Question
Suppose that random variables $Y_1$ and $Y_2$ have joint probability density function given by $$f(y_1,y_2)=\begin{cases}cy_1^2y_2^2,\quad 0\leqslant y_1\leqslant y_2, y_1 + y_2\leqslant 2,\\0,\quad \text{elsewise}\end{cases}$$
a) Calculate the value of c.
b) What is the probability that $Y_1 + Y_2$ is less than $1$?
My answers
a) $$\begin{aligned} \int_{0}^{2} \int_{0}^{2-y_2}\left(cy_1^2y_2^2\right)dy_1dy_2 &= \frac{16}{45}c \\ \implies \frac{16}{45}c & = 1 \\ c & =\frac{45}{16} \end{aligned}$$
b) $$\begin{aligned} \mathbb{P}(Y_1 + Y_2 < 1) & = \mathbb{P}(Y_2 < 1 - Y_1)\\ & = \int_{0}^{1} \int_{0}^{1-y_2}\left(\frac{45}{16} y_1^2y_2^2\right)dy_1dy_2 \\ & = \frac{1}{64} \end{aligned}$$
However, I have just learnt joint PDFs and finding the limits of these double integrals are very confusing to me. May I know whether my approaches and answers to both parts are correct?
Edit
Following the comments, I have made some changes to my working and am wondering whether my answers are correct now.
a) $$\begin{aligned} \int_{0}^{1} \int_{y_1}^{2-y_1}\left(cy_1^2y_2^2\right)dy_2dy_1 &= \frac{8}{45}c \\ \implies \frac{8}{45}c & = 1 \\ c & =\frac{45}{8} \end{aligned}$$
b) $$\begin{aligned} \mathbb{P}(Y_1 + Y_2 < 1) & = \mathbb{P}(Y_2 < 1 - Y_1)\\ & = \int_{0}^{\frac{1}{2}} \int_{y_1}^{1-y_1}\left(\frac{45}{8} y_1^2y_2^2\right)dy_2dy_1 \\ & = \frac{1}{64} \end{aligned}$$