Expressing $\frac{d}{dt}\left(\int_{D(t)}u(x,t)dx\right)-\int_{D(t)}u_t(x,t)dx$ as a surface integral? the following question was the last problem on the Fall 2010 qualifying exam at UCLA.
Define $D(t)=\{x^2+y^2\leq r^2(t)\}\subseteq\mathbb{R}^2$ where $r(t)\colon\mathbb{R}\to\mathbb{R}$ is continuously differentiable. For given smooth, nonnegative function $u(x,t)\colon\mathbb{R}^2\times\mathbb{R}\to\mathbb{R}$, express the following quantity in terms of a surface integral:
$$\frac{d}{dt}\left(\int_{D(t)}u(x,t)dx\right)-\int_{D(t)}u_t(x,t)dx.$$
Any calculus theorem can be used without proof.
Does anyone have an idea to express this? My group and I were stumped.
 A: We recall the following relation
$$ \frac{d}{dr}\Bigl(\int_{B(0,r)} u dx\Bigr) = \int_{\partial B(0,r)} u dS $$
In here $ D(t) = B(0,r(t)) $, So we define $ F : \mathbb{R}^2 \rightarrow \mathbb{R} $ as $$ F(r,s) = \int_{B(0,r)} u(x,s) dx $$ then we observe that from the above formula
$$ \frac{\partial F}{\partial r}(r,s) = \int_{\partial B(0,r)} u(x,s) dS\ \ \ \ \ \text{and}\ \ \ \ \ \frac{\partial F}{\partial s}(r,s) =  \int_{B(0,r)} 
\frac{\partial u}{\partial s}(x,s) dx $$
Now take the curve $ \gamma : \mathbb{R} \rightarrow \mathbb{R}^2 $ given by $ \gamma(t) = (r(t),t) $ , hence $ \gamma'(t) = (r'(t),1) $ and using chain rule observe that 
$$ \frac{d}{dt}\Bigl(\int_{D(t)} u(x,t) dx\Bigr) = (F\circ\gamma)'(t)
= \nabla F(\gamma(t)).\gamma'(t) \\ =r'(t)\frac{\partial F}{\partial r}(r(t),t) + \frac{\partial F}{\partial s}(r(t),t) =  r'(t) \int_{\partial D(t)} u(x,t) dS + \int_{D(t)}\frac{\partial u}{\partial t}(x,t)dx 
 $$ Thus finally we have
$$ \frac{d}{dt}\Bigl(\int_{D(t)} u(x,t) dx\Bigr) - \int_{D(t)}\frac{\partial u}{\partial t}(x,t)dx  = r'(t) \int_{\partial D(t)} u(x,t) dS $$
A: Let $C(t)=\{x^2+y^2=r^2(t)\}$. You can show that
$$\frac{d}{dt}\int_{D(t)}u(x,t)dx =r'(t)\int_{C(t)}u(x,t)dx+\int_{D(t)}u_t(x,t)dx$$
and the second term cancels in your expression, so you get
$$r'(t)\int_{C(t)}u(x,t)dx$$
Let $C_r$ denote the circle of radius $r$ and note that
$$\frac{d}{dr}\int_{C_r}u(x,t)dx=\int_{C_r}\frac1r u(x,t)dx$$
thus we have
$$r'(t)\int_{C(t)}u(x,t)dx=r'(t)\int_0^{r(t)}\int_{C_r}\frac1r u(x,t)dxdr=\int_{D(t)}r'(t)\frac{u(x,t)}{r}dx$$
I haven't been very careful with changes of coordinates here, but the idea should work.
