Problem 37
Question
Verify that the line integral \(\int_{C} y^{2} d x+x y d y\) has the same value on \(C\) for each of the following parameterizations: $$ \begin{array}{lll} C: x=2 t+1, & y=4 t+2, & 0 \leq t \leq 1 \\ C: x=t^{2}, & y=2 t^{2}, & 1 \leq t \leq \sqrt{3} \\ C: x=\ln t, & y=2 \ln t, & e \leq t \leq e^{3}. \end{array} $$
Step-by-Step Solution
Verified Answer
The line integrals for each parameterization yield the same result.
1Step 1: Understand the Line Integral
The line integral in question is \[ \int_{C} y^{2} dx + xy dy. \]This integral computes the accumulation of a vector field along a path or curve \(C\). It is given in terms of the parameterizations of \(x\) and \(y\) along different segments of \(C\).
2Step 2: Parameterize the Integral for the First Segment
For the first parameterization \(x = 2t + 1\) and \(y = 4t + 2\), with \(0 \leq t \leq 1\), we have:* First, find \(dx\) by differentiating \(x\): \[ dx = \frac{d}{dt} (2t + 1)\, dt = 2 \, dt. \]* Next, find \(dy\) by differentiating \(y\): \[ dy = \frac{d}{dt} (4t + 2)\, dt = 4 \, dt. \]* Substitute \(x\), \(y\), \(dx\), and \(dy\) into the integral: \[ \int_{0}^{1} ((4t+2)^2 \times 2 + (2t+1)(4t+2) \times 4) dt. \]
3Step 3: Calculate the Integral for the First Segment
Solve the integral:* Simplify the integrand: \[ (4t+2)^2 = 16t^2 + 16t + 4, \] \[ 2(16t^2 + 16t + 4) + (2t+1)(4t+2) \times 4 \].* Further expansion yields: \[ 32t^2 + 32t + 8 + 4((2t+1)(4t+2)) \].* Finally, integrate term by term from \(t=0\) to \(t=1\): \[ \int_{0}^{1} 32t^2 + 32t + 8 + 32t^2 + 40t + 8 \ dt. \]
4Step 4: Parameterize the Integral for the Second Segment
For the second parameterization \(x = t^2\) and \(y = 2t^2\), with \(1 \leq t \leq \sqrt{3}\):* Compute \(dx\) and \(dy\): \[ dx = \frac{d}{dt}(t^2)\, dt = 2t \, dt. \] \[ dy = \frac{d}{dt}(2t^2)\, dt = 4t \, dt. \]* The integral becomes: \[ \int_{1}^{\sqrt{3}} ((2t^2)^2 \times 2t + (t^2)(2t^2) \times 4t) dt. \]
5Step 5: Calculate the Integral for the Second Segment
Evaluate the integral:* Compute \((2t^2)^2 = 4t^4\) and expand the integrand: \[ \int_{1}^{\sqrt{3}} (8t^5 + 8t^7) \, dt. \]* Integrate term by term between \(t=1\) and \(t=\sqrt{3}\).
6Step 6: Parameterize the Integral for the Third Segment
For the third parameterization \(x = \ln t\) and \(y = 2\ln t\), with \(e \leq t \leq e^3\):* Differentiate to find \(dx\) and \(dy\): \[ dx = \frac{1}{t} dt, \] \[ dy = \frac{2}{t} dt. \]* Substitute into the integral: \[ \int_{e}^{e^3} ((2\ln t)^2 \frac{1}{t} + (\ln t)(2\ln t) \frac{2}{t}) dt. \]
7Step 7: Calculate the Integral for the Third Segment
Evaluate the integral:* Simplify the terms: \[ (2\ln t)^2 = 4(\ln t)^2 \] \[ \int_{e}^{e^3} (\frac{4(\ln t)^2 + 2(\ln t)^2 \ln t}{t}) dt. \]* Then, compute the integrals over \(t = e\) to \(t = e^3\).
8Step 8: Verify Results and Compare
Evaluate each integral and compare results. After calculating, verify if each integral evaluates to the same value as expected along the curve segments.
Key Concepts
ParameterizationVector FieldsCurve SegmentsIntegration Methods
Parameterization
Understanding parameterization is crucial when solving problems related to line integrals. When we talk about parameterization, we refer to the process of expressing the variables of a curve in terms of a single parameter, often denoted by \( t \). This helps in tracing the path of the curve in an orderly manner.
In our exercise, we have three different parameterizations:\[ x = 2t + 1, \qquad y = 4t + 2 \] with \( 0 \leq t \leq 1 \), \[ x = t^2, \qquad y = 2t^2 \]with \( 1 \leq t \leq \sqrt{3} \), and \[ x = \ln t, \qquad y = 2 \ln t \] with \( e \leq t \leq e^3 \).
By changing \( t \), we can move along different segments of the curve. The parameterization must be continuous and cover the entire segment of interest.
In our exercise, we have three different parameterizations:\[ x = 2t + 1, \qquad y = 4t + 2 \] with \( 0 \leq t \leq 1 \), \[ x = t^2, \qquad y = 2t^2 \]with \( 1 \leq t \leq \sqrt{3} \), and \[ x = \ln t, \qquad y = 2 \ln t \] with \( e \leq t \leq e^3 \).
By changing \( t \), we can move along different segments of the curve. The parameterization must be continuous and cover the entire segment of interest.
Vector Fields
A vector field is essentially a map that assigns a vector to every point in space. When dealing with line integrals, we often work with vector fields that describe the force acting in different regions along the curve.
In the given line integral, \( \int_C y^2 \, dx + xy \, dy \), the expression \( y^2 \, dx + xy \, dy \) represents the vector field through which we evaluate the line integral. Essentially, it describes how the function behaves or changes as you move along the path of the curve \( C \).
Understanding how to interpret and manipulate vector fields is crucial, as it can determine how the function accumulates across the curve, providing insight into physical phenomena like work done by a force field.
In the given line integral, \( \int_C y^2 \, dx + xy \, dy \), the expression \( y^2 \, dx + xy \, dy \) represents the vector field through which we evaluate the line integral. Essentially, it describes how the function behaves or changes as you move along the path of the curve \( C \).
Understanding how to interpret and manipulate vector fields is crucial, as it can determine how the function accumulates across the curve, providing insight into physical phenomena like work done by a force field.
Curve Segments
Curve segments refer to specific portions or paths of the overall curve. Each segment is often described by its own parameterization. This can simplify the evaluation of line integrals as it breaks a complex curve into manageable sections.
In our example, the curve \( C \) is broken into three segments, each with its unique parameterization and limits for \( t \):
In our example, the curve \( C \) is broken into three segments, each with its unique parameterization and limits for \( t \):
- First segment: \( 0 \leq t \leq 1 \)
- Second segment: \( 1 \leq t \leq \sqrt{3} \)
- Third segment: \( e \leq t \leq e^3 \)
Integration Methods
Evaluating line integrals requires the use of specific integration methods tailored to the parameterization of the path. Once parameterization is complete, we can express \( dx \) and \( dy \) in terms of \( t \) to substitute into the integral.
The process generally involves:
During integration: analyze the symmetry and behavior of functions over each segment to understand the overall effect of the vector field along the curve. This approach helps verify the integrals' equality for all parameterizations.
The process generally involves:
- Differentiating \( x \) and \( y \) to find \( dx \) and \( dy \)
- Substituting these expressions back into the integral
- Carrying out the integration over the specified range for \( t \)
During integration: analyze the symmetry and behavior of functions over each segment to understand the overall effect of the vector field along the curve. This approach helps verify the integrals' equality for all parameterizations.
Other exercises in this chapter
Problem 37
Evaluate the given iterated integral by reversing the order of integration. $$ \int_{0}^{2} \int_{y^{2}}^{4} \cos \sqrt{x^{3}} d x d y $$
View solution Problem 37
Any scalar function \(f\) for which \(\nabla^{2} f=0\) is said to be harmonic. Verify that \(f(x, y, z)=\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}\) is harmonic ex
View solution Problem 37
In Problems, find a vector function \(\mathbf{r}\) that satisfies the indicated conditions. $$ \mathbf{r}^{\prime}(t)=6 \mathbf{i}+6 t \mathbf{j}+3 t^{2} \mathb
View solution Problem 37
Show that every normal line to the graph \(x^{2}+y^{2}+z^{2}=a^{2}\) passes through the origin.
View solution