Dawson 201-SN3-RE: integral calculus tutoring

1. Find the bounds and the upper curve. The square root requires $x\ge0$. At an intersection, $\sqrt{x}=x/2$. Squaring gives $x=x^2/4$, or $x(x-4)=0$, so the candidates are $x=0$ and $x=4$. Both satisfy the original equation. Between them, $\sqrt{x}>x/2$: for $0<x<4$, squaring these nonnegative expressions reduces the comparison to $x>x^2/4$. 2. Compute planar area. Let $A$ denote the area. A vertical strip has height equal to the upper curve’s value minus the lower curve’s value and thickness $dx$, an infinitesimal horizontal width. Thus $A=\int_0^4(\sqrt{x}-x/2)\,dx=\left[\frac23x^{3/2}-\frac{x^2}{4}\right]_0^4$. The bracket means the upper-end value minus the lower-end value. Evaluating gives $A=(16/3-4)-0=4/3$ square units. 3. Compute volume by washers. Let $V$ denote the volume, $R(x)=\sqrt{x}$ the outer radius and $r(x)=x/2$ the inner radius, both measured from the rotation axis. A cross-section has area $\pi R(x)^2-\pi r(x)^2$; therefore $V=\pi\int_0^4(x-x^2/4)\,dx=\pi\left[\frac{x^2}{2}-\frac{x^3}{12}\right]_0^4=\pi(8-16/3)=8\pi/3$ cubic units. 4. Explain the distinction. Planar area sums strip heights times widths, whereas volume sums circular cross-sectional areas times widths. A washer is an outer disk with an inner disk removed, so its area is $\pi(R^2-r^2)$, not $\pi(R-r)^2$. As a scaling check, doubling both curve heights doubles the planar area but quadruples the rotation volume because the radii are squared.