Double integral. Dear friends, today I’ll show how to evaluate double integrals in integration. Have a look!!

**Double integral**

Suppose I have to integrate a function like

And this is the ‘double integral’.

Now this means that at first, I’ll integrate the function with respect to . And at that time, will act as a constant. Also, component in the function will also be constant. Then I’ll substitute the limits as the lower limit and as the upper limit.

Next, I’ll integrate the function for the second time with respect to . And I’ll repeat the same process.

Anyway, now I’ll give some examples and that will clarify my point.

So have a look at these examples!!

#### Solved examples of double integral

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 are the examples.

**Example 1**

According to Stroud and Booth (2013)*, “Evaluate

”

**Solution**

Now here I have to evaluate the double integral

So let’s say

First of all, I’ll integrate for . Then I’ll integrate for . And, I’ll use the * formulas in integration* for that.

**Step 1**

Therefore it will be

Next, I’ll substitute the limits. Now here the lower limit is . And the upper limit is .

Thus will be

Then I’ll simplify it to get

Now I’ll integrate it for .

**Step 2**

So it will be

Next, I’ll substitute the limits. And that gives

Then I’ll simplify it. So that means

Therefore the value of is

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 (2013)*, “Evaluate

”

**Solution**

Now here I have to evaluate the double integral

So let’s say

First of all, I’ll integrate it for . Then I’ll integrate for . And just like example 1, here also I’ll use the standard formulas in integration.

**Step 1**

Therefore it will be

Next, I’ll substitute the limits. Now here the lower limit is . And the upper limit is .

Thus will be

Then I’ll simplify it to get

Now I’ll integrate it for .

**Step 2**

So it will be

Next, I’ll substitute the limits. And that gives

Then I’ll simplify it to get

Thus the value of is

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

Now I’ll give another example.

**Example 3 **

According to Stroud and Booth (2013)*, “Evaluate

”

**Solution**

Now here I have to evaluate the double integral

As I can see, here I have to integrate with the polar coordinates and and not with the cartesian coordinates and .

So let’s say

First of all, I’ll integrate it for . Then I’ll integrate for . And just like examples 1 and 2, here also I’ll use the standard formulas in integration.

**Step 1**

Since I have to integrate , I’ll use the same method as the * integration of trigonometric functions*.

Therefore it will be

As I know

I can say that

So will be

And this gives

Now I’ll substitute the limits to get

Next, I’ll simplify it. So this gives

And now I’ll integrate it for .

**Step 2**

So it will be

And that means

Next, I’ll substitute the limits. And that gives

Then I’ll simplify it to get

Thus the value of is

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

Dear friends, this is the end of today’s post. 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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