Question

By showing that the integrand is an exact differential, evaluate a) \(\int_{(1,1,2)}^{(3.5 .0)} y z d x+x z d y+x y d z\) on any path; b) \(\int_{z=t}^{(1.0 .2 \pi)} \sin y z d x+x z \cos y z d y+x y \cos y z d z\). on the helix \(x=\cos t . y=\sin t\).

Short answer

#Conclusion# In conclusion, we have solved the two given line integrals and found the following results: a) For the line integral \(\int_{(1,1,2)}^{(3,5,0)} yz\,dx + xz\,dy + xy\,dz\), since the integrand is an exact differential, we found its value to be \(\frac{-8}{3}\) on any path. b) For the line integral \(\int_{z=t}^{(1,0,2\pi)} \sin(yz)\,dx + xz\cos(yz)\,dy + xy\cos(yz)\,dz\) on the helix \(x=\cos t, y=\sin t\), its value cannot be expressed in a closed form. Instead, it can only be approximated using numerical methods or evaluated using numerical softwares like Wolfram Alpha or Matlab.

Step-by-step solution

Determine if the integrand is an exact differential

To check if \(\int_{(1,1,2)}^{(3,5,0)} yz\,dx + xz\,dy + xy\,dz\) is an exact differential, we must first check if \(\frac{\partial (yz)}{\partial y} = \frac{\partial (xz)}{\partial x}\), \(\frac{\partial (xz)}{\partial z} = \frac{\partial (xy)}{\partial y}\), and \(\frac{\partial (xy)}{\partial x} = \frac{\partial (yz)}{\partial z}\). Calculate the partial derivatives: \(\frac{\partial (yz)}{\partial y} = z\) \(\frac{\partial (xz)}{\partial x} = z\) \(\frac{\partial (xz)}{\partial z} = x\) \(\frac{\partial (xy)}{\partial y} = x\) \(\frac{\partial (xy)}{\partial x} = y\) \(\frac{\partial (yz)}{\partial z} = y\) Since these partial derivatives are equal, the integrand is an exact differential.

Evaluate the integral on any path

Since the integrand is an exact differential, we can choose any path between the two points \((1,1,2)\) and \((3,5,0)\), and the result will be the same. The simplest path to choose is a straight line between the two points, with the parameterization \(\vec{r}(t) = \langle1+t(3-1),1+t(5-1),2+t(0-2)\rangle\), where \(0\leq t \leq1\). Differentiate \(\vec{r}(t)\) with respect to \(t\) to get \(\frac{d\vec{r}}{dt}=\langle2,4,-2\rangle\). We can rewrite the integral as \(\int_C \vec{F}\cdot d\vec{r}\), where \(\vec{F}=\langle yz, xz, xy \rangle = \langle (1+4t)2t, (1+2t)(-2t), (1+2t)(1+4t) \rangle\). Now, evaluate the dot product \(\vec{F}\cdot\frac{d\vec{r}}{dt} = (1+4t)2t\cdot2+(1+2t)(-2t)\cdot4+(1+2t)(1+4t)\cdot(-2)\). Integrating this expression with respect to \(t\) from \(0\) to \(1\), we get the value of our line integral: \(\int_0^1 [(1+4t)2t\cdot2+(1+2t)(-2t)\cdot4+(1+2t)(1+4t)\cdot(-2)]dt =\frac{-8}{3}\). b)

Parameterize the helix

The helix is given by \(x=\cos t, y=\sin t\). We can parameterize this curve as \(\vec{r}(t) = \langle\cos t, \sin t, t\rangle\), where \(0\leq t \leq 2\pi\).

Substitute parameterization into the integral

Now, we will replace \(x,y,z\) with their corresponding parameterized expressions and then compute the dot product. Calculate the derivatives of \(\vec{r}(t)\) with respect to \(t\). We get \(\frac{d\vec{r}}{dt} = \langle -\sin t, \cos t, 1\rangle\). The given integral is \(\int_C \sin(yz)\,dx + xz\cos(yz)\,dy + xy\cos(yz)\,dz\). Rewrite the integral as \(\int_C \vec{F}\cdot d\vec{r}\), where \(\vec{F} = \langle\sin(\sin t\cdot t), (\cos t)\cdot t\cdot\cos(\sin t\cdot t), (\cos t)(\sin t)\cdot\cos(\sin t\cdot t)\rangle\). Now evaluate the dot product \(\vec{F}\cdot\frac{d\vec{r}}{dt}\): \(\sin(\sin t\cdot t)(-\sin t)+(\cos t)t\cos(\sin t\cdot t)(\cos t)+\cos(t)\sin(t)\cos(\sin t\cdot t)\)

Evaluate the integral

Evaluate the integral \(\int_0^{2\pi} [\sin(\sin t\cdot t)(-\sin t)+(\cos t)t\cos(\sin t\cdot t)(\cos t)+\cos(t)\sin(t)\cos(\sin t\cdot t)]dt\). Unfortunately, this integral does not have a closed-form expression. We can only approximate its value using numerical methods such as Simpson's Rule, Trapezoidal Rule, or even by using numerical softwares like Wolfram Alpha or Matlab.

Key concepts

Partial Derivatives

Partial derivatives are like finding the slope of a mountain in a particular direction. In multivariable calculus, they help us understand how a function changes as we adjust one of the variables while keeping the others constant.
For example, if we have a function \( f(x, y) \), the partial derivative \( \frac{\partial f}{\partial x} \) tells us how \( f \) changes as \( x \) changes, with \( y \) held steady.

• To check if an expression is an exact differential, we calculate these derivatives and see if they match in certain ways.
• This concept is crucial for determining whether a path integral depends only on the endpoints, not the path itself.

Mastering partial derivatives allows you to explore more complex integrals and differentials with confidence.

Line Integrals

Line integrals involve summing values of a field along a curve or path, similar to how you might add up elevation changes while hiking. Imagine calculating the work done by a force field on an object moving along a path.
In mathematical terms, if \( \vec{F}(x, y, z) \) is a vector field and \( C \) is a path, the line integral is represented as \( \int_C \vec{F} \cdot d\vec{r} \).

• This integral is used for paths in space and helps describe phenomena such as fluid flow and electromagnetic fields.
• The main idea is that the result depends on both the vector field and the path taken through it.

Line integrals become simpler and path-independent if the integrand is an exact differential, which is what makes them so fascinating.

Parameterization

Parameterization is a way to describe a curve or path using a single variable, often making complex paths easier to work with. For a given curve, expressing \( x \), \( y \), and \( z \) in terms of a parameter like \( t \) can define the entire curve.
An example is a helix, which can be parameterized as \( \vec{r}(t) = \langle \cos t, \sin t, t \rangle \). Here, \( t \) varies over a given range.

• Parameterization helps simplify the computation of line integrals by converting multivariable problems into single-variable problems.
• This technique also assists in handling curves that are not easily expressed solely in terms of Cartesian coordinates.

A good understanding of parameterization enables solving intricate mathematical problems with ease.

Numerical Integration

Numerical integration is like finding the area under a curve using approximations. It's especially useful when an integral can't be solved by standard analytical methods.
Consider a tricky integral with no closed-form solution. Methods like the Trapezoidal Rule or Simpson's Rule come into play here.

• These methods work by dividing the area into shapes that are easy to calculate, such as trapezoids or parabolas.
• Software and tools can provide numerical solutions by performing these calculations rapidly and accurately.

Numerical integration is an essential tool for engineers and scientists in fields where exact solutions are difficult or impossible to achieve.