这个问题能不能用中值定理?

不太适合.

$$ \begin{aligned} & \iint_{x^2+y^2 \leq t^2} f\left(t^2-x^2-y^2\right) d x d y \\ = & \int_0^{2 \pi} d \theta \int_0^t r f\left(t^2-r^2\right) d r \\ = & 2 \pi \int_0^t r f\left(t^2-r^2\right) d r \\ = & \pi \int_0^t f\left(t^2-r^2\right) d r^2 \\ = & \pi \int_{t^2}^0 f(u) d\left(t^2-u\right) \\ = & \pi \int_0^{t^2} f(u) d u\end{aligned} $$

其中$u = t^2 - r^2$.

从而

$$ \begin{aligned} 原式 & =\lim_{t \rightarrow 0}\frac{\pi \int_0^{t^2} f(u) d u}{t^4} \\ & =\lim_{t \rightarrow 0}\frac{\pi f\left(t^2\right) \cdot 2 t}{4 t^3} \\ & =\lim_{t \rightarrow 0}\frac{\pi}{2} \cdot \frac{f\left(t^2\right)}{t^2} \\ & =\lim_{t \rightarrow 0}\frac{\pi}{2} \cdot f^{\prime}(0) \\ & =\frac{\pi}{2}\end{aligned} $$

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