Problem 92 - Calculus Volume 3

Question

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 \((-4,0,3), \quad C_{2}\) is a line segment from (-4,0,3) to (0,1,5)

Step-by-Step Solution

Answer

The line integral along the path C is \(\frac{68}{3}.\)

Step 1: Review Parametrization for Curve C1

Curve \(C_1\) lies on the circle \(x^2+y^2=16\) and plane \(z=3\). The path from \((0,4,3)\) to \((-4,0,3)\) suggests using trigonometric functions. A suitable parametrization is \(x=4\cos(t), y=4\sin(t), z=3\) for \(t\) between \(\frac{\pi}{2}\) and \(\pi\).

Step 2: Differentiate Parametrization for C1

Differentiate parametrization with respect to \(t\): \(dx=-4\sin(t)dt\), \(dy=4\cos(t)dt\), and \(dz=0\).

Step 3: Integrate Over C1

Substitute in the line integral expression for \(C_1\): \[ \int_{C_1} z^2 dx + y dy + 2y dz = \int_{\frac{\pi}{2}}^{\pi} 9(-4\sin(t))dt + 4\sin(t)(4\cos(t))dt + 0. \] Compute each integral separately and then sum them.

Step 4: Evaluate the Integrals for C1

The first term is \(-36\int_{\frac{\pi}{2}}^{\pi} \sin(t) dt = 36\left[\cos(t)\right]_{\frac{\pi}{2}}^{\pi} = 36(-1 - 0) = -36\). The second term is \(16\int_{\frac{\pi}{2}}^{\pi} \sin(t)\cos(t) dt = 8\left[\sin^2(t)\right]_{\frac{\pi}{2}}^{\pi} = 8(0 - 1) = -8\). Thus the total integral over \(C_1\) is \(-36 - 8 = -44\).

Step 5: Parametrize C2 as a Line Segment

For curve \(C_2\), parametrize the line from \((-4,0,3)\) to \((0,1,5)\) as \(x = -4 + 4t, y = 0 + t, z = 3 + 2t\) for \(t\) in \([0,1]\).

Step 6: Differentiate Parametrization for C2

Calculate the differentials: \(dx = 4dt\), \(dy = dt\), \(dz = 2dt\).

Step 7: Integrate Over C2

Substitute these into the integral for \(C_2\): \[ \int_{C_2} (z^2 dx + y dy + 2y dz) = \int_{0}^{1} ((3+2t)^2(4) + t(1) + 2t(2))dt. \]

Step 8: Evaluate the Single Integral for C2

Compute \(4\int_0^1 (9 + 12t + 4t^2)dt + \int_0^1 5tdt = 36 + 24 + \frac{8}{3} = 60 + \frac{8}{3}\). Evaluate it as \[64 + \frac{8}{3} = \frac{192}{3} + \frac{8}{3} = \frac{200}{3}.\]

Step 9: Sum the Integrals Over C1 and C2

Add the integrated results from \(C_1\) and \(C_2\): \(-44 + \frac{200}{3}\). Convert \(-44\) to thirds to find the total: \(-132/3 + 200/3 = 68/3.\)

Key Concepts

Parametrization

Parametrization is a fundamental concept in calculus that involves expressing a geometric curve or surface using one or more parameters. This technique is especially useful in computing line integrals, as it allows us to express each coordinate using a single variable. For curve \(C_1\), which is part of a circular path on the cylinder with the plane \(z=3\), we use trigonometric functions to express the path in terms of \(t\).

• \(x = 4\cos(t)\),
• \(y = 4\sin(t)\),
• \(z = 3\)

This representation helps in simplifying the integration process over the path. Parametrization not only allows us to describe paths effectively but also accommodates complexities involving curves and spatial coordinates.

Cylinder

A cylinder in mathematics is a three-dimensional surface composed of all the points equidistant from a line (the axis). The given cylinder's equation is \(x^2 + y^2 = 16\). This implies it is circular in the \(xy\)-plane with a radius of 4. Consider the cylinder's intersection with the plane \(z=3\), creating a circle at a constant height of 3. This is crucial as it simplifies our line integral calculation by keeping \(z\) constant. Cylindrical shapes often make complex spatial problems more manageable, as they allow decomposition into simpler components, such as circles, which are easier to parametrize and manipulate mathematically.

Plane Intersection

The intersection of geometric shapes, such as a cylinder and a plane, leads to simpler curves which can be analyzed with line integrals. In this problem, plane \(z=3\) cuts the cylinder \(x^2+y^2=16\), forming a circular path at \(z=3\). This path forms the curve \(C_1\) from

• (0,4,3)
• to > (-4,0,3)

Intersections are valuable in geometry as they break down complex structures into simple ones, facilitating analysis and computation. Understanding how intersections work helps predict these simple curves and paths within more extensive integrations.

Differentiation

Differentiation brings the change in parameters into account, crucial for computing line integrals. For the curve \(C_1\), we differentiate the parametrized equations with respect to \(t\) as follows:

• \(dx = -4\sin(t)\,dt\)
• \(dy = 4\cos(t)\,dt\)
• \(dz = 0\)

This process gives us the infinitesimal segments \(dx\), \(dy\), \(dz\) to substitute back into the line integral expression. Integrating these differentials over the specified interval of the parameter allows us to calculate the line integral across the defined path. Differentiation of parametrizations ensures that each small segment of the path is accounted for, contributing to the total integral of our original expression.