題目 Problem (10) (10 pts) Let a>0a>0a>0. Evaluate the double integral ∫−aa∫−a2−x2a2−x2cos(x2+y2) dy dx.\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\cos(x^2+y^2)\,dy\,dx.∫−aa∫−a2−x2a2−x2cos(x2+y2)dydx. 解答 解法一 積分區域是半徑 aaa 的圓盤,且被積函數只含 x2+y2x^2+y^2x2+y2,改用極座標最直接: x=rcosθ,y=rsinθ,dA=r dr dθ.x=r\cos\theta,\qquad y=r\sin\theta,\qquad \mathrm{d}A=r\,\mathrm{d}r\,\mathrm{d}\theta.x=rcosθ,y=rsinθ,dA=rdrdθ. 因此 ∬x2+y2≤a2cos(x2+y2) dA= ∫02π∫0acos(r2)r dr dθ= 2π12sin(r2)∣0a= πsin(a2).\begin{align*} \iint_{x^2+y^2\le a^2}\cos(x^2+y^2)\,\mathrm{d}A =&\,\int_0^{2\pi}\int_0^a\cos(r^2)r\,\mathrm{d}r\,\mathrm{d}\theta \\[4mm] =&\,2\pi\left.\frac12\sin(r^2)\right|_0^a \\[4mm] =&\,\boxed{\pi\sin(a^2)}. \end{align*}∬x2+y2≤a2cos(x2+y2)dA===∫02π∫0acos(r2)rdrdθ2π21sin(r2)0aπsin(a2).