Problem 25 - Calculus: Early Transcendental Functions

Question

Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}.\) \(\mathbf{F}(x, y)=\left\langle 3 x^{2} y+1,3 x y^{2}\right\rangle, C\) is the bottom half-circle from \((1,0)\) to \((-1,0)\)

Step-by-Step Solution

Answer

To calculate the line integral of the vector field over the given curve, we first parametrize the curve, then substitute this parameterization into the vector field, after which we take the dot product with \(d\mathbf{r}\) and finally integrate over the given limits of \(t\). The detailed computation would depend on how comfortable you are with computing trigonometric integrals

Step 1: Parametrize the Curve

The curve is a semi-circle and can be parametrized as follows: \(\mathbf{r}(t) = \langle \cos(t), -\sin(t) \rangle\), \(t\) would vary from \(\pi\) to \(2\pi\) because it's the bottom half circle. The differential vector \(d \mathbf{r}\) would be the derivative of \(\mathbf{r}(t)\) with respect to \(t\). Thus, \(d \mathbf{r} = (-\sin(t),-\cos(t))dt\).

Step 2: Calculate Dot Product

Next, substitute \(\mathbf{r}(t)\) into \(\mathbf{F}(x,y)\) to get \(\mathbf{F}(r(t))\). We get \(\mathbf{F}(r(t)) = \langle 3\cos(t)^{2}(-\sin(t))+1,3\cos(t)(-\sin(t))^{2}\rangle\). Now, to evaluate \(\mathbf{F} \cdot d\mathbf{r}\), take the dot product of \(\mathbf{F}(r(t))\) and \(d\mathbf{r}\). This gives: \(\mathbf{F}(r(t)) \cdot d\mathbf{r}= [-3\cos^{2}(t)\sin(t)-\sin(t)+1]*(-\sin(t))dt + [3\cos(t)\sin^{2}(t)]*(-\cos(t))dt= [3\cos^{2}(t)\sin^{2}(t)+\sin^{2}(t)-\sin(t)]dt \).

Step 3: Evaluate Integral

Now integrate from \(t=\pi\) to \(t=2\pi\). So, \(\int_{\pi}^{2\pi} [3\cos^{2}(t)\sin^{2}(t)+\sin^{2}(t)-\sin(t)]dt\). Calculate this integral to find the value.

Key Concepts

Line Integral of Vector Fields

The method of evaluating line integrals of vector fields is essential for understanding various physical concepts, such as work done by a force field. Picture walking along a curve while pushing a box; the force you exert along this path can be described by a vector field. In a mathematical sense, a vector field is a function that assigns a vector to every point in a space.

A line integral extends this idea to find the cumulative effect of a vector field along a path. If we have a curve C and a vector field \(\mathbf{F}(x, y)\), the line integral is written as \(\int_C \mathbf{F} \cdot d\mathbf{r}\), where \(d\mathbf{r}\) represents a tiny bit of the curve. This line integral effectively adds up the dot product of the field vector with the differential path vector over the entire curve.

To evaluate a line integral, you need the exact definition of your path in space, which is where parametrization comes in.

Parametrization of Curves

To compute the line integral, we must first express the path over which we are integrating in a format that we can work with algebraically. This is accomplished through parametrization, which provides us a way to describe a curve with a single parameter, typically t.

In our given problem, the curve C is a semi-circle, parametrized by the function \(\mathbf{r}(t) = \langle \cos(t), -\sin(t) \rangle\), with t ranging from \(\pi\) to \(2\pi\) to capture the bottom half-circle. The position vector \(\mathbf{r}(t)\) essentially 'traces' the curve as t varies.

It's crucial to choose a parametrization that simplifies the process of integration while correctly representing the curve. Once the curve is parametrized, we can proceed to evaluate the line integral using this description.

Dot Product in Line Integrals

After parametrization, the next step involves the dot product. The dot product combines two vectors and results in a scalar. In the context of line integrals, it measures the component of one vector along another.

For the dot product in line integrals, we evaluate \(\mathbf{F}(\mathbf{r}(t))\), which is the vector field along the curve, and then dot it with the differential element of the curve \(d\mathbf{r}\). In our exercise, \(d\mathbf{r}\) is derived from the derivative of the parametrization with respect to t. The resulting expression from the dot product is a scalar function that depends on t, which is then integrated over the interval corresponding to the curve.

The dot product serves a critical role by bridging the vector field and the curve's parametrization, allowing for the evaluation of the line integral along the parametrized curve.