Line integral of a scalar field. Hello friends, today it’s about the line integral of a scalar field.

Line integral of a scalar field_compressed

**The line integral of a scalar field**

**Solved examples of the line integral of a scalar field**

Disclaimer: None of these examples are mine. I have chosen these from some book or books. I have also given the due reference at the end of the post.

So here is the first example.

** Example 1**

According to Stroud and Booth (2011),* “If , evaluate between and along the curve with parametric equations .”

**Solution**

Here the given scalar field is . Also the parametric equations of and are .

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

**Step 1**

So, I’ll substitute and in .

Then it will be

Now I’ll simplify it. So it will be

And this gives the value of as

Next I’ll get the values of and .

Since , the value of is

As I know , the value of is

Also, since , the value of is

Now I have to get the value of .

Thus in terms of , it will be

So this means

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

**Step 2**

Ok, so I have to integrate the scalar field between and . And that means I have to find out the value of corresponding to the points and .

So, I already know that the coordinate of is . Also, the parametric form of is . Then I’ll equate the coordinate of to the parametric form of .

Thus it becomes

Now I’ll check if is also valid for and coordinates.

If I put in , it becomes

So this gives the coordinate as .

Similarly, I put in . And that gives

So here also I get the coordinate as . Thus it means is the lower limit of integration.

In the same way, now I’ll get the value of corresponding to the point . Now the coordinate of is . Therefore I can say that

which gives .

So I have two limits of integration – one is and the other is . Therefore will be

Thus my next step will be to evaluate .

**Step 3**

So I can rewrite as

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

Next, I’ll substitute these limits to get

And this gives

If I simplify it, I’ll get

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

Now I’ll give another example.

** Example 2**

According to Stroud and Booth (2011),* “If , evaluate between and along the path having parametric equations .”

**Solution**

Here the given scalar field is . Also the parametric equations of and are .

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

**Step 1**

So, I’ll substitute and in .

Then it will be

Now I’ll simplify it a bit to get

Now I’ll get the values of and .

Since , the value of is

As I know , the value of is

Also, since , the value of is

Next, I’ll get the value of .

Thus in terms of , it will be

So this means

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

**Step 2**

Ok, so I have to integrate the scalar field between and . And that means I have to find out the value of corresponding to the points and .

So, I already know that the coordinate of is . Also, the parametric form of is . Then I’ll equate the coordinate of to the parametric form of . Thus it becomes

Now I’ll check if is also valid for and coordinates.

If I put in , it becomes

So this gives the coordinate as .

Similarly, I put in . And that gives

So here also I get the coordinate as . Thus it means is the lower limit of integration.

In the same way, now I’ll get the value of corresponding to the point . Now the coordinate of is . Therefore I can say that

which gives .

So I have two limits of integration – one is and the other is . Therefore will be

Thus my next step will be to evaluate .

**Step 3**

So I can rewrite as

Now I’ll integrate this vector in the same way as in example 1.

Hence it will be

Next, I’ll substitute these limits to get

And this gives

If I simplify it, I’ll get

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

Now I’ll give my last example.

** Example 3**

According to Stroud and Booth (2011),* “Evaluate to one decimal place the integral along the curve with parametric equations between and .”

**Solution**

Here the given scalar field is . Also the parametric equations of and are .

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

**Step 1**

So, I’ll substitute and in .

Thus it will be

And this means

Now I’ll get the values of and .

Since , the value of is

As I know , the value of is

Also, since , the value of is

Next, I’ll get the value of .

Thus in terms of , it will be

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

**Step 2**

Ok, so I have to integrate the scalar field between and . And that means I have to find out the value of corresponding to the points and .

So, I already know that the coordinate of is . Also, the parametric form of is . Then I’ll equate the coordinate of to the parametric form of . Thus it becomes

Now I’ll check if is also valid for and coordinates.

If I put in , it becomes

So this gives the coordinate as .

Similarly, I put in . And that gives

So here also I get the coordinate as . Thus it means is the lower limit of integration.

In the same way, now I’ll get the value of corresponding to the point . Now the coordinate of is . Therefore I can say that

which gives .

So I have two limits of integration – one is and the other is . Therefore will be

Thus my next step will be to evaluate .

**Step 3**

So I can rewrite as

where

and

Now I’ll integrate these vectors in the same way as in examples 1 and 2.

Hence it will be

Next, I’ll substitute these limits to get

Then I’ll simplify it to get

(1)

Now I’ll get the value of .

###### Step 4

Here also I’ll do it in the same way as . Therefore it will be

Next, I’ll substitute these limits to get

Then I’ll simplify it to get

(2)

Finally I’ll add equations (1) and (2) to get the value of .

And this means

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

Dear friends, this is the end of today’s post on how to get the line integral of a scalar field. Thank you very much for reading this. Please let me know how you feel about it. Soon I will be back again with a new post. Till then, bye, bye!!

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