Applications of Integration in Areas and Volumes

Introduction

Key Concepts

1. Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concept of differentiation with integration, establishing the foundation for calculating areas under curves and volumes of solids. It consists of two parts:

• First Part: If \( f \) is a continuous function on \([a, b]\) and \( F \) is an antiderivative of \( f \), then: $$ \int_{a}^{b} f(x) dx = F(b) - F(a) $$ This part allows the evaluation of definite integrals using antiderivatives.
• Second Part: It states that the derivative of an integral with a variable upper limit is the original function: $$ \frac{d}{dx} \left( \int_{a}^{x} f(t) dt \right) = f(x) $$ This establishes that differentiation and integration are inverse processes.

Example: To find the area under \( f(x) = x^2 \) from 0 to 3: $$ \int_{0}^{3} x^2 dx = \left[ \frac{x^3}{3} \right]_{0}^{3} = \frac{27}{3} - 0 = 9 $$ Thus, the area is 9 square units.

2. Integration Techniques for Area Calculation

Calculating the area between curves often involves integration. The general approach is to find the integral of the upper function minus the lower function over the interval of interest.

Formula: $$ \text{Area} = \int_{a}^{b} [f(x) - g(x)] dx $$ where \( f(x) \) is the upper function and \( g(x) \) is the lower function.

Example: Find the area between \( f(x) = x^2 \) and \( g(x) = x + 2 \) from \( x = -1 \) to \( x = 2 \).

1. Determine points of intersection: $$ x^2 = x + 2 \Rightarrow x^2 - x - 2 = 0 \Rightarrow x = \frac{1 \pm \sqrt{9}}{2} \Rightarrow x = 2 \text{ or } x = -1 $$
2. Set up the integral: $$ \text{Area} = \int_{-1}^{2} [(x + 2) - x^2] dx $$
3. Evaluate the integral: $$ \int_{-1}^{2} (x + 2 - x^2) dx = \left[ \frac{x^2}{2} + 2x - \frac{x^3}{3} \right]_{-1}^{2} $$ $$ = \left( \frac{4}{2} + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 - \frac{-1}{3} \right) $$ $$ = \left( 2 + 4 - \frac{8}{3} \right) - \left( \frac{1}{2} - 2 + \frac{1}{3} \right) $$ $$ = \left( 6 - \frac{8}{3} \right) - \left( -\frac{3}{2} + \frac{1}{3} \right) $$ $$ = \frac{10}{3} + \frac{7}{6} = \frac{27}{6} = 4.5 $$

The area between the curves is 4.5 square units.

3. Volume Calculation Using the Disk Method

The Disk Method is employed to find the volume of a solid of revolution obtained by rotating a region around an axis. It involves integrating the area of circular disks perpendicular to the axis of rotation.

Formula: $$ V = \pi \int_{a}^{b} [f(x)]^2 dx $$ where \( f(x) \) is the radius of the disk at a given \( x \).

Example: Find the volume of the solid obtained by rotating \( f(x) = \sqrt{x} \) around the x-axis from \( x = 0 \) to \( x = 4 \).

1. Set up the integral: $$ V = \pi \int_{0}^{4} (\sqrt{x})^2 dx = \pi \int_{0}^{4} x dx $$
2. Evaluate the integral: $$ \pi \left[ \frac{x^2}{2} \right]_{0}^{4} = \pi \left( \frac{16}{2} - 0 \right) = 8\pi $$

The volume of the solid is \( 8\pi \) cubic units.

4. Volume Calculation Using the Shell Method

The Shell Method is another technique for finding volumes of solids of revolution, especially useful when the axis of rotation is parallel to the axis of integration.

Formula: $$ V = 2\pi \int_{a}^{b} x f(x) dx $$ where \( x \) is the radius and \( f(x) \) is the height of the cylindrical shell at a given \( x \).

Example: Find the volume of the solid obtained by rotating \( f(x) = x^2 \) around the y-axis from \( x = 0 \) to \( x = 2 \).

1. Set up the integral: $$ V = 2\pi \int_{0}^{2} x (x^2) dx = 2\pi \int_{0}^{2} x^3 dx $$
2. Evaluate the integral: $$ 2\pi \left[ \frac{x^4}{4} \right]_{0}^{2} = 2\pi \left( \frac{16}{4} - 0 \right) = 8\pi $$

The volume of the solid is \( 8\pi \) cubic units.

5. Applications in Real-World Problems

Integration is indispensable in various fields such as engineering, physics, and economics for modeling and solving real-world problems related to areas and volumes.

• Engineering: Designing objects with specific volume and surface area requirements, such as containers and structural components.
• Physics: Calculating quantities like electric charge distributions and mass distributions in continuous bodies.
• Economics: Optimizing resource allocation by modeling cost and revenue functions over specific intervals.

Example: An engineer designing a cylindrical tank needs to determine the volume to ensure it meets storage requirements. Using the Disk Method, they can integrate the area to find the precise volume needed.

6. Integrating Polar Coordinates

Some regions are more naturally expressed in polar coordinates. Integration in polar coordinates involves different formulas for area and volume calculations.

Area Formula: $$ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} [r(\theta)]^2 d\theta $$

Example: Find the area of one loop of the polar curve \( r(\theta) = 1 + \cos(\theta) \).

1. Set up the integral: $$ A = \frac{1}{2} \int_{0}^{\pi} (1 + \cos(\theta))^2 d\theta $$
2. Expand and integrate: $$ \frac{1}{2} \int_{0}^{\pi} (1 + 2\cos(\theta) + \cos^2(\theta)) d\theta $$ $$ = \frac{1}{2} \left[ \theta + 2\sin(\theta) + \frac{\theta}{2} + \frac{\sin(2\theta)}{4} \right]_{0}^{\pi} $$ $$ = \frac{1}{2} \left[ \frac{3\pi}{2} \right] = \frac{3\pi}{4} $$

The area of one loop is \( \frac{3\pi}{4} \) square units.

7. Advanced Applications: Multiple Integrals for Volume

For more complex shapes, multiple integrals are used to calculate volumes. This involves integrating functions of two variables over a defined region.

Double Integral Formula (Cartesian Coordinates): $$ V = \int \int_{D} f(x, y) dA $$ where \( D \) is the region in the \( xy \)-plane and \( f(x, y) \) represents the height function.

Example: Find the volume under \( f(x, y) = x + y \) over the rectangular region \( D = [0, 2] \times [0, 3] \).

1. Set up the double integral: $$ V = \int_{0}^{2} \int_{0}^{3} (x + y) dy dx $$
2. Integrate with respect to \( y \): $$ \int_{0}^{3} (x + y) dy = \left[ xy + \frac{y^2}{2} \right]_{0}^{3} = 3x + \frac{9}{2} $$
3. Integrate with respect to \( x \): $$ \int_{0}^{2} \left( 3x + \frac{9}{2} \right) dx = \left[ \frac{3x^2}{2} + \frac{9x}{2} \right]_{0}^{2} = \left( 6 + 9 \right) = 15 $$

The volume under the surface is 15 cubic units.

Comparison Table

Summary and Key Takeaways