数学试题讲解

设函数

f (t)

连续, 则二次积分

\int_{0}^{\frac{π}{2}} d θ \int_{2 \cos θ}^{2} f (r^{2}) r d r =

\int_{0}^{2} d x \int_{\sqrt{2 x - x^{2}}}^{\sqrt{4 - x^{2}}} \sqrt{x^{2} + y^{2}} f (x^{2} + y^{2}) d y

\int_{0}^{2} d x \int_{\sqrt{2 x - x^{2}}}^{\sqrt{4 - x^{2}}} f (x^{2} + y^{2}) d y

\int_{0}^{2} d y \int_{1 + \sqrt{1 - y^{2}}}^{\sqrt{4 - y^{2}}} \sqrt{x^{2} + y^{2}} f (x^{2} + y^{2}) d x

\int_{0}^{2} d y \int_{1 + \sqrt{1 - y^{2}}}^{\sqrt{4 - y^{2}}} f (x^{2} + y^{2}) d x