1. **State the problem:** We need to find the area of the shaded region bounded by the lines and curve: $$y=4$$ (horizontal line), $$x=\frac{y}{4}$$ (vertical line in terms of y), and $$x=\sqrt{y}$$ (curve). 2. **Understand the boundaries:** - The horizontal line is at $$y=4$$. - The vertical boundary is given by $$x=\frac{y}{4}$$, which can be rewritten as $$y=4x$$. - The curve is $$x=\sqrt{y}$$, or equivalently $$y=x^2$$. 3. **Find the intersection points:** - Intersection of $$y=4$$ and $$x=\sqrt{y}$$: Substitute $$y=4$$ into $$x=\sqrt{y}$$: $$x=\sqrt{4}=2$$. So point is $$(2,4)$$. - Intersection of $$y=4$$ and $$x=\frac{y}{4}$$: Substitute $$y=4$$ into $$x=\frac{y}{4}$$: $$x=\frac{4}{4}=1$$. So point is $$(1,4)$$. - Intersection of $$x=\sqrt{y}$$ and $$x=\frac{y}{4}$$: Set $$\sqrt{y} = \frac{y}{4}$$. Square both sides: $$y = \frac{y^2}{16}$$. Multiply both sides by 16: $$16y = y^2$$. Rearrange: $$y^2 - 16y = 0$$. Factor: $$y(y - 16) = 0$$. So $$y=0$$ or $$y=16$$. For $$y=0$$, $$x=\sqrt{0}=0$$. For $$y=16$$, $$x=\sqrt{16}=4$$. Since the shaded region is bounded by $$y=4$$, the relevant intersection is at $$y=0$$, $$x=0$$. 4. **Set up the integral for the area:** The region is bounded between $$y=0$$ and $$y=4$$. For each $$y$$ in $$[0,4]$$, the horizontal distance between the curves is: $$\text{right boundary} - \text{left boundary} = \sqrt{y} - \frac{y}{4}$$. 5. **Calculate the area:** $$\text{Area} = \int_0^4 \left(\sqrt{y} - \frac{y}{4}\right) dy$$. 6. **Evaluate the integral:** $$\int_0^4 \sqrt{y} dy = \int_0^4 y^{1/2} dy = \left[ \frac{2}{3} y^{3/2} \right]_0^4 = \frac{2}{3} (4)^{3/2} - 0 = \frac{2}{3} \times 8 = \frac{16}{3}$$. $$\int_0^4 \frac{y}{4} dy = \frac{1}{4} \int_0^4 y dy = \frac{1}{4} \left[ \frac{y^2}{2} \right]_0^4 = \frac{1}{4} \times \frac{16}{2} = \frac{1}{4} \times 8 = 2$$. 7. **Subtract to find the area:** $$\text{Area} = \frac{16}{3} - 2 = \frac{16}{3} - \frac{6}{3} = \frac{10}{3}$$. **Final answer:** The area of the shaded region is $$\boxed{\frac{10}{3}}$$ square units.