Chapter 13: Problem 9
Evaluate the integrals in Exercises \(1-10\) $$ \int_{0}^{\pi / 2}\left[\cos t \mathbf{i}-\sin 2 t \mathbf{j}+\sin ^{2} t \mathbf{k}\right] d t $$
Short Answer
Expert verified
\( \mathbf{i} - \mathbf{j} + \frac{1}{4} \pi \mathbf{k} \).
Step by step solution
01
Break Down the Integral
The given vector integral is of the form \( \int_{a}^{b} [f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k}] \, dt \). In this case, the functions are: \( f(t) = \cos t \), \( g(t) = -\sin 2t \), and \( h(t) = \sin^2 t \). We need to evaluate each component of the vector separately.
02
Evaluate the \(\mathbf{i}\) Component
The \(\mathbf{i}\) component is \( \int_{0}^{\pi/2} \cos t \, dt \). Integrate \( \cos t \) to obtain \( \sin t \), then evaluate from \(0\) to \(\pi/2\):\[ \left[ \sin t \right]_{0}^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1. \]
03
Evaluate the \(\mathbf{j}\) Component
The \(\mathbf{j}\) component is \( \int_{0}^{\pi/2} -\sin 2t \, dt \). Recognize that the integral of \(-\sin 2t\) is \(\frac{1}{2} \cos 2t\). Integrate and evaluate:\[ \left[ \frac{1}{2} \cos 2t \right]_{0}^{\pi/2} = \frac{1}{2}(\cos(\pi) - \cos(0)) = \frac{1}{2}(-1 - 1) = -1. \]
04
Evaluate the \(\mathbf{k}\) Component
The \(\mathbf{k}\) component is \( \int_{0}^{\pi/2} \sin^2 t \, dt \). Since \( \sin^2 t = \frac{1-\cos 2t}{2} \), evaluate using this form:\[ \int_{0}^{\pi/2} \frac{1-\cos 2t}{2} \, dt = \frac{1}{2} \left[ t - \frac{1}{2} \sin 2t \right]_{0}^{\pi/2}. \]Solve:\[ \frac{1}{2} \left( \frac{\pi}{2} - 0 - (0) + 0 \right) = \frac{1}{4} \pi. \]
05
Combine Results into a Vector
Combine the results from steps 2, 3, and 4 into one vector:\[ 1 \mathbf{i} - 1 \mathbf{j} + \frac{1}{4} \pi \mathbf{k}. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Vector Integral
Vector integrals are a fundamental concept in vector calculus that involve integrating vector functions over a particular interval. In simpler terms, a vector integral examines how the components of a vector function change as it moves along a parameter, typically time or distance, and adds them up over that interval. This kind of integral is especially useful in physics and engineering when dealing with vector fields.
- When dealing with vector integrals, the integral can be separated into scalar integrals for each of the components of the vector.
- This helps in simplifying the process since each component can be integrated individually using regular techniques.
Integration Techniques
To tackle integrals efficiently, it's important to be familiar with a variety of integration techniques. Depending on the integrands' functions, different methods are applied. Here are a few techniques relevant to the problem at hand:
- Basic Integration: The simplest technique where a known antiderivative is used, such as integrating \( \cos t \) to obtain \( \sin t \).
- Substitution: Useful when dealing with functions like \(-\sin 2t\), which can be rewritten to use a simpler trigonometric identity or rule.
- Trigonometric Identities: Often, trigonometric identities are used to simplify integrals. For instance, \( \sin^2 t \) was rewritten using the identity \( \frac{1-\cos 2t}{2} \), making it easier to integrate.
Trigonometric Integrals
Trigonometric integrals play a crucial role in calculus, particularly when dealing with periodic, wave-like functions. These integrals require a solid understanding of trigonometric identities and their integrals.In the example from the exercise:
- The integral of \( \cos t \) directly yielded \( \sin t \), demonstrating the straightforward process when dealing with basic trigonometric functions.
- Integrating \(-\sin 2t\) made use of recognizing the derivative, which simplifies to \( \frac{1}{2} \cos 2t \) upon integration. This shows the importance of knowing antiderivatives for more complex forms like \( \sin 2t \).
- Finally, the \( \sin^2 t \) was simplified using an identity, making the integration process more manageable. Such identities are pivotal for simplifying otherwise challenging integrals into a format that's easier to work with.
Component-Wise Integration
Component-wise integration focuses on evaluating each component of a vector or function separately. This technique breaks down a potentially complex problem into simpler, more manageable parts.
- In vector calculus, especially, vector functions can be dissected into scalar component functions along each axis represented by unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \).
- By integrating each component individually, as done in the exercise, we can simplify the problem rather than attempting to integrate the entire vector in one go.
- Each component function \(f(t)\), \(g(t)\), \(h(t)\) (corresponding to components \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) respectively) is treated as a separate scalar integration problem.