1. **Problem Statement:** We have a joint probability density function (p.d.f) of random variables $X$ and $Y$ given by: $$f(x,y) = \begin{cases} c x^2 y, & x^2 \leq y \leq 1 \\ 0, & \text{elsewhere} \end{cases}$$ We need to find the constant $c$ such that $f(x,y)$ is a valid p.d.f, and then find $P(X > 1)$. 2. **Key Formula:** For a joint p.d.f, the total probability must be 1: $$\int \int f(x,y) \, dx \, dy = 1$$ 3. **Region of Integration:** Given $x^2 \leq y \leq 1$, for each fixed $y$, $x$ ranges from $-\sqrt{y}$ to $\sqrt{y}$. Also, $y$ ranges from 0 to 1. 4. **Set up the integral to find $c$:** $$\int_0^1 \int_{-\sqrt{y}}^{\sqrt{y}} c x^2 y \, dx \, dy = 1$$ 5. **Integrate with respect to $x$ first:** $$\int_{-\sqrt{y}}^{\sqrt{y}} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-\sqrt{y}}^{\sqrt{y}} = \frac{(\sqrt{y})^3 - (-\sqrt{y})^3}{3} = \frac{y^{3/2} - (-y^{3/2})}{3} = \frac{2 y^{3/2}}{3}$$ 6. **Substitute back into the integral:** $$\int_0^1 c y \cdot \frac{2 y^{3/2}}{3} \, dy = \int_0^1 c \frac{2}{3} y^{5/2} \, dy = 1$$ 7. **Integrate with respect to $y$:** $$c \frac{2}{3} \int_0^1 y^{5/2} \, dy = c \frac{2}{3} \left[ \frac{y^{7/2}}{7/2} \right]_0^1 = c \frac{2}{3} \cdot \frac{2}{7} = c \frac{4}{21}$$ 8. **Solve for $c$:** $$c \frac{4}{21} = 1 \implies c = \frac{21}{4}$$ 9. **Find $P(X > 1)$:** Since $x^2 \leq y \leq 1$, for $X > 1$, $x^2 > 1$, but $y \leq 1$, so the condition $x^2 \leq y$ cannot hold for $x > 1$ because $y$ cannot be greater than 1. Thus, the joint p.d.f is zero for $X > 1$. Therefore, $$P(X > 1) = 0$$ **Final answers:** - $c = \frac{21}{4}$ - $P(X > 1) = 0$