简单计算一下即可，答案如图所示
设x=rcosu,y=rsinu,方程变为r^2=2a^2[(cosu)^2-(sinu)^2],
a>0时r=a√{2[1-2(sinu)^2]},kπ-π/4<=u<=kπ+π/4，k∈Z.
所求面积=∫∫
=2∫<-π/4，π/4>du∫<0,a√{2[1-2(sinu)^2]}rdr
=2a^2*∫<-π/4，π/4>[1-2(sinu)^2]du
=a^2*sin(2u)|<-π/4，π/4>
=2a^2.
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