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‎Advanced_Engineering_Mathematics_Kreyszig/chap10.md

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# Chapter 10
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Example 1
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A line integral is an integral of a function along a curve in
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space. We usually represent the curve by a parametric equation,
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e.g. $\mathbf{r}(t) = [x(t), y(t), z(t)] = x(t)\mathbf{i} +
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y(t)\mathbf{j} + z(t)\mathbf{k}$. So, in general the curve will be a
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vector function, and the function we want to integrate will also be a
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vector function.
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Then, we can write the line integral definition as:
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$$
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\int_C \mathbf{F(r)}\cdot d\mathbf{r} =
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\int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r'}(t) dt
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$$
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where $\mathbf{r'}(t) = \frac{d\mathbf{r}}{dt}$. This integrand is a
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scalar function, because of the dot product.
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The following examples are adapted from Chapter 10 in Advanced
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Engineering Mathematics by Kreysig.
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The first example is the evaluation of a line integral in the
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plane. We want to evaluate the integral of $\mathbf{F(r)}=[-y, -xy]$
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on the curve $\mathbf{r(t)}=[-sin(t), cos(t)]$ from t=0 to t =
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π/2. The answer in the book is given as 0.4521. Here we evaluate this
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numerically, using autograd for the relevant derivative. Since the
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curve has multiple outputs, we have to use the jacobian function to
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get the derivatives. After that, it is a simple bit of matrix
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multiplication, and a call to the quad function.
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```python
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import autograd.numpy as np
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from autograd import jacobian
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from scipy.integrate import quad
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def F(X):
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x, y = X
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return -y, -x * y
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def r(t):
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return np.array([-np.sin(t), np.cos(t)])
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drdt = jacobian(r)
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def integrand(t):
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return F(r(t)) @ drdt(t)
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I, e = quad(integrand, 0.0, np.pi / 2)
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print(f'The integral is {I:1.4f}.')
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```
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```text
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The integral is 0.4521.
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```
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We get the same result as the analytical solution.
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The next example is in three dimensions. Find the line integral along
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$\mathbf{r}(t)=[cos(t), sin(t), 3t]$ of the function
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$\mathbf{F(r)}=[z, x, y]$ from t=0 to t=2 π. The solution is given as
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21.99.
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```python
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import autograd.numpy as np
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from autograd import elementwise_grad, grad, jacobian
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def F(X):
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x, y, z = X
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return [z, x, y]
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def C(t):
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return np.array([np.cos(t), np.sin(t), 3 * t])
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dCdt = jacobian(C, 0)
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def integrand(t):
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return F(C(t)) @ dCdt(t)
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I, e = quad(integrand, 0, 2 * np.pi)
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print(f'The integral is {I:1.2f}.')
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```
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```text
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The integral is 21.99.
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```
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That is also the same as the analytical solution. Note the real
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analytical solution was 7 π, which is nearly equivalent to our answer.
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```python
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print (7 * np.pi - I)
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```
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```text
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3.552713678800501e-15
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```
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As a final example, we consider an alternate form of the line
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integral. In this form we do not use a dot product, so the integral
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results in a vector. This doesn't require anything from autograd, but
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does require us to be somewhat clever in how to do the integrals since
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quad can only integrate scalar functions. We need to integrate each
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component of the integrand independently. Here is one approach where
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we use lambda functions for each component. You could also manually
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separate the components.
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```python
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def F(r):
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x, y, z = r
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return x * y, y * z, z
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def r(t):
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return np.array([np.cos(t), np.sin(t), 3 * t])
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def integrand(t):
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return F(r(t))
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[quad(lambda t: integrand(t)[i], 0, 2 * np.pi)[0] for i in [0, 1, 2]]
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```
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```text
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Out[1]: [-6.845005548807377e-17, -18.849555921538755, 59.21762640653615]
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```
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```python
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[0, -6 * np.pi, 6 * np.pi**2]
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```
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```text
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Out[1]: [0, -18.84955592153876, 59.21762640653615]
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```
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which is evidently the same as our numerical solution.
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Maybe an alternative, but more verbose is this vectorized integrate
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function. We still make temporary functions for integrating, and the
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vectorization is essentially like the list comprehension above, but we
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avoid the lambda functions.
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```python
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@np.vectorize
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def integrate(i):
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def integrand(t):
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return F(r(t))[i]
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I, e = quad(integrand, 0, 2 * np.pi)
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return I
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integrate([0, 1, 2])
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```
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```text
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Out[1]: array([-6.84500555e-17, -1.88495559e+01, 5.92176264e+01])
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```
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‎Advanced_Engineering_Mathematics_Kreyszig/out.html

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<html>
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<head>
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<script type="text/x-mathjax-config">MathJax.Hub.Config({ tex2jax: {inlineMath: [["$","$"] ]}});</script>
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<script type="text/javascript" src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.5/MathJax.js?config=TeX-AMS_HTML-full">
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</script>
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<meta charset='utf-8'>
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</head>
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<body>
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<h1>Chapter 10</h1>
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<p>Example 1</p>
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<p>A line integral is an integral of a function along a curve in
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space. We usually represent the curve by a parametric equation,
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e.g. $\mathbf{r}(t) = [x(t), y(t), z(t)] = x(t)\mathbf{i} +
16+
y(t)\mathbf{j} + z(t)\mathbf{k}$. So, in general the curve will be a
17+
vector function, and the function we want to integrate will also be a
18+
vector function.</p>
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<p>Then, we can write the line integral definition as:</p>
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<p>$$
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\int_C \mathbf{F(r)}\cdot d\mathbf{r} =
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\int_a^b \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r'}(t) dt
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$$</p>
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<p>where $\mathbf{r'}(t) = \frac{d\mathbf{r}}{dt}$. This integrand is a
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scalar function, because of the dot product.</p>
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<p>The following examples are adapted from Chapter 10 in Advanced
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Engineering Mathematics by Kreysig.</p>
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<p>The first example is the evaluation of a line integral in the
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plane. We want to evaluate the integral of $\mathbf{F(r)}=[-y, -xy]$
30+
on the curve $\mathbf{r(t)}=[-sin(t), cos(t)]$ from t=0 to t =
31+
π/2. The answer in the book is given as 0.4521. Here we evaluate this
32+
numerically, using autograd for the relevant derivative. Since the
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curve has multiple outputs, we have to use the jacobian function to
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get the derivatives. After that, it is a simple bit of matrix
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multiplication, and a call to the quad function.</p>
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<pre><code class="python">import autograd.numpy as np
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from autograd import jacobian
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from scipy.integrate import quad
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def F(X):
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x, y = X
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return -y, -x * y
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def r(t):
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return np.array([-np.sin(t), np.cos(t)])
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drdt = jacobian(r)
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def integrand(t):
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return F(r(t)) @ drdt(t)
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I, e = quad(integrand, 0.0, np.pi / 2)
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print(f'The integral is {I:1.4f}.')
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</code></pre>
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<pre><code class="text">The integral is 0.4521.
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</code></pre>
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<p>We get the same result as the analytical solution.</p>
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<p>The next example is in three dimensions. Find the line integral along
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$\mathbf{r}(t)=[cos(t), sin(t), 3t]$ of the function
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$\mathbf{F(r)}=[z, x, y]$ from t=0 to t=2 π. The solution is given as
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21.99.</p>
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<pre><code class="python">import autograd.numpy as np
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from autograd import elementwise_grad, grad, jacobian
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def F(X):
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x, y, z = X
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return [z, x, y]
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def C(t):
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return np.array([np.cos(t), np.sin(t), 3 * t])
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dCdt = jacobian(C, 0)
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def integrand(t):
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return F(C(t)) @ dCdt(t)
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I, e = quad(integrand, 0, 2 * np.pi)
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print(f'The integral is {I:1.2f}.')
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</code></pre>
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<pre><code class="text">The integral is 21.99.
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</code></pre>
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<p>That is also the same as the analytical solution. Note the real
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analytical solution was 7 π, which is nearly equivalent to our answer.</p>
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<pre><code class="python">print (7 * np.pi - I)
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</code></pre>
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<pre><code class="text">3.552713678800501e-15
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</code></pre>
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<p>As a final example, we consider an alternate form of the line
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integral. In this form we do not use a dot product, so the integral
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results in a vector. This doesn't require anything from autograd, but
97+
does require us to be somewhat clever in how to do the integrals since
98+
quad can only integrate scalar functions. We need to integrate each
99+
component of the integrand independently. Here is one approach where
100+
we use lambda functions for each component. You could also manually
101+
separate the components.</p>
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<pre><code class="python">def F(r):
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x, y, z = r
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return x * y, y * z, z
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def r(t):
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return np.array([np.cos(t), np.sin(t), 3 * t])
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def integrand(t):
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return F(r(t))
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[quad(lambda t: integrand(t)[i], 0, 2 * np.pi)[0] for i in [0, 1, 2]]
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</code></pre>
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<pre><code class="text">Out[1]: [-6.845005548807377e-17, -18.849555921538755, 59.21762640653615]
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</code></pre>
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<pre><code class="python">[0, -6 * np.pi, 6 * np.pi**2]
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</code></pre>
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<pre><code class="text">Out[1]: [0, -18.84955592153876, 59.21762640653615]
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</code></pre>
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<p>which is evidently the same as our numerical solution.</p>
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<p>Maybe an alternative, but more verbose is this vectorized integrate
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function. We still make temporary functions for integrating, and the
127+
vectorization is essentially like the list comprehension above, but we
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avoid the lambda functions.</p>
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<pre><code class="python">@np.vectorize
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def integrate(i):
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def integrand(t):
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return F(r(t))[i]
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I, e = quad(integrand, 0, 2 * np.pi)
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return I
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integrate([0, 1, 2])
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</code></pre>
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<pre><code class="text">Out[1]: array([-6.84500555e-17, -1.88495559e+01, 5.92176264e+01])
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</code></pre>
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</body>
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</html>
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