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Calculus - Further areas between curves – Volumes (HL)

Grade 11IB_AA

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

🔑Concepts

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The area between two curves y=f(x)y = f(x) and y=g(x)y = g(x) from x=ax=a to x=bx=b is given by the integral of the upper function minus the lower function: ∫ab(f(x)−g(x)) dx\int_{a}^{b} (f(x) - g(x)) \, dx.

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When a region bounded by a curve y=f(x)y = f(x), the xx-axis, and the lines x=ax=a and x=bx=b is rotated 360∘360^\circ about the xx-axis, a solid of revolution is formed.

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The volume of revolution about the xx-axis is calculated using the disk method: V=π∫aby2 dxV = \pi \int_{a}^{b} y^2 \, dx.

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For rotations about the yy-axis, the function must be expressed as x=g(y)x = g(y), and the volume is V=π∫cdx2 dyV = \pi \int_{c}^{d} x^2 \, dy.

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The volume of the solid generated by rotating the area between two curves f(x)f(x) and g(x)g(x) (where f(x)≥g(x)≥0f(x) \ge g(x) \ge 0) about the xx-axis is found using the washer method: V=π∫ab([f(x)]2−[g(x)]2) dxV = \pi \int_{a}^{b} ([f(x)]^2 - [g(x)]^2) \, dx.

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In IB HL, you may also encounter volumes where the cross-section is not a circle (e.g., squares or triangles), though rotation remains the primary focus for 'Volumes of Revolution'.

📐Formulae

A=∫ab∣f(x)−g(x)∣ dxA = \int_{a}^{b} |f(x) - g(x)| \, dx

Vx=π∫aby2 dx=π∫ab[f(x)]2 dxV_x = \pi \int_{a}^{b} y^2 \, dx = \pi \int_{a}^{b} [f(x)]^2 \, dx

Vy=π∫cdx2 dy=π∫cd[g(y)]2 dyV_y = \pi \int_{c}^{d} x^2 \, dy = \pi \int_{c}^{d} [g(y)]^2 \, dy

Vbetween=π∫ab(youter2−yinner2) dxV_{between} = \pi \int_{a}^{b} (y_{outer}^2 - y_{inner}^2) \, dx

💡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 the intersection points: x2=x  ⟹  x4=x  ⟹  x(x3−1)=0x^2 = \sqrt{x} \implies x^4 = x \implies x(x^3 - 1) = 0. Thus, x=0x = 0 and x=1x = 1.
  2. Identify the upper curve: For x∈[0,1]x \in [0, 1], x≥x2\sqrt{x} \ge x^2.
  3. Set up the integral: A=∫01(x−x2) dxA = \int_{0}^{1} (\sqrt{x} - x^2) \, dx
  4. Integrate: A=[23x3/2−13x3]01=(23−13)−(0)=13A = \left[ \frac{2}{3}x^{3/2} - \frac{1}{3}x^3 \right]_{0}^{1} = \left( \frac{2}{3} - \frac{1}{3} \right) - (0) = \frac{1}{3}

Explanation:

The area is found by integrating the difference between the top function x\sqrt{x} and bottom function x2x^2 between their intersection points.

Problem 2:

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

Solution:

  1. Use the formula V=π∫aby2 dxV = \pi \int_{a}^{b} y^2 \, dx.
  2. Substitute the function: V=π∫01(ex)2 dx=π∫01e2x dxV = \pi \int_{0}^{1} (e^x)^2 \, dx = \pi \int_{0}^{1} e^{2x} \, dx
  3. Integrate: V=π[12e2x]01V = \pi \left[ \frac{1}{2}e^{2x} \right]_{0}^{1}
  4. Evaluate: V=π2(e2−e0)=π2(e2−1)V = \frac{\pi}{2} (e^2 - e^0) = \frac{\pi}{2}(e^2 - 1) cubic units.

Explanation:

To find the volume of revolution, square the function, integrate with respect to xx, and multiply by π\pi.

Problem 3:

Find the volume of the solid generated when the region bounded by y=x2y = x^2 and y=2xy = 2x is rotated 360∘360^\circ about the xx-axis.

Solution:

  1. Find intersections: x2=2x  ⟹  x(x−2)=0x^2 = 2x \implies x(x - 2) = 0, so x=0x = 0 and x=2x = 2.
  2. Identify outer and inner radii: For x∈[0,2]x \in [0, 2], 2x≥x22x \ge x^2. So R(x)=2xR(x) = 2x and r(x)=x2r(x) = x^2.
  3. Set up washer method: V=π∫02((2x)2−(x2)2) dx=π∫02(4x2−x4) dxV = \pi \int_{0}^{2} ((2x)^2 - (x^2)^2) \, dx = \pi \int_{0}^{2} (4x^2 - x^4) \, dx
  4. Integrate: V=π[4x33−x55]02V = \pi \left[ \frac{4x^3}{3} - \frac{x^5}{5} \right]_{0}^{2}
  5. Evaluate: V=π(4(8)3−325)=π(323−325)=π(160−9615)=64π15V = \pi \left( \frac{4(8)}{3} - \frac{32}{5} \right) = \pi \left( \frac{32}{3} - \frac{32}{5} \right) = \pi \left( \frac{160 - 96}{15} \right) = \frac{64\pi}{15}

Explanation:

When rotating the area between two curves, we subtract the volume of the inner solid from the volume of the outer solid. This is why we square the individual functions first before subtracting.