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Calculus - Applications of Integration (Area between curves, Volumes of Revolution)

Grade 12IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Area Under a Curve: The definite integral ∫abf(x)dx\int_{a}^{b} f(x) dx calculates the 'signed' area between the function f(x)f(x) and the x-axis. Visually, areas where the graph is above the x-axis are positive, while areas below the x-axis are negative. To find the total physical area, one must split the integral at the x-intercepts or integrate the absolute value ∣f(x)∣|f(x)|.

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Area Between Two Curves: This is found by integrating the difference between the 'top' function and the 'bottom' function: ∫ab(f(x)−g(x))dx\int_{a}^{b} (f(x) - g(x)) dx. Visually, this represents the shaded region sandwiched between two graphs. The limits aa and bb are usually the x-coordinates of the points where the two graphs intersect.

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Intersection Points: To find the boundaries of a region enclosed by two curves, set f(x)=g(x)f(x) = g(x) and solve for xx. Visually, these points are the 'corners' or vertices of the area you are calculating. If the curves intersect more than twice, you must set up separate integrals for each section to account for which curve is on top.

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Volume of Revolution (x-axis): When a region is rotated 360∘360^{\circ} around the x-axis, it forms a 3D solid. Visually, the cross-sections of this solid perpendicular to the x-axis are circular disks with radius r=f(x)r = f(x). The total volume is the sum of the areas of these infinitely thin disks: π∫r2dx\pi \int r^2 dx.

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Volume of Revolution (y-axis): If the area is rotated around the y-axis, the function must be expressed in terms of yy, such that x=g(y)x = g(y). Visually, the solid's cross-sections are horizontal disks with radius r=g(y)r = g(y), and we integrate with respect to yy between the lower vertical limit cc and the upper vertical limit dd.

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The Washer Method: When the area between two curves f(x)f(x) and g(x)g(x) is rotated around an axis, it creates a solid with a hole in the middle, known as a 'washer'. Visually, the cross-section is an annulus (a ring). The volume is calculated by subtracting the inner volume from the outer volume: π∫(Router2−rinner2)dx\pi \int (R_{outer}^2 - r_{inner}^2) dx.

📐Formulae

Total Area: A=∫ab∣f(x)∣dxA = \int_{a}^{b} |f(x)| dx

Area between curves f(x)f(x) and g(x)g(x): A=∫ab(f(x)−g(x))dxA = \int_{a}^{b} (f(x) - g(x)) dx, where f(x)≥g(x)f(x) \ge g(x)

Volume of Revolution (x-axis): V=π∫ab[f(x)]2dxV = \pi \int_{a}^{b} [f(x)]^2 dx

Volume of Revolution (y-axis): V=π∫cd[g(y)]2dyV = \pi \int_{c}^{d} [g(y)]^2 dy

Volume between curves (x-axis rotation): V=π∫ab([f(x)]2−[g(x)]2)dxV = \pi \int_{a}^{b} ([f(x)]^2 - [g(x)]^2) dx

Area with respect to y: A=∫cd(xright−xleft)dyA = \int_{c}^{d} (x_{right} - x_{left}) dy

💡Examples

Problem 1:

Find the area of the region enclosed by the curves y=x2y = x^2 and y=xy = \sqrt{x}.

Solution:

  1. Find intersection points: Set x2=x  ⟹  x4=x  ⟹  x(x3−1)=0x^2 = \sqrt{x} \implies x^4 = x \implies x(x^3 - 1) = 0. The curves intersect at x=0x=0 and x=1x=1.
  2. Determine which curve is on top: In the interval (0,1)(0, 1), 0.5≈0.707\sqrt{0.5} \approx 0.707 and 0.52=0.250.5^2 = 0.25. Since x>x2\sqrt{x} > x^2, the top curve is y=xy = \sqrt{x}.
  3. Set up the integral: A=∫01(x−x2)dxA = \int_{0}^{1} (\sqrt{x} - x^2) dx.
  4. Integrate: [23x3/2−13x3]01=(23(1)−13(1))−(0)=13[\frac{2}{3}x^{3/2} - \frac{1}{3}x^3]_0^1 = (\frac{2}{3}(1) - \frac{1}{3}(1)) - (0) = \frac{1}{3}. The area is 13\frac{1}{3} square units.

Explanation:

We first find the boundaries by setting the functions equal. By observing the values between 0 and 1, we identify the upper boundary to correctly subtract the lower boundary, ensuring a positive area result.

Problem 2:

Calculate the volume of the solid formed when the region bounded by y=exy = e^x, the x-axis, x=0x=0, and x=1x=1 is rotated 360∘360^{\circ} about the x-axis.

Solution:

  1. Identify the radius: The radius of the solid is r=y=exr = y = e^x.
  2. Use the volume formula: V=π∫01(ex)2dxV = \pi \int_{0}^{1} (e^x)^2 dx.
  3. Simplify the integrand: V=π∫01e2xdxV = \pi \int_{0}^{1} e^{2x} dx.
  4. Evaluate the integral: V=π[12e2x]01V = \pi [\frac{1}{2}e^{2x}]_0^1.
  5. Substitute limits: V=π2(e2(1)−e2(0))=π2(e2−1)V = \frac{\pi}{2} (e^{2(1)} - e^{2(0)}) = \frac{\pi}{2}(e^2 - 1). The volume is π(e2−1)2\frac{\pi(e^2 - 1)}{2} cubic units.

Explanation:

To find the volume of revolution about the x-axis, we square the function, multiply by π\pi, and integrate over the given interval. Remember that (ex)2=e2x(e^x)^2 = e^{2x} using exponent laws.