Determine The Volume Of The Shaded Region

Determinethe Volume of the Shaded Region: A Step‑by‑Step Guide

When faced with a diagram that highlights a particular area—often called the shaded region—and asks for the volume generated by that region, the task is essentially to translate a geometric picture into a mathematical integral. Whether the region is rotated about an axis, extruded into a prism, or sliced into infinitesimal slabs, the core idea remains the same: find the cross‑sectional area as a function of a single variable and integrate that area over the appropriate interval. Below is a comprehensive walkthrough that covers the theory, practical steps, illustrative examples, and common pitfalls, all written to help you confidently determine the volume of any shaded region you encounter.

1. Introduction: What Does “Volume of the Shaded Region” Mean?

In many calculus and geometry problems, a figure shows two or more curves, lines, or surfaces that enclose a planar area. That area is shaded to draw attention to it. The problem then asks for the volume of the solid obtained when this shaded area is:

• revolved around a line (the washer or shell method),
• extruded perpendicular to the plane (forming a prism or cylinder), or
• used as a base for a solid with known cross‑sections (the slice method).

The phrase “determine the volume of the shaded region” therefore signals that you must set up an integral whose integrand represents the area of a typical slice of the solid, and whose limits of integration correspond to the extremes of the shaded region in the chosen direction.

2. General Strategy for Determining the Volume

Regardless of the specific method (disk, washer, shell, or slicing), the process follows a consistent pattern:

1. Identify the axis of revolution or direction of extrusion.
   This tells you which variable will be used for integration (usually x or y).

2. Sketch the shaded region and label its boundaries.
   Write down the equations of the top, bottom, left, and right curves as functions of the chosen variable.

3. Determine the shape of a typical cross‑section.

   • If the region is revolved, the cross‑section is a disk or washer (radius = distance from the axis).
   • If the region is extruded, the cross‑section is simply the shaded area itself (constant thickness).
   • If cross‑sections are known shapes (triangles, squares, semicircles), express their area in terms of the variable.

4. Set up the integral:
   [ V = \int_{a}^{b} A(x),dx \quad \text{or} \quad V = \int_{c}^{d} A(y),dy ] where (A(x)) or (A(y)) is the area of a typical slice.

5. Evaluate the integral.
   Perform antiderivatives, apply the Fundamental Theorem of Calculus, and simplify.

6. Interpret the result.
   Include proper units (cubic units) and verify that the answer is positive and reasonable.

3. Detailed Examples

Below are three classic scenarios that illustrate how to apply the general strategy. Each example walks through the steps in depth, highlighting where decisions are made and why.

3.1 Example 1: Washer Method – Revolving Around the x-Axis

Problem Statement:
The shaded region lies between the curves (y = x^2) and (y = \sqrt{x}) from (x = 0) to (x = 1). Determine the volume of the solid obtained when this region is revolved about the x-axis.

Solution Walkthrough:

1. Axis of revolution: x-axis → integrate with respect to x.

2. Boundaries:

   • Top curve: (y_{\text{top}} = \sqrt{x})
   • Bottom curve: (y_{\text{bot}} = x^{2})
   • Intersection points: solve (\sqrt{x}=x^{2}) → (x=0) or (x=1). Hence limits (a=0), (b=1).

3. Cross‑section shape: Revolving a vertical strip produces a washer.

   • Outer radius (R(x) = y_{\text{top}} = \sqrt{x})
   • Inner radius (r(x) = y_{\text{bot}} = x^{2})

4. Area of a washer:
   [ A(x) = \pi\big[R(x)^{2} - r(x)^{2}\big] = \pi\big[(\sqrt{x})^{2} - (x^{2})^{2}\big] = \pi\big[x - x^{4}\big] ]

5. Integral:
   [ V = \int_{0}^{1} \pi\big[x - x^{4}\big],dx = \pi\left[\frac{x^{2}}{2} - \frac{x^{5}}{5}\right]_{0}^{1} = \pi\left(\frac{1}{2} - \frac{1}{5}\right) = \pi\left(\frac{5-2}{10}\right)=\frac{3\pi}{10} ]

6. Result: The volume is (\displaystyle \frac{3\pi}{10}) cubic units.

3.2 Example 2: Shell Method – Revolving Around the y-Axis

Problem Statement:
The same shaded region (between (y = x^{2}) and (y = \sqrt{x}), (0\le x\le1)) is now revolved about the y-axis. Find the volume.

Solution Walkthrough:

1. Axis of revolution: y-axis → integrate with respect to x using cylindrical shells (radius = x, height = difference of functions).

2. Height of a shell:
   [ h(x) = y_{\text{top}} - y_{\text{bot}} = \sqrt{x} - x^{2} ]

3. Radius of a shell: (r(x) = x).

4. Volume of a thin shell:
   [ dV = 2\pi \cdot (\text{radius}) \cdot (\text{height}) \cdot (\text{thickness}) = 2\pi x\big(\sqrt{x} - x^{2}\big)dx ]

5. Integral:
   [ V = \int_{0}^{1} 2\pi x\big(\sqrt{x} - x^{2}\big)dx = 2\pi\int_{0}^{1}\big(x^{3/2} - x^{3}\big)dx = 2\pi\left[\frac{2}{5}x^{5/2} - \frac{x^{4}}{4}\right]_{0}^{1} = 2\pi\left(\frac{2}{5} - \frac

… ( \frac{x^{4}}{4}\Big|_{0}^{1} = 2\pi\Big(\frac{2}{5} - \frac{1}{4}\Big) = 2\pi\Big(\frac{8}{20} - \frac{5}{20}\Big) = 2\pi\Big(\frac{3}{20}\Big) = \frac{3\pi}{10}).

Result: The volume generated by revolving the region about the y-axis is (\displaystyle \frac{3\pi}{10}) cubic units. This matches the washer‑method result, confirming that the two approaches are consistent; the value is positive and, given the modest size of the region (bounded by (0\le x\le1) and (0\le y\le1)), a volume of roughly (0.94) cubic units is reasonable.

3.3 Example 3: Disk Method – Revolving Around a Horizontal Line (y = -1)

Problem Statement:
Consider the region bounded by (y = x^{2}), the x-axis, and the vertical line (x = 2). Find the volume of the solid obtained when this region is revolved about the horizontal line (y = -1).

Solution Walkthrough:

1. Axis of revolution: The line (y = -1) is parallel to the x-axis, so we integrate with respect to x using the disk (or washer) method.

2. Outer radius: The distance from the curve (y = x^{2}) to the axis (y = -1) is
   [ R(x) = x^{2} - (-1) = x^{2} + 1. ]

3. Inner radius: The region touches the x-axis ((y = 0)), whose distance to the axis is
   [ r(x) = 0 - (-1) = 1. ]

4. Cross‑sectional area: Each perpendicular slice yields a washer with area
   [ A(x) = \pi\big[R(x)^{2} - r(x)^{2}\big] = \pi\big[(x^{2}+1)^{2} - 1^{2}\big] = \pi\big[x^{4} + 2x^{2} + 1 - 1\big] = \pi\big(x^{4} + 2x^{2}\big). ]

5. Limits of integration: The region extends from (x = 0) (where the curve meets the x-axis) to (x = 2). Thus (a = 0), (b = 2).

6. Integral for volume:
   [ V = \int_{0}^{2} \pi\big(x^{4} + 2x^{2}\big),dx = \pi\left[\frac{x^{5}}{5} + \frac{2x^{3}}{3}\right]_{0}^{2} = \pi\left(\frac{32}{5} + \frac{16}{3}\right). ]

   Combining the fractions: [ \frac{32}{5} + \frac{16}{3} = \frac{96}{15} + \frac{80}{15} = \frac{176}{15}. ]

   Hence [ V = \frac{176\pi}{15}\ \text{cubic units}. ]

7. Positivity and reasonableness: The integrand (x^{4}+2x^{2}) is non‑negative on ([0,2]), guaranteeing a positive volume. Numerically, [ V \approx \frac{176 \times 3.1416}{15} \approx 36.

.84\ \text{cubic units}. ]

Result: The volume of the solid formed by revolving the region about (y = -1) is (\displaystyle \frac{176\pi}{15}) cubic units. This value is positive and, given the extent of the region (extending from (x = 0) to (x = 2) and curving upward as (y = x^{2})), a volume of about (37) cubic units is consistent with the geometry.

4. Conclusion

The disk and washer methods provide systematic ways to compute volumes of solids of revolution by slicing the solid into thin disks or washers perpendicular to the axis of rotation. The key steps are:

1. Identify the axis of revolution and determine whether to integrate with respect to (x) or (y).
2. Express the outer and inner radii as functions of the integration variable, measuring distances from the curve(s) to the axis.
3. Set up the integral using (V = \pi\int (R^2 - r^2),dx) (or (dy)), where (R) and (r) are the outer and inner radii.
4. Evaluate the integral over the appropriate bounds.
5. Check the result for positivity and reasonableness in the context of the region’s size.

These methods are versatile, applying to revolutions about both the (x)-axis and (y)-axis, as well as to horizontal or vertical lines offset from the axes. Mastery of the disk and washer methods equips you to tackle a wide variety of volume problems in calculus and beyond.