We want to find the shaded area of the region R that is bounded by the functions
[tex]f(x)=\sqrt{x}\text{ and }f(x)=\frac{x}{4}[/tex]First, we find the point of intersection, this occurs at a point where the functions are equal:
[tex]\begin{gathered} \sqrt{x}=\frac{x}{4} \\ 4\sqrt{x}=x \end{gathered}[/tex]Squaring both sides we get:
[tex]\begin{gathered} 16x=x^2 \\ x^2-16x=0 \\ x(x-16)=0 \end{gathered}[/tex]This says us that they intersect at the point x and when x=16. For finding the shaded area, we calculate the following integral:
[tex]\begin{gathered} \int_0^{16}\int_{\frac{x}{4}}^{\sqrt{x}}1dydx=\int_0^{16}\sqrt{x}-\frac{x}{4}dx \\ =\int_0^{16}x^{1/2}-\frac{x}{4}dx \\ =\frac{2}{3}x^{\frac{3}{2}}-\frac{x^2}{8}|_0^{16} \\ =\frac{2}{3}(16)^{\frac{3}{2}}-\frac{16^2}{8} \\ =\frac{32}{3} \end{gathered}[/tex]Then, the shaded area is 32/3 units, or 10.6666... The rotations do not affect the area, as they are isometries.
