곡면 S S S x 2 + y 2 + z 2 = 2 x^2 + y^2 + z^2 = 2 x 2 + y 2 + z 2 = 2 x 2 + y 2 = 1 x^2 + y^2 = 1 x 2 + y 2 = 1 z ≥ 0 z \ge 0 z ≥ 0 S S S
(a) r ( ϕ , θ ) = 2 sin ϕ cos θ ı ^ + 2 cos ϕ ȷ ^ + 2 sin ϕ sin θ k ^ \mathbf{r}(\phi,\theta) = 2\sin \phi\cos\theta\,\hat{\boldsymbol{\imath}} +2\cos \phi\,\hat{\boldsymbol{\jmath}}
+2\sin \phi\sin\theta\,\hat{\mathbf{k}} r ( ϕ , θ ) = 2 sin ϕ cos θ ^ + 2 cos ϕ ^ + 2 sin ϕ sin θ k ^
0 ≤ ϕ ≤ π 4 0\le\phi\le\frac{\pi}{4} 0 ≤ ϕ ≤ 4 π 0 ≤ θ ≤ 2 π 0\le\theta\le2\pi 0 ≤ θ ≤ 2 π r ( x , y ) = x ı ^ − y ȷ ^ + 2 − x 2 − y 2 k ^ \mathbf{r}(x,y) = x\,\hat{\boldsymbol{\imath}} -y\,\hat{\boldsymbol{\jmath}} +\sqrt{2-x^2-y^2}\,\hat{\mathbf{k}} r ( x , y ) = x ^ − y ^ + 2 − x 2 − y 2 k ^
x 2 + y 2 ≤ 1 x^2+y^2\le 1 x 2 + y 2 ≤ 1 r ( u , θ ) = u sin θ ı ^ + u cos θ ȷ ^ + 2 − u 2 k ^ \mathbf{r}(u,\theta) = u\sin\theta\,\hat{\boldsymbol{\imath}} +u\cos \theta\,\hat{\boldsymbol{\jmath}}
+\sqrt{2-u^2}\,\hat{\mathbf{k}} r ( u , θ ) = u sin θ ^ + u cos θ ^ + 2 − u 2 k ^
0 ≤ u ≤ 2 0\le u\le 2 0 ≤ u ≤ 2 0 ≤ θ ≤ 2 π 0\le\theta\le2\pi 0 ≤ θ ≤ 2 π r ( ϕ , θ ) = 2 sin ϕ cos θ ı ^ + 2 sin ϕ sin θ ȷ ^ + 2 cos ϕ k ^ \mathbf{r}(\phi,\theta) = \sqrt{2}\sin \phi\cos\theta\,\hat{\boldsymbol{\imath}}
+\sqrt{2}\sin \phi\sin\theta\,\hat{\boldsymbol{\jmath}}
+\sqrt{2}\cos \phi\,\hat{\mathbf{k}} r ( ϕ , θ ) = 2 sin ϕ cos θ ^ + 2 sin ϕ sin θ ^ + 2 cos ϕ k ^
0 ≤ ϕ ≤ π 4 0\le\phi\le\frac{\pi}{4} 0 ≤ ϕ ≤ 4 π 0 ≤ θ ≤ 2 π 0\le\theta\le2\pi 0 ≤ θ ≤ 2 π r ( ϕ , z ) = − 2 − z 2 sin ϕ ı ^ + 2 − z 2 cos ϕ ȷ ^ + z k ^ \mathbf{r}(\phi,z) = -\sqrt{2-z^2}\sin \phi\,\hat{\boldsymbol{\imath}}
+\sqrt{2-z^2}\cos \phi\,\hat{\boldsymbol{\jmath}}
+z\,\hat{\mathbf{k}} r ( ϕ , z ) = − 2 − z 2 sin ϕ ^ + 2 − z 2 cos ϕ ^ + z k ^
0 ≤ ϕ ≤ 2 π 0\le\phi\le2\pi 0 ≤ ϕ ≤ 2 π 1 ≤ z ≤ 2 1\le z\le\sqrt{2} 1 ≤ z ≤ 2
다 풀었으면 정답 확인 (a) 아니요. 0 < ϕ ≤ π 4 0 <\phi\le\frac{\pi}{4} 0 < ϕ ≤ 4 π π < θ < 2 π \pi < \theta <2\pi π < θ < 2 π z = sin ϕ sin θ z=\sin\phi\sin \theta z = sin ϕ sin θ
(b) 예. x 2 + ( − y ) 2 + ( 2 − x 2 − y 2 ) 2 = 2 x^2+\big(-y\big)^2+\big(\sqrt{2-x^2-y^2}\big)^2=2 x 2 + ( − y ) 2 + ( 2 − x 2 − y 2 ) 2 = 2 x 2 + y 2 ≤ 1 x^2+y^2\le 1 x 2 + y 2 ≤ 1 x 2 + ( − y ) 2 ≤ 1 x^2+(-y)^2\le 1 x 2 + ( − y ) 2 ≤ 1 2 − x 2 − y 2 ≥ 0 \sqrt{2-x^2-y^2}\ge 0 2 − x 2 − y 2 ≥ 0
(c) 아니요. 1 < u ≤ 2 1<u\le 2 1 < u ≤ 2 ( u sin θ ) 2 + ( u cos θ ) 2 = u 2 > 1 (u\sin\theta)^2+(u\cos\theta)^2=u^2>1 ( u sin θ ) 2 + ( u cos θ ) 2 = u 2 > 1 2 < u ≤ 2 \sqrt{2}<u\le 2 2 < u ≤ 2 2 − u 2 \sqrt{2-u^2} 2 − u 2
(d) 예. 다음에 주목한다.
( 2 sin ϕ cos θ ) 2 + ( 2 sin ϕ sin θ ) 2 + ( 2 cos ϕ ) 2 = 2 \big(\sqrt{2}\sin \phi\cos\theta\big)^2
+\big(\sqrt{2}\sin \phi\sin\theta\big)^2
+\big(\sqrt{2}\cos \phi\big)^2=2 ( 2 sin ϕ cos θ ) 2 + ( 2 sin ϕ sin θ ) 2 + ( 2 cos ϕ ) 2 = 2 0 ≤ ϕ ≤ π 4 0\le\phi\le\frac{\pi}{4} 0 ≤ ϕ ≤ 4 π z = 2 cos ϕ > 0 z=\sqrt{2}\cos \phi>0 z = 2 cos ϕ > 0 ϕ \phi ϕ 0 0 0 π 4 \frac{\pi}{4} 4 π r ( ϕ ) = 2 sin ϕ r(\phi)=\sqrt{2}\sin\phi r ( ϕ ) = 2 sin ϕ 0 0 0 1 1 1 ϕ \phi ϕ 0 0 0 π 4 \frac{\pi}{4} 4 π θ \theta θ 0 0 0 2 π 2\pi 2 π ( x = r ( ϕ ) cos θ , y = r ( ϕ ) sin θ ) \big(x=r(\phi)\cos\theta\,,\,y=r(\phi)\sin\theta\big) ( x = r ( ϕ ) cos θ , y = r ( ϕ ) sin θ ) x 2 + y 2 ≤ 1 x^2+y^2\le1 x 2 + y 2 ≤ 1
(e)
예. 다음에 주목한다.
( − 2 − z 2 sin ϕ ) 2 + ( 2 − z 2 cos ϕ ) 2 + ( z ) 2 = 2 \big(-\sqrt{2-z^2}\sin \phi\big)^2
+\big(\sqrt{2-z^2}\cos \phi\big)^2
+\big(z\big)^2=2 ( − 2 − z 2 sin ϕ ) 2 + ( 2 − z 2 cos ϕ ) 2 + ( z ) 2 = 2 1 ≤ z ≤ 2 1\le z\le\sqrt{2} 1 ≤ z ≤ 2 z > 0 z>0 z > 0 z z z 1 1 1 2 \sqrt{2} 2 r ( z ) = 2 − z 2 r(z)=\sqrt{2-z^2} r ( z ) = 2 − z 2 1 1 1 0 0 0 z z z 1 1 1 2 \sqrt{2} 2 ϕ \phi ϕ 0 0 0 2 π 2\pi 2 π ( x = − r ( z ) sin ϕ , y = r ( z ) cos ϕ ) \big(x=-r(z)\sin\phi\,,\,y=r(z)\cos\phi\big) ( x = − r ( z ) sin ϕ , y = r ( z ) cos ϕ ) x 2 + y 2 ≤ 1 x^2+y^2\le1 x 2 + y 2 ≤ 1