Calculus-Master

Area and Volume of Solids

Calculating the area between curves and finding volumes of solids using the Disk, Washer, or Cross-section methods.

Core Theorem
Area Between Curves: A=∫ab[ftop(x)−gbottom(x)]dxA = \int_{a}^{b} [f_{top}(x) - g_{bottom}(x)] dx.
 Disk Method: V=π∫ab[R(x)]2dxV = \pi \int_{a}^{b} [R(x)]^2 dx.
 Washer Method: V=π∫ab([Rout(x)]2−[Rin(x)]2)dxV = \pi \int_{a}^{b} ([R_{out}(x)]^2 - [R_{in}(x)]^2) dx.
Known Cross-sections: V=∫abA(x)dxV = \int_{a}^{b} A(x) dx.
Step-by-Step SOP
  1. 1

    1. Sketch and Identify

    Graph the functions and identify the region. Determine the axis of revolution or the base of the solid to identify the 'Top' and 'Bottom' functions.
  2. 2

    2. Find Boundaries

    Solve the system of equations f(x)=g(x)f(x) = g(x) to find the points of intersection. These values serve as the limits of integration [a,b][a, b].
  3. 3

    3. Set up the Integral

    Define the radius R(x)R(x) for rotation or the area formula A(x)A(x) for known cross-sections (e.g., Side2Side^2 for squares).
  4. 4

    4. Evaluate

    Apply the appropriate formula, ensuring π\pi is included for solids of revolution, and compute the definite integral.

Practice Exercises

Example 01Easy
Find the volume of the solid generated by revolving y=xy=\sqrt{x} about the x-axis from x=0x=0 to x=4x=4.
NEED A HINT?
The region is flush against the x-axis, so use the Disk Method with R(x)=xR(x) = \sqrt{x}.
SHOW DETAILED EXPLANATION

Step 1: Set up

V=π∫04(x)2dx=π∫04xdxV = \pi \int_{0}^{4} (\sqrt{x})^2 dx = \pi \int_{0}^{4} x dx.

Step 2: Solve

π[12x2]04=π(12(16)−0)=8π\pi [\frac{1}{2}x^2]_{0}^{4} = \pi (\frac{1}{2}(16) - 0) = 8\pi.
Example 02Medium
Find the area of the region bounded by y=x2y = x^2 and y=2x+3y = 2x + 3.
NEED A HINT?
Find intersections first, then subtract the parabola from the line.
SHOW DETAILED EXPLANATION

Step 1: Find Intersections

Set the functions equal: x2=2x+3x^2 = 2x + 3. Rearrange to x2−2x−3=0x^2 - 2x - 3 = 0. Factoring gives (x−3)(x+1)=0(x-3)(x+1) = 0, so the limits of integration are a=−1a = -1 and b=3b = 3.

Step 2: Set up the Integral

By testing a value (e.g., x=0x=0), we see y=2x+3y=2x+3 is the 'Top' function. Area A=∫−13[(2x+3)−(x2)]dxA = \int_{-1}^{3} [(2x + 3) - (x^2)] dx.

Step 3: Integrate and Evaluate

Integral: [x2+3x−13x3]−13[x^2 + 3x - \frac{1}{3}x^3]_{-1}^{3}.
Evaluation at 3: (32+3(3)−13(3)3)=(9+9−9)=9(3^2 + 3(3) - \frac{1}{3}(3)^3) = (9 + 9 - 9) = 9.
Evaluation at -1: ((−1)2+3(−1)−13(−1)3)=(1−3+13)=−53((-1)^2 + 3(-1) - \frac{1}{3}(-1)^3) = (1 - 3 + \frac{1}{3}) = -\frac{5}{3}.
Total Area: 9−(−53)=273+53=3239 - (-\frac{5}{3}) = \frac{27}{3} + \frac{5}{3} = \frac{32}{3}.
Example 03Medium
The region bounded by y=xy = x and y=xy = \sqrt{x} is revolved about the x-axis. Find the volume.
NEED A HINT?
Use the Washer Method. Between x=0x=0 and x=1x=1, x\sqrt{x} is the outer radius.
SHOW DETAILED EXPLANATION

Step 1: Identify Radii and Limits

Intersections: x=x  ⟹  x=x2  ⟹  x=0,1\sqrt{x} = x \implies x = x^2 \implies x=0, 1.
Outer radius R(x)=xR(x) = \sqrt{x}, Inner radius r(x)=xr(x) = x.

Step 2: Set up the Integral

Using Washer Method: V=π∫01([x]2−[x]2)dx=π∫01(x−x2)dxV = \pi \int_{0}^{1} ([\sqrt{x}]^2 - [x]^2) dx = \pi \int_{0}^{1} (x - x^2) dx.

Step 3: Integrate and Evaluate

Integral: π[12x2−13x3]01\pi [\frac{1}{2}x^2 - \frac{1}{3}x^3]_{0}^{1}.
Evaluation: π[(12(1)2−13(1)3)−(0)]=π(12−13)=π(36−26)=π6\pi [(\frac{1}{2}(1)^2 - \frac{1}{3}(1)^3) - (0)] = \pi (\frac{1}{2} - \frac{1}{3}) = \pi (\frac{3}{6} - \frac{2}{6}) = \frac{\pi}{6}.
Common Pitfalls
  • ⚠
    Incorrect Squaring in Washer MethodA common mistake is writing π∫(Rout−Rin)2dx\pi \int (R_{out} - R_{in})^2 dx. The correct form must be the difference of squares: Rout2−Rin2R_{out}^2 - R_{in}^2.
  • ⚠
    Missing PiStudents often forget to multiply by π\pi when using the Disk or Washer methods, which are based on circular cross-sections (πr2\pi r^2).
  • ⚠
    Intersection CrossoversIf the functions cross within the interval [a,b][a, b], you must split the integral into sections to ensure the area remains positive.
  • ⚠
    Radius vs. DiameterIn cross-section problems involving semi-circles, the distance between curves (ftop−gbottom)(f_{top} - g_{bottom}) is often the diameter. Remember to divide by 2 to get the radius before squaring.

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Gary Chang

Gary Chang

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5+ Years of Calculus Teaching Experience | AP Calculus Specialist. Dedicated to helping students master calculus through step-by-step logic and clear visualizations.

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