Volume by Washers: Region Between y = x^2 and y = x Revolved about x = 3 (Washer and Shells cross-check)

Goal: Find the volume of the solid formed by revolving the region between the curves around the vertical line x = 3.
Region described in the transcript appears to be bounded by the parabola and the line: y = x^2 and y = x (i.e., the area between the parabola and the line).
The axis of revolution is the line x = 3 (a vertical line).
The approach described uses cross-sections that are washers (disks with a hole), i.e., the washer method, which uses horizontal slices (dy) when revolving about a vertical line.
Key idea: For each fixed y, the cross-section is a washer whose outer radius R(y) comes from the distance from the axis to the far boundary of the region, and whose inner radius r(y) comes from the distance from the axis to the near boundary.

Curves bounding the region: y = x (a straight line through (0,0) and (1,1)) and y = x^2 (a parabola opening upward with vertex at (0,0)).
Intersection points: Solve x^2 = x → x(x − 1) = 0 ⇒ x = 0 or x = 1. Corresponding y-values are y = 0 and y = 1. Thus, the region sits between x = 0 and x = 1 and between y = x^2 (below) and y = x (above).
For a given y ∈ [0, 1], the x-coordinates on the region satisfy:
- Left boundary: from the line, x_left = y (since y = x ⇔ x = y).
- Right boundary: from the parabola, x_right = √y (since y = x^2 ⇔ x = √y for x ≥ 0 in this region).
Therefore, at a fixed y between 0 and 1, the horizontal cross-section of the region extends from x = y to x = √y.

Distance from the axis x = 3 to the left boundary x_left = y gives the outer radius:
- R(y) = 3 − x_left = 3 − y.
Distance from the axis x = 3 to the right boundary x_right = √y gives the inner radius:
- r(y) = 3 − x_right = 3 − √y.
The cross-sectional area at height y is a washer with area:
- A(y) = π [R(y)^2 − r(y)^2] = π [(3 − y)^2 − (3 − √y)^2].

Volume is the integral of the cross-sectional area over y from the bottom of the region to the top:
- V = ∫_{y=0}^{1} A(y) dy
- V = ∫_{0}^{1} π [(3 − y)^2 − (3 − √y)^2] dy.
Simplified integrand (optional step):
- (3 − y)^2 = 9 − 6y + y^2
- (3 − √y)^2 = 9 − 6√y + y
- A(y) = π [ (9 − 6y + y^2) − (9 − 6√y + y) ] = π [ y^2 − 7y + 6√y ].
- Therefore, V = π ∫_{0}^{1} [ y^2 − 7y + 6√y ] dy.

Compute the integral term-by-term:
- ∫{0}^{1} y^2 dy = [ y^3 / 3 ]{0}^{1} = 1/3
- ∫{0}^{1} (−7y) dy = −7 [ y^2 / 2 ]{0}^{1} = −7/2
- ∫{0}^{1} 6√y dy = 6 ∫{0}^{1} y^{1/2} dy = 6 [ (2/3) y^{3/2} ]_{0}^{1} = 4
Sum inside the brackets: 1/3 − 7/2 + 4 = 1/3 + (4 − 3.5) = 1/3 + 1/2 = 5/6
Therefore, V = π × (5/6) = (5/6) π.

If using shells with vertical slices (dx) around a vertical axis, the radius is (3 − x) and the height is the vertical extent of the region at that x, which is h(x) = ytop − ybottom = x − x^2 for 0 ≤ x ≤ 1.
Shells setup:
- V = 2π ∫{x=0}^{1} [radius × height] dx = 2π ∫{0}^{1} (3 − x)(x − x^2) dx.
Expand the integrand: (3 − x)(x − x^2) = 3x − 3x^2 − x^2 + x^3 = 3x − 4x^2 + x^3.
Integrate: ∫{0}^{1} (3x − 4x^2 + x^3) dx = [ (3/2)x^2 − (4/3)x^3 + (1/4)x^4 ]{0}^{1} = 3/2 − 4/3 + 1/4 = 5/12.
Multiply by 2π: V = 2π × (5/12) = (5/6)π, which matches the washer result.

Volume by washers around a vertical axis uses horizontal slices (dy) when the axis is vertical.
Outer radius: R(y) = distance from axis to the left boundary of the region; inner radius: r(y) = distance from axis to the right boundary.
For the region between y = x and y = x^2, bounded for 0 ≤ y ≤ 1, the left and right x-boundaries are xleft = y and xright = √y, giving R(y) = 3 − y and r(y) = 3 − √y.
Washer area: A(y) = π [R(y)^2 − r(y)^2].
Volume: V = ∫ A(y) dy over the vertical extent of the region.
Alternative shells method corroborates results: V = 2π ∫ (radius)(height) dx with radius = 3 − x and height = x − x^2 for 0 ≤ x ≤ 1.

The transcript includes a digression about the period of the sine function:
- It states questions about the period and attempts to relate endpoints like −π/2 and π/4 to the period, mentioning π and π/2 in the discussion.
- Correct fact: The period of sin(x) is 2π, and one full cycle occurs over an interval of length 2π. A half-cycle corresponds to π. The interval from −π/2 to π/4 has length 3π/4, which is not a full half-cycle. The transcript’s casual notes mix these ideas and should be treated as a side remark rather than part of the volume calculation.

Volume of the solid obtained by revolving the region bounded by y = x and y = x^2 about x = 3:
- $V = \frac{5}{6}\pi.$