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The Line Integral Of A Scalar Field

EXAMPLE 1

According to Stroud and Booth (2011),* “If V = x^3y + 2xy^2 + yz, evaluate \int_c V~\text{d}\textbf{r}between A (0, 0, 0) and B (2, 1, -3) along the curve with parametric equations x = 2t, y = t^2, z = - 3t^3.”

SOLUTION

Here the given scalar field is V = x^3y + 2xy^2 + yz. Also the parametric equations of x, y and z are x = 2t, y = t^2, z = - 3t^3.

Now my first step will be to write V in terms of t.

STEP 1

So, I’ll substitute x = 2t, y = t^2 and z = - 3t^3 in V.

Then it will be

\[V = (2t)^3(t^2) + 2 (2t)(t^2)^2 + (t^2)(-3t^3).\]

Now I’ll simplify it. So it will be

\[V = 8t^5 + 4t^5 - 3t^5.\]

And this gives the value of V as

\[V = 9t^5.\]

Next I’ll get the values of \text{d}x, \text{d}y and \text{d}z.

Since x = 2t, the value of \text{d}x is

\[ \boxed{\text{d}x = 2 \text{dt}.} \]

As I know y = t^2, the value of \text{d}y is

\[ \boxed{\text{d}y = 2 t\text{dt}.} \]

Also, since z = - 3t^3, the value of \text{d}z is

\[ \boxed{\text{d}z = - 9 t^2\text{dt}.} \]

Now I have to get the value of \int_c V~\text{d}\textbf{r}.

Thus in terms of t, it will be

\[\int_c V~\text{d}\textbf{r} = \int_c 9t^5 (\textbf{i}2\text{dt} + \textbf{j}2t\text{dt} + \textbf{k}(-9t^2)\text{dt}).\]

So this means 

\[ \boxed{\int_c V~\text{d}\textbf{r} = 9 \int_c (\textbf{i} 2t^5 \text{dt} + \textbf{j} 2t^6 \text{dt} - \textbf{k} 9t^7\text{dt}) .} \]

Now my next step is to find out the limits of integration.

STEP 2

Ok, so I have to integrate the scalar field V between A (0, 0, 0) and B (2, 1, -3). And that means I have to find out the value of t corresponding to the points Aand B.

So, I already know that the x- coordinate of A is 0. Also, the parametric form of x is x = 2t. Then I’ll equate the x- coordinate of A to the parametric form of x.

Thus it becomes

\[2t = 0 \Rightarrow t = 0.\]

Now I’ll check if t = 0 is also valid for y- and z- coordinates.

If I put t = 0 in y, it becomes

\[y = (0)^2.\]

So this gives the y- coordinate as 0.

Similarly, I put t = 0 in z. And that gives

\[z = -3(0)^3.\]

So here also I get the z- coordinate as 0. Thus it means t = 0 is the lower limit of integration.

In the same way, now I’ll get the value of t corresponding to the point B. Now the x- coordinate of B is 2. Therefore I can say that

\[2t = 2\]

which gives t = 1.

So I have two limits of integration – one is t = 0 and the other is t = 1. Therefore \int_c V~\text{d}\textbf{r} will be

\begin{equation*} \int_c V~\text{d}\textbf{r} = 9 \int_{0}^{1} (\textbf{i} 2t^5 \text{dt} + \textbf{j} 2t^6 \text{dt} - \textbf{k} 9t^7\text{dt}). \end{equation*}

Thus my next step will be to evaluate \int_c V~\text{d}\textbf{r}.

STEP 3

So I can rewrite \int_c V~\text{d}\textbf{r} as

\[\int_c V~\text{d}\textbf{r} = 9 \left[ \int_{0}^{1} \textbf{i} 2t^5 \text{dt} + \int_{0}^{1} \textbf{j} 2t^6 \text{dt} - \int_{0}^{1} \textbf{k} 9t^7\text{dt}\right].\]

Now I’ll integrate this vector in the same way as the integration of a vector field. Also, I’ll follow the same rules as the rules for integration.

Hence it will be

\[\int_c V~\text{d}\textbf{r} = 9 \left[ \frac{2t^6}{6} \textbf{i} + \frac{2t^7}{7} \textbf{j} - \frac{9t^8}{8}\textbf{k}\right]_{0}^{1}.\]

Next, I’ll substitute these limits to get

\[\int_c V~\text{d}\textbf{r} = 9 \left[ \frac{2}{6} \textbf{i} + \frac{2}{7} \textbf{j} - \frac{9}{8}\textbf{k}\right].\]

And this gives

\[\int_c V~\text{d}\textbf{r} = \left[ \frac{9 \times 2}{6} \textbf{i} + \frac{9 \times 2}{7} \textbf{j} - \frac{9 \times 9}{8}\textbf{k}\right].\]

If I simplify it, I’ll get

\[\int_c V~\text{d}\textbf{r} = 3 \textbf{i} + \frac{18}{7} \textbf{j} - \frac{81}{8}\textbf{k}.\]

Hence I can conclude that this is the answer to the given example.