題目

Problem

(a) $\displaystyle\iint_D y\,dA = \underline{\quad(15)\quad}$ , where $D$ is the region in the first quadrant that lies between the circles $x^2+y^2=4$ and $(x-1)^2+y^2=1$ .

(b) The volume of the solid $R$ lying below the plane $z=3-2y$ and above the paraboloid $z=x^2+y^2$ is $\underline{\quad(16)\quad}$ .

解答

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(a) 在極座標中， $x^2+y^2=4$ 給出 $r=2$ ，而

(x-1)^2+y^2=1 \quad\Longrightarrow\quad r=2\cos\theta.

因此

\begin{align*} \iint_Dy\,\mathrm{d}A =&\,\int_0^{\pi/2}\int_{2\cos\theta}^2(r\sin\theta)r\,\mathrm{d}r\,\mathrm{d}\theta\\[2mm] =&\,2. \end{align*}

故

\boxed{(15)=2}.

(b) 交線由 $x^2+y^2=3-2y$ 決定，亦即

x^2+(y+1)^2=4.

令 $u=x,v=y+1$ ，則兩曲面高度差為 $4-u^2-v^2$ ，投影區域為 $u^2+v^2\le4$ 。所以

\begin{align*} V =&\,\int_0^{2\pi}\int_0^2(4-r^2)r\,\mathrm{d}r\,\mathrm{d}\theta\\[2mm] =&\,8\pi. \end{align*}

因此

\boxed{(16)=8\pi}.