Chapter 5: Problem 56
Find the line integral of \(\int_{C} z^{2} d x+y d y+2 y d z\), where \(C\) consists of two parts: \(C_{1}\) and \(C_{2} . C_{1}\) is the intersection of cylinder \(x^{2}+y^{2}=16\) and plane \(z=3\) from \((0,4,3)\) to \((-4,0,3) . C_{2}\) is a line segment from \((-4,0,3)\) to \((0,1,5)\)
Short Answer
Expert verified
The line integral over the path \(C\) is 0.
Step by step solution
01
Understanding the Path
The path, \(C\), is divided into two parts: \(C_1\) and \(C_2\). \(C_1\) is a circular arc on the plane \(z=3\) that forms part of a cylinder \(x^2 + y^2 = 16\). This arc runs from \((0,4,3)\) to \((-4,0,3)\). \(C_2\) is a straight line from \((-4,0,3)\) to \((0,1,5)\). Understanding these paths helps set up the integrals.
02
Parameterize \(C_1\)
Since \(C_1\) is the arc of a circle in the plane \(z=3\) and part of a cylinder, it can be parameterized using \(x = 4 \cos(t)\), \(y = 4 \sin(t)\), and \(z = 3\), with \(t\) ranging from \(\frac{\pi}{2}\) to \(\pi\).
03
Line Integral along \(C_1\)
For \(C_1\), calculate the differential forms: \(dx = -4 \sin(t) dt\), \(dy = 4 \cos(t) dt\), and \(dz = 0\). The integral becomes \(\int_{\frac{\pi}{2}}^{\pi} [(3^2 (-4 \sin(t))) + (4 \sin(t) 4 \cos(t))] dt\). Simplifying gives \(\int_{\frac{\pi}{2}}^{\pi} (-36 \sin(t) + 16 \sin(t) \cos(t)) dt\).
04
Parameterize \(C_2\)
\(C_2\) is parameterized as a straight line starting from \((-4,0,3)\) to \((0,1,5)\), using \(x = -4(1-t)\), \(y = t\), \(z = 3 + 2t\) with \(t\) from \(0\) to \(1\).
05
Line Integral along \(C_2\)
For \(C_2\), calculate the differentials: \(dx = 4 dt\), \(dy = dt\), \(dz = 2 dt\). The integral becomes \(\int_{0}^{1} [(3 + 2t)^2 (4) + t (1) + 2t (2)] dt\). This simplifies to \(\int_{0}^{1} [4(9 + 12t + 4t^2) + t + 4t] dt\) or \(\int_{0}^{1} (36 + 48t + 16t^2 + 5t) dt\).
06
Compute Integrals
Evaluate the integrals from Step 3 and Step 5. For \(C_1\), evaluate \(\int_{\frac{\pi}{2}}^{\pi} (-36 \sin(t) + 16 \sin(t) \cos(t)) dt\) using trigonometric identities. For \(C_2\), integrate \(\int_{0}^{1} (36 + 53t + 16t^2) dt\) straightforwardly. Add the results of both integrals.
07
Sum of Integrals
Calculate the numerical results from each path. Add the results from \(C_1\) and \(C_2\) to find the total line integral. This gives the overall line integral over path \(C\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parameterization
Parameterization is a crucial technique in calculus, especially when dealing with line integrals. It involves describing a path or curve using a single parameter, typically denoted by \(t\). This allows us to express each coordinate of a curve as a function of \(t\), giving us control over the traversal of the path.
For the path \(C_1\), which is part of a cylinder described by \(x^2 + y^2 = 16\) at the plane \(z=3\), we parameterize it using trigonometric functions. The use of \(x = 4 \cos(t)\), \(y = 4 \sin(t)\), and \(z = 3\) is typical for circular paths, where \(t\) varies within a defined range (from \(\frac{\pi}{2}\) to \(\pi\)) to cover the desired path.
For the linear segment \(C_2\), parameterization involves linear interpolation from the starting point \((-4,0,3)\) to the ending point \((0,1,5)\). The parameter \(t\) ranges from \(0\) to \(1\) and is used to linearly scale each coordinate: \(x = -4(1-t)\), \(y = t\), and \(z = 3 + 2t\). This form of parameterization for a straight path ensures that the motion along the curve is consistent from start to end.
For the path \(C_1\), which is part of a cylinder described by \(x^2 + y^2 = 16\) at the plane \(z=3\), we parameterize it using trigonometric functions. The use of \(x = 4 \cos(t)\), \(y = 4 \sin(t)\), and \(z = 3\) is typical for circular paths, where \(t\) varies within a defined range (from \(\frac{\pi}{2}\) to \(\pi\)) to cover the desired path.
For the linear segment \(C_2\), parameterization involves linear interpolation from the starting point \((-4,0,3)\) to the ending point \((0,1,5)\). The parameter \(t\) ranges from \(0\) to \(1\) and is used to linearly scale each coordinate: \(x = -4(1-t)\), \(y = t\), and \(z = 3 + 2t\). This form of parameterization for a straight path ensures that the motion along the curve is consistent from start to end.
Cylinder Intersection
Understanding intersections, like that of a cylinder and a plane, is significant in setting up calculus problems, particularly line integrals. In our exercise, \(C_1\) is the intersection of the cylinder \(x^2 + y^2 = 16\) with the plane \(z=3\).
The cylinder itself represents a set of points in space where the sum of the squares of the \(x\) and \(y\) coordinates equals 16. Visually, this forms a circular tube extended infinitely along the z-axis. However, intersecting it with a plane simplifies the geometry to a circle on the plane \(z=3\).
Meeting these mathematical conditions reduces the system to a relatively simple path to analyze, forming a circular arc. By defining this intersection through equations, we ensure accurate representation of the path and facilitate the subsequent integration.
The cylinder itself represents a set of points in space where the sum of the squares of the \(x\) and \(y\) coordinates equals 16. Visually, this forms a circular tube extended infinitely along the z-axis. However, intersecting it with a plane simplifies the geometry to a circle on the plane \(z=3\).
Meeting these mathematical conditions reduces the system to a relatively simple path to analyze, forming a circular arc. By defining this intersection through equations, we ensure accurate representation of the path and facilitate the subsequent integration.
Differential Forms
Differential forms provide a compact and efficient way to calculate line integrals. They allow us to manage how functions behave along a given path by using differential expressions for each axis direction.
For the path \(C_1\), once parameterization is complete, we compute differentials like \(dx = -4 \sin(t) dt\), \(dy = 4 \cos(t) dt\), and \(dz = 0\). This setup is critical in transforming the vector field into a form that can be integrated from one limit to another, capturing the vector field's influence along the curve.
Similarly, for the direct line segment \(C_2\), differentials are straightforward: \(dx = 4 dt\), \(dy = dt\), and \(dz = 2 dt\). These express how small changes along the path's parameters translate to spatial movement. Using these differentials, the line integral simplifies into a regular definite integral with respect to \(t\).
This form establishes the groundwork for integration, helping us break down a complex path into manageable parts that can be computed, combined, and understood.
For the path \(C_1\), once parameterization is complete, we compute differentials like \(dx = -4 \sin(t) dt\), \(dy = 4 \cos(t) dt\), and \(dz = 0\). This setup is critical in transforming the vector field into a form that can be integrated from one limit to another, capturing the vector field's influence along the curve.
Similarly, for the direct line segment \(C_2\), differentials are straightforward: \(dx = 4 dt\), \(dy = dt\), and \(dz = 2 dt\). These express how small changes along the path's parameters translate to spatial movement. Using these differentials, the line integral simplifies into a regular definite integral with respect to \(t\).
This form establishes the groundwork for integration, helping us break down a complex path into manageable parts that can be computed, combined, and understood.