1. **State the problem:** We need to define the region bounded by the curve $$y^2 = 4x$$, the line $$y = 4$$, the y-axis $$x=0$$, and the vertical line $$x=4$$. 2. **Analyze the curve:** The curve $$y^2 = 4x$$ can be rewritten as $$x = \frac{y^2}{4}$$. This is a parabola opening to the right. 3. **Boundaries:** - Left boundary: $$x=0$$ (the y-axis). - Right boundary: $$x=4$$. - Upper boundary: $$y=4$$. - Lower boundary: the parabola $$y^2=4x$$ or equivalently $$x=\frac{y^2}{4}$$. 4. **Determine the range of y:** Since the parabola and the line intersect where $$y^2=4x$$ and $$y=4$$, substitute $$y=4$$ into the parabola: $$4^2 = 4x \implies 16 = 4x \implies x=4$$. So the intersection point is at $$(4,4)$$. 5. **Region description:** - Horizontally, $$x$$ ranges from $$0$$ to $$4$$. - Vertically, for each $$x$$ in $$[0,4]$$, $$y$$ ranges from the parabola $$y = -2\sqrt{x}$$ (since $$y^2=4x$$ implies $$y=\pm 2\sqrt{x}$$) up to the line $$y=4$$. 6. **Since the shaded region is at the bottom-left part and bounded above by $$y=4$$ and below by $$y^2=4x$$, and between $$x=0$$ and $$x=4$$, the region is:** $$\{(x,y) \mid 0 \leq x \leq 4, \quad 2\sqrt{x} \leq y \leq 4 \}$$ Note: The lower boundary is the positive branch $$y=2\sqrt{x}$$ because the shaded region is above the parabola and below $$y=4$$. **Final answer:** $$\boxed{\{(x,y) \mid 0 \leq x \leq 4, \quad 2\sqrt{x} \leq y \leq 4 \}}$$