Problem 16 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 20 , evaluate the line integral over the given curve. \(\int_{c} y^{2} d x+z^{2} d y+x^{2} d z ; C: \mathbb{R}(t)=(t-1) \mathbf{i}+(t+1) \mathbf{j}+t^{2} \mathbf{k}, 0 \leq t \leq 1\)

Step-by-Step Solution

Answer

2.7

Step 1 - Parameterize the curve

The curve is given by \(\mathbf{r}(t)=(t-1)\mathbf{i}+(t+1)\mathbf{j}+t^{2}\mathbf{k}, \ 0 \leq t \leq 1.\) So, \(x=t-1, \ y=t+1, \ z=t^{2}.\)

Step 2 - Compute the derivatives

Calculate the derivatives of each component with respect to t: \(\frac{dx}{dt}=1, \ \frac{dy}{dt}=1, \ \frac{dz}{dt}=2t.\)

Step 3 - Substitute the parameterized variables and their derivatives into the line integral

Rewrite the integral in terms of t: \(\int_{c} y^{2}dx+ z^{2}dy+ x^{2}dz = \int_{0}^{1}((t+1)^{2}(1) + (t^{2})^{2}(1) + (t-1)^{2}(2t)) \, dt.\) So the expression becomes: \(\int_{0}^{1}(t^{2}+2t+1+t^{4}+2t(t^{2}-2t+1)) \, dt.\)

Step 4 - Simplify the integrand

Combine like terms and simplify: \(\int_{0}^{1}(t^{2}+2t+1+t^{4}+2t^{3}-4t^{2}+2t) \, dt.\) Which simplifies to: \(\int_{0}^{1}(t^{4}+2t^{3}-3t^{2}+4t+1) \, dt.\)

Step 5 - Integrate with respect to t

Now integrate term by term: \(\int_{0}^{1}(t^{4}) \, dt + \int_{0}^{1}(2t^{3}) \, dt + \int_{0}^{1}(-3t^{2}) \, dt + \int_{0}^{1}(4t) \, dt + \int_{0}^{1}(1) \, dt.\)

Step 6 - Evaluate the definite integrals

1. \(\int_{0}^{1}t^{4} \, dt = \left[ \frac{t^{5}}{5} \right]_{0}^{1} = \frac{1}{5}.\) 2.\(\int_{0}^{1}2t^{3} \, dt = \left[ \frac{2t^{4}}{4} \right]_{0}^{1} = \frac{1}{2}.\) 3. \(\int_{0}^{1}-3t^{2} \, dt = \left[ -t^{3} \right]_{0}^{1}=-1.\)4.\(\int_{0}^{1}4t \, dt = \left[ 2t^{2} \right]_{0}^{1} = 2.\) 5. \(\int_{0}^{1}1 \, dt = \left[ t \right]_{0}^{1} = 1.\) Summing these values: \(\frac{1}{5} + \frac{1}{2} - 1 + 2 + 1 = \frac{1}{5} + \frac{1}{2} - 1 + 3 = \frac{1}{5} + \frac{1}{2} + 2 = \frac{2}{10} + \frac{5}{10} + 2 = \frac{7}{10} + 2 = \frac{7}{10} + \frac{20}{10} = \frac{27}{10} = 2.7\)

Key Concepts

Parameterization of Curves

Parameterization is the process of defining a curve using a parameter, often denoted as \(t\). In this exercise, the curve \(\textbf{r}(t)=(t-1)\mathbf{i}+(t+1)\mathbf{j}+t^{2}\mathbf{k}\) is given in parameterized form, which means each component \(x, y, z\) is expressed as functions of \(t\). For our specific problem:

• \(x=t-1\)
• \(y=t+1\)
• \(z=t^{2}\)

Determining these forms is the first step in evaluating the line integral. This parameterization simplifies substitution during the integration process and sets the path for further calculation.
The parameter \(t\) ranges between 0 and 1 for this curve, defining the limits of our integration.

Integrating Vector Fields

Vector fields describe a vector quantity at every point in space. Line integrals in vector fields help find the total of this quantity along a curve. Before integrating, the vector field components must be expressed in terms of the parameter \(t\). In our problem, the integral \(\int_{c} y^{2} dx+ z^{2} dy+ x^{2} dz\) is transformed using derivatives of the parameterized curve:

• \(\frac{dx}{dt}=1\)
• \(\frac{dy}{dt}=1\)
• \(\frac{dz}{dt}=2t\)

Substituting back into the integral, we get:
\(\int_{0}^{1} ( (t+1)^{2}(1) + (t^{2})^{2}(1) + (t-1)^{2}(2t) ) \, dt\)
Simplify and integrate each term separately to evaluate the contribution of each vector field component along the curve.

Definite Integrals in Calculus

Definite integrals are used to find the net area under a curve from one point to another. In vector calculus, they help compute quantities like work done by a force field along a curve. After obtaining the integrand in our problem as \(\int_{0}^{1} (t^{4}+2t^{3}-3t^{2}+4t+1) \, dt\), split the integral into simpler parts:
\(\int_{0}^{1} t^{4} \, dt + \int_{0}^{1} 2t^{3} \, dt + \int_{0}^{1} -3t^{2} \, dt + \int_{0}^{1} 4t \, dt + \int_{0}^{1} 1 \, dt\)
Calculate each term individually:

• \(\int_{0}^{1} t^{4} \, dt = [\frac{t^{5}}{5}]_{0}^{1} = \frac{1}{5}\)
• \(\int_{0}^{1} 2t^{3} \, dt = [\frac{2t^{4}}{4}]_{0}^{1} = \frac{1}{2}\)
• \(\int_{0}^{1} -3t^{2} \, dt = [ -t^{3} ]_{0}^{1} = -1\)
• \(\int_{0}^{1} 4t \, dt = [ 2t^{2} ]_{0}^{1} = 2\)
• \(\int_{0}^{1} 1 \, dt = [ t ]_{0}^{1} = 1\)

Sum these results: \(\frac{1}{5} + \frac{1}{2} -1 +2 +1 = 2.7\). This final value represents the evaluated line integral.