Problem 26 - Calculus Early Transcendentals

Question

Evaluate the line integral along the curve C. $$ \begin{array}{l}{\int_{C} x^{2} d x+x y d y+z^{2} d z} \\ {C: x=\sin t, y=\cos t, z=t^{2} \quad(0 \leq t \leq \pi / 2)}\end{array} $$

Step-by-Step Solution

Answer

The solution to the integral is \(\frac{\pi^6}{192}\).

Step 1: Substitute Parametric Equations

The curve \( C \) is given in terms of \( t \) as \( x = \sin t, \ y = \cos t, \ z = t^2 \). We'll substitute these into the integrand.

Step 2: Express Differential Elements

Differentiate each parametric equation with respect to \( t \) to get the differentials: \( dx = \cos t \, dt \), \( dy = -\sin t \, dt \), and \( dz = 2t \, dt \).

Step 3: Substitute and Set Up Integral

Substitute the expressions for \( x, y, z, dx, dy, \) and \( dz \) into the line integral: \[ \int_{0}^{\pi/2} (\sin^2 t \cdot \cos t + \sin t \cos t \cdot -\sin t + (t^2)^2 \cdot 2t) \, dt. \]

Step 4: Simplify the Integral

Simplify the terms in the integral: \( \sin^2 t \cos t - \sin^2 t \cos t + 2t^5 = 2t^5 \). The integral becomes \( \int_{0}^{\pi/2} 2t^5 \, dt \).

Step 5: Evaluate the Integral

Integrate \( 2t^5 \): \[ \int 2t^5 \, dt = \frac{2}{6}t^6 = \frac{1}{3}t^6. \] Evaluate from 0 to \( \pi/2 \): \[ \frac{1}{3} \left(\left(\frac{\pi}{2}\right)^6 - 0\right). \]

Step 6: Compute Final Result

Calculate \( \left(\frac{\pi}{2}\right)^6 \) to get the final result of the integral. \( (\pi/2)^6 = \pi^6 / 64 \), so the answer is \( \frac{1}{3} \cdot \frac{\pi^6}{64} = \frac{\pi^6}{192} \).

Key Concepts

Parametric Equations and Their Role

Parametric equations provide a way to describe curves using one or more parameters, instead of relying on traditional Cartesian coordinates. For our curve \( C \), the parametric equations are given by \( x = \sin t \), \( y = \cos t \), and \( z = t^2 \). Here, \( t \) is the parameter that ranges from \( 0 \) to \( \pi/2 \). By expressing the curve in this form, it becomes easier to integrate, since all variables are provided as functions of \( t \).
In physics and engineering, these equations allow us to track objects or define paths that can't be neatly described with \( x \), \( y \), and \( z \) alone. They extend the possibilities of integration, making calculations of line integrals like this possible.

Understanding Differential Elements

Differential elements are small pieces of the curve that we consider for our integration process. They represent infinitesimal changes along the path, and are crucial for setting up the integral. Here, we find the differentials \( dx \), \( dy \), and \( dz \) by differentiating each of the parametric equations with respect to \( t \).
Specifically, we calculate:

• \( dx = \cos t \, dt \)
• \( dy = -\sin t \, dt \)
• \( dz = 2t \, dt \)

These expressions allow us to transform the integral from one that involves \( x \), \( y \), and \( z \), to one that involves \( t \). This alignment simplifies the problem significantly and sets us for the final stages of solving the integral.

Integral Simplification Techniques

Simplification is a key step in making integrals easy to solve. After substituting into the integral, we have:
\[ \int_{0}^{\pi/2} (\sin^2 t \cdot \cos t + \sin t \cos t \cdot -\sin t + (t^2)^2 \cdot 2t) \, dt. \]
By carefully observing each term, we simplify as follows:

• The first and second terms: \( \sin^2 t \cos t - \sin^2 t \cos t \) cancel each other out.
• The third term simplifies to \( 2t^5 \).

Thus, our integral becomes much simpler: \( \int_{0}^{\pi/2} 2t^5 \, dt \). Simplifying an integral like this reduces complexity and makes the task of integration much more straightforward.

Integral Evaluation and Final Result

Evaluating the integral is the final step. We have the integral \( \int_{0}^{\pi/2} 2t^5 \, dt \). Here, we apply basic rules of integration. The indefinite integral of \( 2t^5 \) is:
\[ \int 2t^5 \, dt = \frac{2}{6}t^6 = \frac{1}{3}t^6. \]
Next, we use the limits \( 0 \) and \( \pi/2 \) to find the definite integral:
\[ \frac{1}{3} \left(\left(\frac{\pi}{2}\right)^6 - 0\right). \]
Finally, evaluating \( \left(\frac{\pi}{2}\right)^6 \) gives \( \pi^6 / 64 \), and our result is:
\[ \frac{1}{3} \cdot \frac{\pi^6}{64} = \frac{\pi^6}{192}. \] This process illustrates the power of breaking down complex equations into manageable parts and solving them step-by-step.