Question

Evaluate the line integral using the Fundamental Theorem of Line Integrals. Use a computer algebra system to verify your results. \(\int_{C}-\sin x d x+z d y+y d z\) \(C:\) smooth curve from (0,0,0) to \(\left(\frac{\pi}{2}, 3,4\right)\)

Short answer

The value of the line integral is 23.5.

Step-by-step solution

Find the Potential Function

A vector field \((M,N,P)\) is conservative and hence has a potential function if and only if it satisfies the following relationships: \(\frac{\partial M}{\partial x} = \frac{\partial N}{\partial y}\) and \(\frac{\partial N}{\partial z} = \frac{\partial P}{\partial y}\). For \((M,N,P) = (-sin(x), z, y)\), we have \(\frac{\partial M}{\partial x} = -cos(x)\) and \(\frac{\partial N}{\partial y} = 0\). Also, \(\frac{\partial N}{\partial z} = 1\) and \(\frac{\partial P}{\partial y} = 1\). As these relationships hold, it follows that this vector field is conservative and has a potential function F(x,y,z). We find F(x,y,z) by integrating: \(F(x,y,z) = -\int sin(x) dx + \int z dy + \int y dz = cos(x) + yz + f(y, z)\). Here, f(y, z) is a function of y and z whose partial derivative with respect to z is z and with respect to y is y. One function satisfying this condition is \(\frac{1}{2}(y^2 + z^2)\). So the potential function is \(F(x,y,z) = cos(x) + yz + \frac{1}{2}(y^2 + z^2)\).

Apply the Fundamental Theorem of Line Integrals

We know that the Fundamental Theorem of Line Integrals states that if \(F\) is the potential function of a vector field and \(C\) is a curve from point \(a\) to point \(b\), then \(\int_{C} Fdr = F(b) - F(a)\). Hence, we evaluate \(F\) at points \(\left(0,0,0\right)\) and \(\left(\frac{\pi}{2}, 3, 4\right)\). We get \(F\left(0,0,0\right) = 1 + 0 + 0 = 1\) and \(F\left(\frac{\pi}{2}, 3, 4\right) = 0 + 3*4 + \frac{1}{2}(3^2 + 4^2) = 12 + 12.5 = 24.5\).

Calculate the Line Integral

Then, by the Fundamental Theorem of Line Integrals, the line integral of \(F\) from \(\left(0,0,0\right)\) to \(\left(\frac{\pi}{2}, 3, 4\right)\) is \(F\left(\frac{\pi}{2}, 3, 4\right) - F\left(0,0,0\right) = 24.5 - 1 = 23.5\).

Key concepts

Line Integral

When we talk about line integrals, we're essentially looking at a way to sum up the effects of a vector field along a specific path or curve. Imagine walking through a hilly terrain with varying slopes. If you want to calculate the work done while walking along a particular path, you need to consider the force (coming from the vector field) and the direction you're walking (the path, or 'line').

In mathematics, a line integral along a curve C from point A to B is represented as \(\int_C \mathbf{F} \cdot d\mathbf{r}\), where \mathbf{F}\ is a vector field and \mathbf{r}\ is a vector that traces out the curve. This concept is central to many areas of physics and engineering, where fields such as gravity or electromagnetism are involved.

Vector Field

Imagine a weather map showing wind direction and strength at different points; that's essentially a vector field. Mathematically, a vector field on a plane or in space assigns a vector to every point in that region. For example, the vector field \(\mathbf{F}(x, y, z) = (M, N, P)\) indicates that at any point \( (x,y,z) \) in space, there's a corresponding vector with components M, N, and P.

These fields can represent real forces, like magnetic or gravitational fields, or abstract ones, such as in fluid flow simulations. They're fundamental to understanding physical phenomena and their mathematical representations.

Potential Function

A potential function offers a neat way to simplify computing line integrals. If a vector field \(\mathbf{F}\) is conservative, which means its behavior can be captured completely by a single function, then that function is the potential function, usually denoted by \(F(x,y,z)\).

It's like having a topographical map that tells you the elevation at any point. Knowing just the elevation (the potential function), you can figure out the slope (the vector field) everywhere. For gravitational force, the potential function represents gravitational potential energy. Finding a potential function allows us to compute line integrals much more easily, as we'll see with the Fundamental Theorem of Line Integrals.

Conservative Vector Field

A conservative vector field is one where the line integral between two points does not depend on the path taken, only on the end points. This type of field has the special property that it can be expressed as the gradient of a potential function.

In the context of our problem, we check if the given vector field is conservative by verifying certain conditions involving the partial derivatives of its components. These conditions, like curl being zero, ensure that the field's behavior is consistent and path-independent — a hallmark of conservation. Environments like a central gravitational field or an electrostatic field without time-varying magnetic fields are examples of conservative vector fields.

Partial Derivatives

Partial derivatives are a way to see how a function changes as you tweak just one variable at a time while keeping others fixed. Think about adjusting the volume on your TV with one remote and the brightness with another; changing the volume affects the sound but not the picture.

In the exercise, the conditions for a vector field to have a potential function are expressed in terms of partial derivatives. By finding partial derivatives of the vector field's components and checking their equality, we identify if there's a potential function. This step is crucial for verifying whether the vector field is conservative and for applying the Fundamental Theorem of Line Integrals to calculate our line integral with ease.