Iterated Integral With Odd Upper-Bound Let $f:[0,1]\rightarrow \mathbb{R}$ be an Lebesgue integrable function on $[0,1]$.  I would like to compute the following integral but I believe that I may be wrong when applying Fubini's theorem
$$
\int_0^{\sqrt{t}}\int_0^{\sqrt{v}} f(u)dudv.
$$
Here's what I have computed (using the change of variables formulate formula and Fubini I "computed" that)
$$
\begin{aligned}
\int_0^{\sqrt{t}}\int_0^{\sqrt{v}} f(u)dudv = &
\int_0^{\sqrt{t}}\int_0^{v} f(\sqrt{u}) \frac1{2\sqrt{u}}dudv\\
= &
\int_0^{t} \int_0^{\sqrt{u}}f(\sqrt{v}) \frac1{2\sqrt{v}}dvdu\\
= &
\frac1{2}\int_0^{t} \int_0^{u}f(\sqrt{\sqrt{v}}) \frac1{\sqrt{\sqrt{v}}} \frac1{2\sqrt{v}}dvdu\\
= &
\frac1{4}\int_0^{t} \int_0^{u}f(\sqrt[4]{v}) v^{-\frac{3}{4}}dvdu.
\end{aligned}
$$
Is this correct?  I feel as though something fishy happened when interchanging the integrals wrt. the bounds of integration.  
 A: There's a problem here:

$$
\int_0^{\sqrt{t}}\int_0^{v} f(\sqrt{u}) \frac1{2\sqrt{u}}dudv
= 
\int_0^{t} \int_0^{\sqrt{u}}f(\sqrt{v}) \frac1{2\sqrt{v}}dvdu
$$

The region of integration is the triangle whose vertices are $(u=0,v=0)$, $(u=0,v = \sqrt{t})$, and $(u=\sqrt{t},v=\sqrt{t})$. To cover this triangle using the other order of integration, you have to integrate $v$ from $u$ to $\sqrt{t}$, then $u$ from $0$ to $\sqrt{t}$. Additionally, $f$ is still a function of $u$ regardless of the order of integration. So you should have
\begin{multline}
\int_0^{\sqrt{t}}\!\!\!\int_0^{\sqrt{v}}\! f(u)dudv = \int_0^{\sqrt{t}}\!\!\!\int_0^{v}\! \frac{ f(\sqrt{u})}{2\sqrt{u}}dudv
= 
\int_0^{\sqrt{t}}\!\!\!\int_u^{\sqrt{t}}\!\frac{f(\sqrt{u}) }{2\sqrt{u}}dvdu \\ = 
\int_0^{\sqrt{t}}\!\frac{f(\sqrt{u}) }{2\sqrt{u}}\left[\int_u^{\sqrt{t}}\!dv\right]du = \int_0^{\sqrt{t}}\!\frac{f(\sqrt{u})}{2\sqrt{u}}(\sqrt{t}-u)du = \int_0^{t^{1/4}}\!f(x)(\sqrt{t}-x^2)dx
\end{multline}
And here's an example showing the equality of the two integrals.
