求下列曲线所围成的图形的面积.y=dfrac (1) (2)(x)^2与(x)^2+(y)^2=8(两部分都要计算).

如图，由$y=\dfrac {1} {2}{x}^{2}$与${x}^{2}+{y}^{2}=8$联立解得$A(-2,2),B(2,2)$.

$\left ( {1} \right )$两曲线围成的上半部分区域为：圆弧$ACB$和抛物线所围成区域由两部分构成，上半部分是由圆弧$ACB$和线段$AB$围成的弓形，下半部分是由线段$AB$和抛物线围成的区域，其面积为$S=\int ^{2}_{-2} {(\sqrt {{x}^{2}-8}}-\dfrac {1} {2}{x}^{2}){dx}$$=\int ^{2}_{-2} {(\sqrt {{x}^{2}-8}}{-2)dx}$$+\int ^{2}_{-2} {(2-\dfrac {1} {2}{x}^{2})}{dx}$.

$\int ^{2}_{-2} {(\sqrt {{x}^{2}-8}}{-2)dx}={S}_{1}$的值等于扇形$OACB$的面积减去$\triangle OAB$的面积,即${S}_{1}=\dfrac {1} {4}\times \pi \times 8-\dfrac {1} {2}\times 4\times 2=2\pi -4$;

$\int ^{2}_{-2} {(2-\dfrac {1} {2}{x}^{2})}{dx}={S}_{2}=(2x-\dfrac {1} {6}{x}^{3}){|}^{2}_{-2}$$=(2\times 2-\dfrac {1} {6}\times {2}^{3})-[2\times (-2)-\dfrac {1} {6}\times {(-2)}^{3}]=\dfrac {16} {3}$.

所以，$S=\int ^{2}_{-2} {(\sqrt {{x}^{2}-8}}-\dfrac {1} {2}{x}^{2}){dx}=2\pi -4+\dfrac {16} {3}=2\pi +\dfrac {4} {3}$.

$\left ( {2} \right )$两曲线围成的下半部分区域为：圆的面积减去上半部分的面积，其值为$S'=8\pi -(2\pi +\dfrac {4} {3})=6\pi -\dfrac {4} {3}$.

综上，由$y=\dfrac {1} {2}{x}^{2}$与${x}^{2}+{y}^{2}=8$所围成的区域的面积分别为$2\pi +\dfrac {4} {3}$和$6\pi -\dfrac {4} {3}$.