frac{\partial f}{\partial y}\,\hat{\boldsymbol{\jmath}}+\hat{\mathbf{k}}\Big]\,\mathrm{d}x\,\mathrm{d}y
=\Big[-2y\,\hat{\boldsymbol{\jmath}}+\hat{\mathbf{k}}\Big]\,\mathrm{d}x\,\mathrm{d}y\\
{ \boldsymbol{\nabla} }\times\mathbf{F}&=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in]
\frac{\partial }{\partial x} &
\frac{\partial }{\partial y} &
\frac{\partial }{\partial z} \\[0.05in]
x^2-y & y^2+x & 1 \end{matrix}\right]
=2\hat{\mathbf{k}}\\
{ \boldsymbol{\nabla} }\times\mathbf{F}\cdot\hat{\mathbf{n}}\,dS&=2\,\mathrm{d}x\,\mathrm{d}y\\
\oint_C\mathbf{F}\cdot \mathrm{d}\mathbf{r} &=\iint _S{ \boldsymbol{\nabla} }\times\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S
=2\iint _{x^2+y^2\le 1}\,\mathrm{d}x\,\mathrm{d}y\\
&=2\pi
\end{aligned} n ^ d S ∇ × F ∇ × F ⋅ n ^ d S ∮ C F ⋅ d r = [ − ∂ x ∂ f ^ − ∂ y ∂ f ^ + k ^ ] d x d y = [ − 2 y ^ + k ^ ] d x d y = det ^ ∂ x ∂ x 2 − y ^ ∂ y ∂ y 2 + x k ^ ∂ z ∂ 1 = 2 k ^ = 2 d x d y = ∬ S ∇ × F ⋅ n ^ d S = 2 ∬ x 2 + y 2 ≤ 1 d x d y = 2 π