When convolving two functions that are constants in a region and 0 everywhere else, where does the integration start?
For simplicity, let's talk about one-dimensional convolutions only.

You need to understand that the result of a convolution is a function and the alleged convolution integral is actually an infinitude of integrals, one for each time instant.  By this I mean that to compute the output $y(t)$ of a system with impulse response 
$h(t)$ to an input signal at time $t = 5$, we write
$$\begin{align*}
y(5) &amp;= \int_{-\infty}^\infty h(\tau)x(5-\tau)\mathrm d\tau,\\
\end{align*}$$
and compute $y(5)$.  For $t = 6.5$, we write
$$\begin{align*}
y(6.5) &amp;= \int_{-\infty}^\infty h(\tau)x(6.5-\tau)\mathrm d\tau,\\
\end{align*}$$
and compute $y(6)$.  For $t = 10.2$, we write
$$\begin{align*}
y(10.2) &amp;= \int_{-\infty}^\infty h(\tau)x(10.2-\tau)\mathrm d\tau,\\
\end{align*}$$
and compute $y(10.2)$.and so on.  After doing this a while, some smart-ass 
said, "Hey Ma, I am beginning to see a pattern here!  Lookit, I can
write
$$\begin{align*}
y(t) &amp;= \int_{-\infty}^\infty h(\tau)x(t-\tau)\mathrm d\tau, ~\text{for all} ~t\\
\end{align*}$$
and the world was never the same again.  

Suppose that $h(t)$ is $0$ for $t &lt; 0$ or $t &gt; 12$.  Then all the integrals
exhibited above simplify in the sense that the limits can be changed to
$0$ and $12$ since the integrand, being a multiple of $h(\tau)$, is $0$
when $\tau &lt; 0$ or $\tau &gt; 12$.  Suppose in addition that $x(t)$ is nonzero
only when $|t| &lt; 3$.  Now look at the first integral (for $y(5)$ with the
limits changed to $0$ and $12$).  At $\tau = 0$, the integrand is
$h(0)x(5-0) = 0$ since $x(5) = 0$.  As $\tau$ increases towards $12$, the
argument of $x$ in the integrand decreases from $5$ downwards till the 
argument hits $-7$ at $\tau = 12$.  Thus, the integrand is nonzero
only for $\tau \in (2,8)$, having value $h(2)x(5-2) = h(2)x(3)$
at one end of the interval and value $h(8)x(5-8) = h(8)x(-3)$ at
the other.  We conclude that
$$y(5) = \int_{-\infty}^\infty h(\tau)x(5-\tau)\mathrm d\tau
= \int_{0}^{12} h(\tau)x(5-\tau)\mathrm d\tau
= \int_{2}^{8} h(\tau)x(5-\tau)\mathrm d\tau.$$
Similarly,
$$y(6.5) = \int_{-\infty}^\infty h(\tau)x(6.5-\tau)\mathrm d\tau
= \int_{0}^{12} h(\tau)x(6.5-\tau)\mathrm d\tau
= \int_{3.5}^{9.5} h(\tau)x(6.5-\tau)\mathrm d\tau.$$
However,
$$y(10.2) = \int_{-\infty}^\infty h(\tau)x(10.2-\tau)\mathrm d\tau
= \int_{0}^{12} h(\tau)x(10.2-\tau)\mathrm d\tau
= \int_{7.2}^{12} h(\tau)x(10.2-\tau)\mathrm d\tau$$
(can you see why?)

So, there is not a simple answer to your question about the
limits of integration; you have to work them out for each 
value of $t$ for which you want to evaluate the convolution
integral. But you don't have to do an infinite amount of
work!  There are Look Ma moments here and there.  Can you
combine two of the results above to write that for each $t$,
$3 \leq t \leq 9$, we have the useful pattern
$$y(t) = \int_{t-3}^{t+3} h(\tau)x(t-\tau)\mathrm d\tau$$
shown above?

I hope this will help you gain the understanding that
you need in order to complete your homework.