Let $T_1$ be the time between a car accident and the claim reported to the insurance company. Let $T_2$ be the time between the report of the claim and payment of the claim. The joint density function of $T_1$ and $T_2$, $f(t_1, t_2)$, is constant over the region $0<t_1, t_2<6, t_1+t_2<10$ and zero otherwise. Determine $f(t_1, t_2)$.
My attempt: Let $f(t_1, t_2) = k$. Then, $$\int_0^6 \int_0^{10-y} k \; dx\; dy = \int_0^6 k(10-y) \; dy = 1 \implies k\left[60-\frac{35}{2}\right] = 1$$
which ultimately gives $k = \frac{1}{42}$, however, the correct answer is $k = \frac{1}{34}$. Where have I gone wrong?
The problem with your method is that your integral is incorrect. Your limits correspond to the region $0 < t_1< 10, 0<t_2<6, t_1 + t_2 < 10$.
Here's an alternative approach. Let $R$ denote the region in question. First of all, note that $$ \iint_R k\,dx\,dy = k\cdot \iint_R dx\,dy = 1, $$ where $R$ denotes the region of interest. We can rearrange the above to find that that $k = 1/A$, where $A = \iint_R 1\,dx\,dy$ is the area of $R$. To compute the area, subtract the area of a triangle from the area of a square. In particular, define $$ R_1 = \{x,y : 0 < x < 6, 0 < y < 6\},\\ R_2 = \{x,y: 0 < x < 6, 0 < y < 6 \text{ and } x + y \geq 10\}. $$ Verify that $R = R_1 \setminus R_2$, which is to say that $R$ contains the points of $R_1$ not included in $R_2$. $R_1$ is a square with area $6 \times 6 = 36$, and $R_2$ is a triangle with area $\frac 12 2 \times 2 = 2$. Thus, the area of $R$ is $36 - 2 = 34$.
Thus, we arrive at the correct answer $k = 1/A = 1/34$.